MANISH KUMAR
MATHEMATICS
HERON’S FORMULA
INTRODUCTION We are familiar with the shapes of many pane closed figures such as squares, rectangles, quadrilaterals, right triangles, equilateral triangles, isosceles triangles, scalene triangles, etc. We know the rules to find the perimeters and area of some of these figures. For example, a rectangle with length 12 m and with 8 m has perimeter equal to 2 (12m +8 ) = 40m. The area of this rectangle is equal to (12 × 8) m 2 = 96 m2. A square having each side of length 10 m has perimeter equal to 4 × 10 m = 40 and area equal to 102 m2 = 100 m2. Unit of measurement for length or breadth is taken as meter (m) on centimeter (cm) etc. Unit of measurement for area of any plane figure is taken as square metre (m2) or square centimetre (cm2) etc. In this section, we shall find the areas of some triangles.
AREA OF A TRIANGLE USING BASE AND HIGHT From you earlier classes, you know that : Area of a triangle
1 base × Corresponding Height 2
Any side of the triangle may be taken as base and the length o perpendicular from the opposite vertex to the base is the corresponding height.
1 In given figure, Area of ABC BC AD sq. units. 2
A
B
C
D
For example, a triangle having base = 10 m height = 6 m has its area equal to
1 × (10 m × 6 m) = 30 m2. 2
Area of a Right Triangle :- When the triangle is right angles, we can directly apply the above motioned formula by using two sides containing the right angle as base and height.
1 In given figure, Area of ABC = BC AB sq. units. 2
A Hypotenuse
Height
B
Baase
C
MANISH KUMAR
MATHEMATICS
For exampe, a right angles triangle having two sides of length 3 m and 7 m (other than the hypotenuse), has its area =
1 × (3 m + 7 m) = 10.5 m2. 2
Area of an Equilateral Triangle :- Let ABC be equilateral triangle with side a and AD be the perpendicular from A on BC. Then D is the mid point of BC i.e. BD =
a . 2
A a
a B In right-angles ABD, by Pythagoras theorem, we have
a 2
D
a 2
C
AD2 = AB2 - BD2 2
a2 3 2 a AD2 = a2 - = a2 a 4 4 2
AD =
3 a. 2
So, area of ABC =
1 1 3 3 2 × BC × AD = ×a× a= a. 2 2 2 4
3 2 Area of equilateral triangle with side a units a sq. units. 4 For example, an equilateral triangle having side 8 cm, has its area
3 (8)2 cm2 = 16 3 cm2. 4
Area of an Isosceles Triangle :- Let ABC be an isosceles triangle with AB = AC = a and BC = b, and AD be the perpendicular from A on BC. Then, D is the mid-point of BC, i.e. BD =
b . 2
In right angled ABD, by Pythagoras theorem we have : AD2 = AB2 - BD2 2
4a2 b2 b AD2 = a2 - = a2 4 2
AD =
4a2 b2 2
So, are of ABC =
1 1 × BC × AD = ×b× 2 2
4a2 b 2 1 b 4a2 b 2 2 4
MANISH KUMAR
MATHEMATICS
1 Area of isosceles ABC with AB = AC a units and BC = b units = b 4a2 b2 sq. units. 4 For example, an isosceles triangle having equal sides of length 5 cm an unequal side of length 8 cm, has its area =
1 8 4(5)2 82 = 12 cm2. 4
AREA OF A TRIANGLE BY USING HERON’S FORMULA In A scalene triangle, if the length of each side is given but its height is not known and it cannot be obtained easily, we take the help of Heron’s formula or Hero’s formula given by Heron to find the area of such a triangle. Heron’s formula : If a,b,c denote the lengths of the sides of a triangle ABC. Then, Area of ABC = where
s
s ( s a )( s b)( s c)
abc is the semi-perimeter of ABC. 2
Remark : This formula is applicable to all types of triangles whether it is right-angled o equilateral or Ex.1.
isosceles. Find the area of a triangle whose sides are 13 cm, 14 cm and 15 cm
Sol.
Let a,b,c be the sides of the given triangle and s be its semi-perimeter such that a = 13 cm, b = 14 cm and c = 15 cm
1 1 (a + b + c) = (13 + 14 + 15) = 21 2 2 s - a = 21 - 13 = 8, s - b = 21 - 14 = 7 and s - c = 21 - 15 = 6
Now, s =
Hence,
Area of given triangle
=
s(s a)(s b)(s c ) =
Ex.2
21 8 7 6 7 3 8 7 2 3 7 2 4 4 32
Find the area of a triangle, two sides of which are 8 cm and 11 cm and 11 cm the perimeter is 32 cm. [NCERT]
Sol.
Let a,b,c be the sides of the given triangle and 2s be its perimeter such that a = 8 cm, b = 11 cm and 2s = 32 m i.e. s = 16 cm Now,
a + b + c = 2s
8 + 11 + c = 32
c = 13.
s - a = 16 - 8 = 8, s - b = 16 - 11 = 5 and s - c = 16 - 13 = 3
Hence, Area of given triangle =
ss as bs c 16 8 5 3 8 8 30 8 30 cm2
Ex.3
The perimeter of a triangular field is 450 m and its sides are in the ratio 13 : 12 : 5. Find the area of triangle.
Sol.
It is given that the sides a,b,c of the triangle are in the ratio 13 : 12 : 5 i.e., a : b : c = 13 : 12 : 5 a = 13x, b = 12x and x = 5x
Perimeter = 450 13x + 12x + 5x = 450 30x = 450 x = 15
So, the sides of the triangle are
MANISH KUMAR
MATHEMATICS
a = 13 × 15 = 195 m, b = 12 × 15 = 180 m and c = 5 × 15 = 75 m It is given that perimeter = 450 2s = 450 s = 225 Hence,
Area =
ss a s b s c 225225 195225 180225 75
Area =
225 30 45 150 5 2 3 2 3 5 2 3 2 5 5 2 2 3
Area =
56 36 2 2 53 33 2 6750 m2
MANISH KUMAR
MATHEMATICS
Ex.4
The lengths of the sides of a triangle are 5 cm, 12 cm and 13 cm. Find the length of perpendicular from the
Sol.
opposite vertex to the side whose length is 13 cm. Here, a = 5, b = 12 and c = 13.
s=
1 1 (a + b + c) = (5 + 12 + 13) = 15 2 2
Let A be the area of the given triangle. Then,
A=
ss a s bs c 1515 515 1215 13
A=
15 10 3 2 30 cm2
….(i)
Let p be the length of the perpendicular from vertex A on the side BC. Then, A=
1 × (13) × p 2
From (i) and (ii), we get = Ex.5
1 60 × 13 × p = 30 p = cm. 2 13
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22m and 120 m. The advertisements yield an earning of Rs. 5000 per m 2 per year. A company hired both walls for 3 months. How much rent did it pay?
Sol.
The lengths of the sides of the walls are 122 m, 22 m and 120 m. We have, 1222 = 1202 + 202 So, walls are in the form of a right triangle.
1 Area of two walls = 2 Base Heigth 2
1 Area of two walls 2 x 120 22 2640 m2 2
We have, Yearly rent = Rs 5000 per m2
5000 Monthly rent = Rs. per m2 12
[NCERT]
MANISH KUMAR
MATHEMATICS
5000 Hence, rent paid by the company for 3 months = Rs. 3 2640 = Rs. 3300000. 12
APPLICATIONS OF HERON’S FORMULA IN FINDING AREA OF A QUADRILATERAL Heron’s formula can be applied to find the area of a quadrilateral by dividing the quadrilateral into two triangular parts. If we join any of the two diagonals of the quadrilateral, then we get two triangles. Area of each triangle is calculated and the sum of two areas is the area of the quadrilateral.
Ex.6
Find the area of the quadrilateral ABCD, in which AB = 7 cm, BC = 6 cm, CD = 12 cm, DA = 15 cm and AC = 9 cm.
Sol.
The diagonal AC divides the quadrilateral ABCD into two triangles ABC and ACD.
Area of quad. ABCD = Area of ABC + Area of ACD.
We have,
s
679 11 cm 2 Area of ABC =
ss as b s c
Area of ABC =
1111 611 7 11 9
Area of ABC =
11 5 4 2 440 sq. cm
Area of ABC = 20.98 cm2
For ACD, we have
s
9 12 15 18 cm2 2
Area of ACD =
1818 9 18 12 18 15
Area of ACD =
18 9 6 3
Area of ACD =
18 9 6 3
Hence, Area of quad. ABCD
MANISH KUMAR
MATHEMATICS =(20.98 + 54) cm2 = 74.98 cm2
MANISH KUMAR Ex.7
MATHEMATICS
In fig. ABCD is a field in the form of a quadrilateral whose sides are indicated in the figure. If DAB = 900, find the area of the field.
Sol.
Clearly, DAB is a right-angled triangle. Therefore,
DB2 = DA2 + AB2
DB2 = 92 + 402
DB =
[Using Pythagoras Theorem]
81 1600 = 41m
In DAB, we have 2s = DA + AB + BD = (9 + 40 + 41) = 90m
s = 45 m
A1 = Area of DAB =
45 ( 45 9) ( 45 40) ( 45 41)m 2
A1 = Area of DAB =
45 36 5 4 m2
A1 = Area of DAB =
5 9 36 5 4 m2 52 32 6 2 2 2 m2
A1 = Area of DAB = (5 × 3 × 6 × 2) m = 180 m 2
2
In DCB, we have 2s = DC + CB + BD 2s = 28 + 15 + 41 = 84 s = 42 m.
A2 = Area of DCB =
42 (42 - 28) (42 - 15) (42 - 41) m2
A2 = Area of DCB =
42 14 27 1 m2
A2 = Area of DCB =
7 2 3 7 2 3 3 3 m2
A2 = Area of DCB =
7 2 2 2 34 m2
A2 = Area of DCB = (7 × 2 × 3 ) m = 126 m 2
2
2
Hence, Area of the field = A1 + A2 = (180 + 126) m2 = 306 m2 Ex.8
Find the area of trapezium whose parallel sides 25 cm, 13 cm and other sides are 15 cm and 15 cm.
MANISH KUMAR Sol.
MATHEMATICS
Let ABCD be the given trapezium in which AB = 25 cm, CD = 13 cm, BC = 15 cm and AD = 15 cm. Draw CE || AD. Now, ADCE is a parallelogram in which AD || CE and AE || CD. AE = DC = 13 cm and BE = AB - AE = 25 - 13 = 12 cm In BCE, we have
s
15 15 12 21 cm 2
Area of BCE =
ss a s b s c
Area of BCE =
21( 21 15)( 21 15)( 21 12)
Area of BCE =
21 6 6 9 18 21 sq. cm
...(i)
Let h be the height of BCE, then Area of BCE =
1 1 (Base × Height) = × 12 × h = 6h 2 2
...(ii)
From (i) and (ii), we have, 6h = 18 21 h = 3 21 cm Clearly, the height of trapezium ABCD is same as that of BCE.
Ex.9
Area of trapezium =
1 (AB + CD) × h 2
Area of trapezium =
1 (25 + 13) × 3 21 cm2 = 57 21 cm2 2
Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops to suffice the needs of their family. She divided the land in to equals parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get ? [NCERT]
Sol.
Let ABCD be the field which is divided by the diagonals BD = 160 m into two equal parts. Since ABCD is a rhombus of perimeter 400 m. Therefore, AB = BC = CD = DA =
400 m = 100 m 4
Let s be the semi - perimeter of BCD. Then, s
BC CD BD 100 100 160 m 180 m 2 2
Area of BCD =
180 180 100 180 100 180 160 m 2
MANISH KUMAR
MATHEMATICS =
180 80 80 20 m 2 4800 m 2
Hence, each of the two children will get an area of 4800 m 2
THINGS OF REMEMBER 1.
Area of a triangle, whose base and height are known, is calculated by using the formula :
2.
1 × Base × Height 2 Area of a right angled triangle, whose base is b and perpendicular is p, is calculated by using the formula:
3.
1 bp 2 Area of an equilateral triangle, each of whose side is a, is calculated by using the formula :
4.
3 2 a 4 Area of an isosceles triangle, each of whose equal sides is a and the unequal side is b, is calculated by
Area of triangle =
Area of triangle =
Area of triangle
using the formula:
b 4a 2 b 2 4 Area of a triangle with its sides as a,b, and c is calculated by using Heron’s formula : Area of triangle =
5.
Area of triangle s(s a)(s b)(s c)
a bc 2 Area of a quadrilateral whose sides and one diagonal are given, can be calculated by CBSE Based Some Important Questions Where
6.
s
MANISH KUMAR
MATHEMATICS CBSE BASED SOME IMPORTANT QUESTIONS
Q.1
One side of a right triangle measures 126 m and the difference in lengths of its hypotenuse and other side is 42 m. Find the measures of its two unknown sides and calculate its area. Verify the result using Heron’s Formula
[Hint : C - b = 42 c = 42 + b & c2 = a2 + b2
(42 + b)2 = (126)2 + b2] Q.2
Ans. 168 m, 210 m, 10584 m2
Using Heron’s Formula, find the area of an equilateral triangle the length of one side is a. Ans.
Q.3
Find the area of an isosceles triangle, the measure of one of its equal sides being b and the third side a. Ans.
Q.4
3a 2 4
a 4b 2 a 2 4
Find the area of a right angled triangle if the radius of its circumcircle is 3 m and altitude drawn to they hypotenuse is 2 cm.
[Hint :
Let ABC be the right angled triangle right angled at B. Let O be the centre of the circumcircle. Them by geometry O is the mid-point of the hypotenuse AC.] Q.5
Ans. 6 cm2
A regular hexagon has a side 8 cm. Determine its perimeter and area.
[Hint :
Area of hexagon = 6 x area of equilateral triangle OAB] Q.6
Ans. 48 cm, 96 3 cm 2
The perimeter of right triangle is 90 cm. Its hypotenuse is 41 cm. Find the other two sides and the area of the triangle.
[Hint :
a + b + 41 = 90 a + b = 49 cm
MANISH KUMAR Also, a2 + b2 = (41)2 or (49 - b)2 + b2 = (41)2]
MATHEMATICS Ans. 40 cm, 9 cm, 180 cm2
MANISH KUMAR Q.7
2
An isosceles right triangle has an area 200 cm . What is the length of its hypotenuse?
[Hint :
Q.8
MATHEMATICS
a2 200 a 20cm] 2
Ans. 20 2 cm
Radha made a picture of an aeroplane with coloured paper as shown in figure, Find the total area of the paper used. [NCERT]
Ans. 19.3 cm2
Q.9
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28cm and 30 cm, and the parallelogram on the base 28 cm, find the height of the parallelogram.
Ans. h =12 cm
Q.10
In figure, OPQR is a rhombus, three of whose verticles lie on the circle with centre O. If the area of the rhombus is 32 3cm 2 , find the radius of the circle.
[Hint :
OS SQ
X & OP QR RS X 2
Area of equitateral OQR + Area of equitateral OPQ = 32 3]
Ans. 8 cm
MANISH KUMAR Q.11
MATHEMATICS
A parallelogram, the measures of whose sides are 25 cm and 15 cm has one diagonal 20 m long . Find its altitude on the side 25 cm. Ans. 12 cm
MANISH KUMAR Q.12
MATHEMATICS
The base of a triangular field is three times its height. If the cost of cultivqting the field at Rs. 300 per m2 is Rs. 181250 , find its base and height. [NCERT] [Hint : If height = h,. then 3h m. Area of field = Rs.
Q.13
181250 ] 300
Ans. 45 m. 15
Find the square of the radius of te circle whose area is the sum of the area of two triangles whose sides 35, 53, 66 and 33, 56,65 measured in centimeters. (Take = 22/7)
Ans. 588 cm2
Q.14
Find the area of a quadrilateral field whose diagonals 48 m and 32 m and intersect each other at right angles. Find also the cost of the land at the rate of Rs/ 7000 per square metre.
[Hint :
1 2
Required area = AO BD
1 OC BD 2
1 BD( AO OC ) 2 1 BD AC ] 2 Q.15
Ans. 768 m2 , Rs. 5376000
A trapezium PBCQ, with parallel sides QC and PB in the ratio of 7 : 5, is cut off from a rectangle ABCD as shown in the following figure. If the area of the trapezium is
4 part of the area of the 7
rectangle. find the lengths of QC and PB.
Ans. QC = 15 m PB = 10 m Q.16
The perimeter of an equilateral triangle measure 3 times metres as the area of the equilateral triangle measures square metres. Find the length of its side. [ Hint : Let the length of the side be m. Then 3x =
Q.17 Q.18
3 2 3 x ] 4
Ans. 4 m
In an equilateral triangle of side 2 a units. Find the length of its altitude. Ans. 3a units A like in the shape of a square with a diagonal 40 cm and an idosceles triangle of base 10 cm and sides 13 cm each is to be made of three different shades as shown in fig. How paper of each shade has been used in making the kite?
MANISH KUMAR
MATHEMATICS Ans. 400 cm2 ; 400 cm2; 60 cm2
MANISH KUMAR
MATHEMATICS
EXERCISE SUBJECTIVE TYPE QUESTIONS (A) 1. 2. 3. 4. 5. 6. 7. (B) 1.
VERY SHORT ANSWER TYPE QUESTIONS : Write the area of a triangle having 5 cm base and height 6 cm. Write the area of an equilateral triangle whose side is 6 cm. State Heron’s Formula for area of a triangle. In ABC, BC = a, CA = b and AB = c. Write the semiperimeter s. Find the area of isosceles triangle ABC in which AB = AC = 5 cm and BC = 8 cm. Find the area of isosceles triangle having each side of length a cm. Find the area of triangle having three sides given as 5 cm, 6 m and 7 cm. SHORT ANSWER TYPE QUESTIONS : A triangular park in a city has dimensions 100 m × 90 m × 110 m. A contract is given to a company for planting
2.
grass in the park at the rate of Rs. 4000 per hectare. Find the amount to be paid to the company. (Take 2 = 1.414) (one hectare = 10,000 m2) There is a slide in a children park. The front side of the slide has been painted and a message “ONLY FOR CHILDREN” is written on it as whose in fig. If the sides of the triangular front wall of the slide are 9 m. 8 m and 3 m, then find the area which is paint in colour.
3. 4. 5. 6. 7.
The perimeter of a triangular park is 180 m and its sides are in the ratio 5 : 6 : 7. Find the area of the park. A triangle has sides 35 mm, 54 mm and 61 mm long. What is its area. Find also the smallest altitude of the triangle. The perimeter of a right triangle is 12 cm and its hypotenuse is of length 5 cm. Find the other two sides and calculate its area. Verify the result using Heron’s Formula. Using heron’s Formula, find the area of an isosceles triangle, the measure of one of its equal sides being a units and the third side 2b units. The sides of triangle are 39 cm, 42 cm, and 45 c m. A parallelogram stands on the greatest sides of the triangle and has the same area as that the triangle. Find the height of the parallelogram.
8.
From a point in the interior of an equilateral triangle perpendiculars drawn to the three sides are 8 cm, 10 cm and
9.
11 cm respectively. Find the area of the triangle to the nearest cm. (use 3 1.73 ) A municipal corporation wall on road side has dimensions as shown in fig. The wall is to be used for advertisements and it yields an earning or Rs. 400 per m2 in a year. Find the total amount of revenue earned in a year.
MANISH KUMAR 10.
MATHEMATICS
ABCD is quadrilateral such that AB = 5 cm, BC = 4 cm, CD = 7 cm, AD = 6 cm and diagonal BD = 5 cm. prove
that the area of the quadrilateral ABCD is 4 3 6 cm2. 11.
Find the area of the quadrilateral ABCD in which AB = 7 cm, BC = 6 cm, CD = 12 cm, DA = 15 cm and AC = 9 cm. (Take
110 = 10.5 approx.)
12.
A rhombus has perimeter 64 m and one of the diagonals is 22 m. Prove that the area of the rhombus is 66 15 m 2
13.
ABCD is a trapezium in which AB }} CD ; BC and AD are non-parallel sides. It is given that AB = 75 cm, BC = 42 cm, CD = 30 and AD = 39 cm. Find the area of the trapezium.
14.
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O. If the radius of the circle is 10 cm. find the area of the rhombus.
15.
The cross-section of a canal is in the shape of a trapezium. If the canal is 12 m wide at the top and 8 m wide at the bottom and the area of its cross-section is 84 m2, determine its depth.
16.
Students of a school stages a rally for cleanliness campaigns. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA ; while the other through AC, CD and DA. Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m. CD = 15 m, DA = 28 m and B = 900, which group cleaned more area and by how much ? Find the total area cleaned by the students.
17.
Find the perimeter of a square, the sum of lengths of whose diagonals is 144 cm.
MANISH KUMAR 18.
MATHEMATICS
Find the area of a quadrilateral piece of ground one of whose diagonals is 60 metres long and the perpendiculars from the other two vertices are 38 and 22 metres respectively.
MANISH KUMAR
MATHEMATICS
(C)
LONG ANSWER TYPE QUESTIONS :
1.
In figure, AB = 28 m, AC = 24 m, BC = 20 m, CG = 32 m, AG = 40 m and D is mid-point of AG. Find the area of the quadrilateral ABCD.
2.
White and grey coloured triangular plastic sheets are used to make a toy as shown in fig. Find the total areas of white and grey coloured sheets for making the toy.
3.
Suman made an arrangements with white and black coloured paper sheets as showing in fig. Find the total areas of the white and black paper sheets used in making the arrangement.
4.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangular tiles are 26 cm, 20 cm, and 10 cm. The tiles are polished at the rate of 20 p per cm2. Find the cost of polishing the tiles. (Take 3.74)
14 =
MANISH KUMAR 5.
MATHEMATICS
Suman made a picture with some white paper and a single coloured paper as showing in fig. White paper is a available at her home and free of cost. The cost of coloured paper used is at the rate of 10 p per cm 2. Find the total cost of the coloured paper used. (Take
3 = 1.732 and
11 = 3.31)
6.
In figure, P and Q are two lamp posts. If the area of the PBC is same as that of the rectangle ABCD, find the distance between the two lamp posts.
7.
A triangle and parallelogram has same base and same area. If the sides of the triangle are 20 cm, 25 cm and 35 cm, and the base side is 25 cm for the triangle as well as the parallelogram, find the vertical height of the parallelogram. A triangle and a parallelogram has a common side and are of equal areas. The triangle having sides 26 cm, 28 cm and 30 cm stands on the parallelogram. The common side of the triangle and the parallelogram is 28 cm. Find the vertical height of the triangle and that of the parallelogram. A farmer has two triangular fields in the form of ABC and ACD in which the side AC is common as shown in figure. AB = 840 m, BC = 600 m, AC = 480 m, AD = 800 m, AD = 800 m and CD = 640 m. He has marked midpoints E and F on the sides AB and AD respectively. By joining CE and CF, he has made a field in the shape of quadrilateral AECF. He grew in the quadrilateral plot AECF, potatoes in CFD and onions in BEC. How much
8.
9.
are has been used for each crop ? (Take
6 = 2.45 ; one hectare = 10000 m2).
MANISH KUMAR 10.
MATHEMATICS
A field in the form of quadrilateral ABCD whose sides taken in order are respectively equal to 192, 576, 288 and 480 dm has the diagonal equal to 672 dm. Find its area to the nearest square metre.
11.
A trapezium with its parallel sides in the ratio 16 : 15 is cut from a rectangle whose sides measure 63 m and 5 m respectively. The area of the trapezium is
4 of the area of the rectangle. Find the lengths of the parallel sides of 15
the trapezium. 12.
Find the cost, at Rs. 25 per 10 square metres, of turning a plot of land in the form of parallelogram whose adjacent sides and one of the diagonals measure 39 m, 25 m and 56 m respectively.
(D)
NCERT QUESTIONS :
1.
There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
2.
Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42 cm.
3.
An isosceles triangle has perimeter 30cm and each of the equal sides is 12cm. Find the area of the triangle.
4.
A park, in the shape of a quadrilateral ABCD, has C = 900, AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy ?
5.
Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5CM.
6.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30m and its longer diagonal in 48m, how much area of grass field will each cow be getting ?
7.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella ?
8.
A kite in the shape of square with a diagonal 32 cm and an isosceles triangle of base 8cm and sides 6cm each is to be made of three different shades as shown in fig. How much paper of each shade has been used in it ?
MANISH KUMAR OBJECTIVE TYPE QUESTIONS 1. 2.
3.
The base of a right angles triangle Is - 5 metres and hypotenuse is 13 metres. Its area will be (A) 25 m2 (B) 28 m2 (C) 30 m2 (D) None of these The sides of a triangular board are 13 metres, 14 metres and 15 metres. The cost of painting it at the rate of Rs. 8.75 per m2 is (A) Rs. 688.80 (B) Rs. 735 (C) Rs. 730.80 (D) Rs. 722.50 The area of an equilateral triangle whose side is 8 cms, is (A) 64 cm2
4.
5.
6.
8.
10.
11.
12.
(C) 21.3 cm2
3 4 3 If x is the length of a median of an equilateral triangle, then its area is : (A) 4 cm
(B)
(A) x2
(B)
(D) 4 3 cm2
4
x2 3 2
(C)
(C)
x2 3 3
(D) 3 cm
(D)
x2 2
The altitude of an equilateral triangle of side 2 3 cm is :
3 cm 2
(B)
1 cm 2
(C)
3 cm 4
(D) 3 cm
In a triangle ABC, BC = 5 cm, AC = 2 cm and AB = 13 cm. The length of the altitude drawn from B on AC is: (A) 4 cm (B) 5 cm (C) 2 cm (D) 7 cm A triangle of area 9 xy cm2 has been drawn such that its area is equal to the area of an equilateral triangle of side 6 cm. The, the value of y is (A)
9.
2
(B) 16 3 cm
The length of each side of an equilateral triangle having an area of 4 3 cm2, is :
(A) 7.
MATHEMATICS
2 cm
(B)
3 cm
(C) 2 cm
(D) 3 cm
In PQR, side QR = 10 cm and height PM = 4.4 cm. If PR = 11 cm, then altitude QN equals : (A) 4 cm (B) 5 cm (C) 5.5 cm (D) 5.6 cm The area of a right angles triangle is 30 cm2 and the length of its hypotenuse is 13 cm. The length of the shorter leg is : (A) 4 cm (B) 5 cm (C) 6 cm (D) 7 cm Area of a square with side x is equal to the area of a triangle with base x. The altitude of the triangle is : x (A) (B) x (C) 2x (D) 4x 2
13.
A plot of land is in the shape of a right angled isosceles triangle. The length of the hypotenuse is 50 2 m. The cost of fencing it at Rs. 3 per metre will be : (A) less than Rs. 300 (B) less than Rs. 400 (C) more than Rs. 500 (D) More than Rs. 600 The perimeter of an isosceles triangle is equal to 14 cm, the lateral side is to the base in the ratio 5 : 4. The area of the triangle is 1 3 2 2 2 2 (A) (B) (C) 21 cm (D) 2 21 cm 21 cm 21 cm 2 2
14.
If the area of an equilateral triangle is 24 3 sq. m, then its perimeter is :
15.
(A 96 m (B) 12 6 m (C) 4 6 m (D) 2 6 m The ratio of the area of square of side a and equilateral triangle of side a, is :
16.
(A) 2 : 1 (B) 2 : 3 (C) 4 : 3 (D) 4 : If every side of a triangle is doubled, then increase in the area of the triangle is :
3
MANISH KUMAR
MATHEMATICS
(A) ( 2 100)%
(B) 200 %
(C) 300 %
(D) 400 %
17.
If the altitude of a equilateral triangle is
6 , then its area is :
18.
(A) 3 3 (B) 2 3 (C) 2 2 (D) 6 2 A surveyor in his field book has drawn the plot as shown in the given figure. The area of the plot is :
1 1 (az + by + ct + dx) (B) (bt + cx + ay + az) 2 2 1 1 1 (C) (cx + bt + by + az) (D) (d + t) (c + x) (a + b) (y + z) 2 2 2 If an equilateral triangle of area X and a square of area Y have the same perimeter, then : (A) X > Y (B) X = Y (C) X < Y D) X Y (A)
19. 20.
A square and an equilateral triangle have equal perimeters. If the diagonals of the square is 12 2 cm, then the area of the triangle is : (A) 24 2 cm
2
2
2
B) 24 3 cm
(A)
VERY SHORT ANSWERS TYPE QUESTIONS :
(B)
abc 2 SHORT ANSWER TYPE QUESTIONS : 2. 9 3 cm 2
4.
3 2 a 4
6.
7. 6 6 cm 2
Rs. 1696.8
2. 2 35 m 2
3. 600 6 m 2
4. 420 5 mm 2 ,
840 5 mm 61
5.
3 cm. 4 cm. 6 cm2
6. b a 2 b 2 sq. units
7. 16.8 cm
8. 485 cm2
9. Rs. 50400
11. 75 cm (approx)
13. 1764 m 2
(D)
5. 12cm2
EXERCISE
1.
2
(C)
D) 64 3 cm
ANSWER KEY
HERON’S F ORMULAR
1. 15 cm2
2
C) 48 3 cm
2
14. 50 3 cm 2
2
16. I group cleaned more area by 54 m ; 306 m LONG ANSWER TYPE QUESTIONS :
17. 144 2 cm
1. 96 ( 6 2) m2
3.
2. 16 2 cm2 ; 16 2 cm2
15. 8.4 m 18. 1800 m2
3 9 55 cm2 ; 55 cm 2 4 4
4. Rs. 287.23
5. Rs. 14.92 6. 2 14cm 7. 4 6cm 8. 24 cm ; 12 cm 9. Area for wheat = 14.736 hectares, Area for potatoes = 7.68 hectares, Area for onions = 7.056 hectares. 10. 1111 m2 11. 25.6 m, 8 m 12. Rs. 2100 NCERT QUESTIONS : 1. 20 2m 2
3. 9 15cm 2
2. 21 11cm 2
4. 65.5 m2 (approx)
5. 15.2 cm2 (approx)
6. 48 m2 7. 1000 6 cm2, 1000 6 cm2 8. Area of shade I = Area of shade II, = 256 cm2 and area of shade III = 17.92 cm2
ANSWER KEY Que.
1
2
3
4
5
6
7
8
9
10
MANISH KUMAR
MATHEMATICS
Ans.
C
B
B
A
C
D
B
B
A
B
Que.
11
12
13
14
15
16
17
18
19
20
Ane.
C
C
D
B
D
C
B
B
C
D
MANISH KUMAR
MATHEMATICS
SURFACE AREAS & VOLUMS
INTRODUCTIONS Up till now we have been dealing with figures that can be drawn on the page of our notebook or on the blackboard. These are called plane figures. In this chapter, we shall study about some solid figures like cuboids, cube, cylinder and sphere. These figures are three dimensional figures. We shall also learn to find the surface areas and volumes of these figures.
UNITS OF MEASUREMENT OF AREA AND VOLUME The inter-relationships between various units of measurement of length, are and - volume are listed below for ready reference :
LENGTH 1 Centimetre (cm) 1 Decimetre (dm) 1 Metre (m) 1 Decametre (dam) 1 Hectometre (hm) 1 Kilometre (km) 1 Myriametre
= = = = = = =
10 milimietre (mm) 10 centimetre 10 dm = 100 cm = 1000 mm 10 m = 1000 cm 10 dam = 100 m 1000 m = 100 dam = 10 hm 10 Kilometre
AREA 2
1 cm = 1 cm × 1 cm = 10 mm × 10 mm = 100 mm2 1 dm2 = dm × 1 dm = 10 cm × 10 cm = 100 cm2 2 1m =1m×1m = 10 dm × 10 dm = 100 dm2 2 1 dam = 1 dam × 1 dam = 10 m × 10 m = 100 m2 1 hm2 = 1 hectare = 1 hm × 1 hm = 100 m × 100 m = 10000 m 2 = 100 dm2 1 km2 = 1 km × 1 km = 10 hm × 10 hm = 100 hm2 or 100 hectare
VOLUME 3
1 cm 1 litre 1 m3 1 dm3 1 m3 1 km3
= 1 ml = 1 cm × 1 cm × 1 cm = 10 mm × 10 mm × 10 mm = 1000 mm3 = 1000 ml = 1000 cm3 = 1 m × 1 m × 1 m = 100 cm × 100 cm × 100 cm = 106 cm3 = 1000 litre = 1 kilolitre. = 1000 cm3 = 1000 dm3 = 109 m3
CUBOID A rectangular solid bounded by six rectangular plane faces is called a cuboids. A match box, a tea-packet, a brick, a book, etc., are all examples of a cuboid. A cuboids has 6 rectangular faces, 12 edges and 8 vertices.
MANISH KUMAR
MATHEMATICS
The following are some definitions of terms related to a cuboid : (i) The space enclosed by a cuboid is called its volume. (ii)
The line joining opposite corners of a cuboid is called its diagonal. A cuboid has four diagonals. A diagonal of a cuboid is the length of the longest rod that ban be placed in the cuboid.
(iii)
The sum of areas of all the six faces of a cuboid is known as its total surface area.
(iv)
The four faces which meet the base of a cuboid are called the lateral faces of the cuboid.
(v)
The sum of areas of the four walls of a cuboid is called its lateral surface area.
Formula For a cuboid of length = units, breadth = b units and height = h units, we have : Sum of length of all edges = 4 ( + b + h) units. Diagonal of cuboid =
b 2 h 2 units
Total Surface Area of cuboid =( b + bh + h ) sq. units Lateral Surface Area of a cuboid = [2( b + b) × h] sq. units Area of four walls of a room = [2( + b) × h] sq. units. Volume of cuboid = ( × b × h) cubic units REMARK : For the calculation of surface area, volume etc. of a cuboid, the length, breath and height must be expressed in the same units.
CUBE A cuboid whose length, breadth and height are all equal is called Ice - cubes, Sugar cubes, Dice etc. are all examples of a cube. Each edge of a cube is called its side. Formulae For a cube of edge = a units, we have : Sum of lengths of all edges = 12 a units. Diagonal of cube = (a 3 )units. Total Surface Area of cube = (6a2) sq. units. Lateral Surface Area of a cube = (4a2) sq. units. Volume of cube = a3 cubic units.
CROSS SECTION A cut which is made through a solid perpendicular to its length is called its cross section. If he cut has the same shape and size at very point of its length, then it is called uniform cross-section. Volume of a solid with uniform cross section = (Area of its cross section) × (length). Lateral Surface Area of a solid with uniform cross section = (Perimeter of cross section) × (length).
MANISH KUMAR Ex.1
MATHEMATICS
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
Sol.
[NCERT]
For shelter length, = 4 m, breadth, b = 3, height, h 2.5 m
Total surface area of the shelter 2( + b) h + b = 2 (4 + 3) (2.5) + (4) (5) = 2 (7) (2.5) + 12 = 47 m2
Hence, 47 m2 of tarpaulin will be required. Ex.2
Find the surface area of a cube whose edge is 15 cm.
Sol.
The edge of the cube = 15 cm, i.e., a = 15 cm. Surface area of the cube = 6a2 = 6 × (15)2 = 1350 cm2.
Ex.3
The paint in a certain container is sufficient to paint an area equal to 9.375 m 2. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container ? [NCERT]
Sol.
For a brick length, = 22.5 cm, breadth, b = 10 cm, height, h = 7.5 cm
Total surface area of a brick = 2( b + bh + h ) = 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) = 2(225 + 75 + 168.75) = 2(468.75) = 937.5 cm2 = 09375 m2
Number of brick that can be painted out =
9.375 = 100 .09375
Hence, 100 bricks can be pained out of the given container. Ex.4
A small indoor greenhouse is made entirely of glass sheets (including the base) held together with tape. It is 40 cm long, 30 cm wide and 30 cm high. Find (i) the area of the glass sheet required and (ii) the total length of the tape required for all the 12 edges.
Sol.
The dimensions of the greenhouse are as under : Length ( ) = 40 cm, Width (b) 30 cm, Height (h) = 30 cm The area of the glass sheet required =
The total outer (or inner) surface area of the greenhouse
=
2 [ × b + b × h + h ×]
=
2 [40 × 30 + 30 × 30 + 30 × 40] cm2
=
2 [1200 + 900 + 1200] cm2
=
2 × 3300 cm2 = 6600 cm2
Hence, 6600 cm2 of glass sheet is required. Length of the tap required = Sum of the length of the 12 edges. =
4 ( + b + h)
=
4 × [40 + 30 + 30] cm = 400 cm
MANISH KUMAR Hence, 400 cm of the tape is required.
MATHEMATICS
MANISH KUMAR Ex.5
MATHEMATICS
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Sol.
Volume of a matchbox = 4 × 2.5 × 1.5 cm3 = 15 cm3
Ex.6
Volume of a packet containing 12 such boxes = 15 × 12 cm3 = 180 cm3.
A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the walls is 24 cm. If this wall is to be built up with bricks of dimensions 24 cm × 12 cm × 8 cm, then find the number of bricks which are required.
Sol.
We know that, the volume of the wall and the sum of the volumes of the required number of bricks is same. Length of the wall = 10 × 100 cm = 1000 cm Breadth or the thickness of the wall = 24 cm Height of the wall = 4 × 100 cm = 400 cm The wall is in the shape of a cuboid and its volume = 1000 × 24 × 400 cm3 Now, a brick is also a cuboid having length = 24 cm, breadth = 12 cm and height = 8 cm. Volume of one brick = 24 × 12 × 8 cm3 The required number of bricks =
Volume of the wall 1000 24 400 5000 4166.6 Volume of one brick 24 12 8 12
Hence, the required number of bricks = 4167. Ex.7
Akriti playing with plastic building blocks which are of identical cubical shapes. She makes a structure as shown in fig. If the edge of each cube is 5 cm, then find the volume of the structure. built by Aakriti.
Sol.
In fig. the structure is made with 10 cubes. Length of an edge of each block = 5 cm. Volume of one block = (5)3 cm3 = 125 cm3 Volume of the structure = Volume of the ten blocks = 10 × 125 cm3 = 1250 cm3 Hence, the volume of the structure is 1250 cm3.
RIGHT CIRCULAR CYLINDER Solid like circular pillars, circular pipes, circular pencils, measuring
jars road rollers and gas cylinders, etc., are said to be in cylindrical shape. In mathematical term, a right circular cylinder is a solid generated by the revolution of a rectangle about its sides.
MANISH KUMAR
MATHEMATICS
Let the rectangle ABCD revolve about its side AB, so as to describe a right circular cylinder as shown in the figure. You must have observed that the cross-section of a right circular cylinder are circles congruent and parallel to each other.
Cylinder Not Right Circular There are two cases when the cylinder is not a right circular cylinder. Case - I: In the following figure, we see a cylinder, which is certainly circular, but is not at right angles to the base, So we cannot say it is a right circular cylinder.
Case-II: In the following figure, we see a cylinder with a non-circular base as the base is not circular. So we cannot call it a right circular cylinder.
REMARK : cylinder.
Unless stated otherwise, here in this chapter the word cylinder would mean a right circular
The following are definitions of some terms related to a right circular cylinder: (i) The radius on any circular end is called the radius of the right circular cylinder. (ii)
Thus, in the above figure, AD as well as BC in a radius of the cylinder. The line joining the centres of circular ends of the cylinder, is called the axis of the right circular cylinder.
In the above figure, the line AB is the axis of the cylinder. Clearly, the axis is perpendicular to the circular ends. RAMERK: If the line joining the centres of circular ends of a cylinder is not perpendicular to the circular ends, then the cylinder is not a right circular cylinder. (iii) (iv)
The length of the axis of the cylinder is called the height or length of the cylinder. The curved surface joining the two bases of a right circular cylinder is called its lateral surface.
Formulae For a right circular cylinder of radius = r units & height = h units, we have :
MANISH KUMAR
MATHEMATICS Area of each circular end = r 2 sq. units. Curved (Lateral) Surface Area = ( 2rh ) sq. units. Total Surface Area = Curved Surface Area + Area of two circular ends. = ( 2rh 2r 2 ) s. units. = [ 2r (h + r)] sq. units. Volume of cylinder = r 2 h cubic units.
The above formulae are applicable to solid cylinder only.
Hollow Right Circular Cylinders Solids like iron pipes, rubber tubes, etc, are in the shape of hollow cylinders. A solid bounded by two coaxial cylinders of the same height and different radii is called a hollow cylinder
Formulae For a hollow cylinder of height h and with external and internal radii R and r respectively, we have Thickness of cylinder = (R - r) units. = ( R 2 πr 2 ) sq. nits. = ( R2 - r2) sq. units. Curved (Lateral) Surface Area = (External Curbed Surface Area) + (Internal Curbed Surface Area) = ( 2Rh 2rh ) q. units = 2h (R + r) sq. units. Total Surface Area = (Curbed Surface Area) + 2 (Area of Base Ring) Area of a cross-section
= [ ( 2Rh 2rh )+2( R 2 r 2 ) ] sq. units = 2 (Rh + rh + R2 - r2) sq. units. Volume of Material = (R2 - r2) h cubic units Volume of Hollow region = r 2 h cubic units Ex.8 Sol.
In a hot water heating system, there is a cylindrical pipe of length 28 m, and diameter 5 m. Find the total radiating surface in he system. [NCERT] 5 5 5 1 Here the length, h of the cylindrical pipe = 28 m and radius, r = cm = m= m= m 2 2 100 200 40 22 1 Total radiating surface in the system = 2rh 2 28 4.4 m 2 . 7 40
MANISH KUMAR Ex.9
Sol.
MATHEMATICS
A cylindrical block of wood has radius 70 cm and length 2 m is to be painted with blue coloured enamel. 22 The cost of painting is Rs. 1.25 per 100 cm2. Find the cost of painting the block. Take . 7 Here, the radius r of the cylindrical block of wood = 70 cm and the length h = 200 cm The total surface area of the cylindrical block = 2r(r h)
22 × 70 × (70 + 200)cm2 = 440 × 270 cm2 = 11800 cm2 7 5 Cost of painting 100 cm2 = Rs. 1.25 = Rs. 4 5 Cost of painting 1 cm2 = Rs. 400 5 5 Then, cost of painting 118800 cm2 = Rs. × 118800 = Rs. × 1188 = Rs. 1485. 4 400 Hence, cost of painting the block of wood = Rs. 1485. =2×
MANISH KUMAR Ex.10
MATHEMATICS
The curbed surface area of a right circular cylinder of height 14 cm in 88 cm 2. Find the diameter of the base of the cylinder.
Sol.
[NCERT]
Let the radius of the base of the cylinder be r cm. height, h = 14 cm Curbed surface area = 88 cm2
2rh = 88
2×
22 × r × 14 = 88 7
r=
88 7 r=1 2 22 14
2=2
Hence, the diameter of the base of the cylinder is 2 cm. Ex.11
A cylindrical vessel, without lid, has to be tin-coated including both of its sides. If the radius of its base is
1 m and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs. 50 per 1000 cm 2 (Use = 2 3.14) Sol.
Radius of the base (r) =
1 1 m= × 100 cm = 50 cm 2 2
Height (h) = 1.4 m = 1.4 × 100 cm = 140 cm Surface area to be tin-coated = 2( 2rh r 2 ) = 2(2 × 3.14 × 50 × 140 + 3.14 × (50)2]
Cost of tin-coating at the rate of Rs. 50 per 1000 cm2 = Rs.
50 × 103620 = Rs.5181. 1000
Hence, the cost of tin-coating in Rs. 5181. Ex.12
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m 2.
Sol.
Radius of the roller (r) =
[NCERT]
84 cm = 42 cm 2
length of the roller (h) = 120 cm
Area of the playground leveled in taking 1 complete revolution = 2rh = 2×
Area of the playground 2
= 31680 × 500 = 15840000 cm = Hence, the area of the playground is 1584 m2.
22 × 42 × 120 = 31680 cm2 7
15840000 2 m 1584m 2 . 100 100
MANISH KUMAR
MATHEMATICS
MANISH KUMAR Ex.13
MATHEMATICS
The pillars of a temple are cylindrical shaped. If each pillar has a circular base of radius 20 cm, and height 7m, then find the quantity of concrete mixture used to build 20 such pillars. Also find the cost of the 3
concrete mixture at the rate of Rs. 200 per m .
22 Take 7
Concrete mixture used for making each pilar = Volume of each = r h 2
Sol.
Radius of base of pillar, r=20 cm =
20 1 m m 100 5
Height of mixture used for 1 pillar Volume of mixture used for 1 pillar 2 22 1 22 3 7m 3 m 7 5 25
Volume of mixture used for 20 pillars 22 88 20 m 3 m 3 17.6m 3 25 5 3
Hence, volume of mixture required to make 20 pillars = 17.6 m 3 Now , cost of mixture at the rate of Rs. 200 per m = Rs. 200 17.6 = Rs. 3520. Hence, the cost of concrete mixture is Rs. 3520. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of 3 water can it hold? (1000 cm = 1 l) Let the base radius of the cylindrical vessel be r cm. Then circumference of the base of the cylindrical vessel = 2r cm.
Ex.14 Sol.
2
22 132 7 r 132 r 21cm 7 2 22
Height of the cylindrical vessel. H = 25 cm
22 (21) 2 (25)cm 3 7 34650 3 = 34650 cm l 34.65l 1000
Capacity of the cylindrical vessel = r2h =
Hence. the cylindrical vessel can hold 34.65 l of water. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe 35 cm. Find the mass of the pipe. If 1 cm3 of wood has a mass of 0.6 g. [NCERT]
Ex.15 Sol.
Inner diameter = 24 cm
Innar radius (r) =
Outer diameter = 28 cm
Outer radius (R) =
24 cm = 12 cm 2
28 cm = 14 cm 2
Length of the pipe (h) = 35 cm Outer volume = R2h =
22 (14) 2 35 21560cm 3 7
MANISH KUMAR
MATHEMATICS
Inner volume = r h = 2
22 (12) 2 35 15840cm 3 7
Volume of the wood used = Outer volume – Inner volume = 21560 cm3 – 15840 cm3 = 5720 cm3 Mass of the wood used = Outer volume – Inner volume = 21560 cm3 = 5720 cm3 Hence, the mass of the pipe is 3.432 kg.
MANISH KUMAR
MATHEMATICS
RIGHT CIRCULAR CONE Solids like an ice-cream cone, a conical tent, a conical vessel, a clow’n cap etc. are said to’ in conical shape. In mathematical terms, a riht circular cone is a solid generated by revolving a right-angled triangle about one of the sides containing the right angle. Let a triangle AOC revolve about it’s side OC, so as to describe a right circular cone, as shown in the figure.
Cones Not Right Circular There are two cases when we cannot call a cone a right circular cone. Case-I : The figure shown below is not a right circular cone because the line joining its vertex to the centre of its base is not at right angle to the base.
Case-II :
The figure shown below is not a right circular cone because the base is not circular.
MANISH KUMAR
MATHEMATICS
REMARK : Unless stated otherwise, by ’cone’ in this chapter, we shall mean’a’ right circular cone’
MANISH KUMAR
MATHEMATICS
The following area definitions of some terms related to right circular cone : (i) The fixed point O is called it vertex of the cone. (ii)
The fixed line OC is called the axis of the cone.
(iii)
A right circular cone has a plane end, which is in circular shape. This is called the base of the cone. The vertex of a right circular cone is farthest from its base.
(iv)
The length of the line segment joining the vertex to the centre of the base is called the height of the cone. Length OC is the height of the cone.
(v)
The length of the line segment joining the vertex to any point on the circular edge of the base, is called the slant height of the cone. Length OA is slant height of the cone.
(vi)
The radius AC of the base circle is called the radius of the cone.
Relation Between Slant Height, Radius and Vertical Height. Let us take a right circular cone with vertex at O, vertical height h, slant height and radius r. A is any point on the rim of the base of the cone and C is the
centre of the base. Here, OC = h, AC = r and OA =
The cone is right circular and therefore, OC is at right angle to the base of the cone. So, we have OC CA, i.e., OCA is right angled at C. Then by Pythagoras theorem, we have :
2 r 2 h2
Formulae For a right circular cone of Radius = r, Height = h & Slant Height = , we have :
Area of the curved (lateral) surface = (r) sq. units = r h 2 r 2 sq. units Total Surface Area of cone = (Curved surface Area + Area of Base) = (r + r) sq2 units = r ( r ) sq. units.
MANISH KUMAR 1 Volume of cone r 2 h cubic units. 2
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Hollow Right Circular Cone Suppose a sector of a circle is folded to make the radii coincide, then we get a hollow right circular. In such a cone;
(i) (ii)
Ex.16
Centre of the circle is vertex of the cone. Radius of the circle is slant height of the cone.
(iii) Length of arc AB is the circumference of the base of the cone. (iv) Area of the sector is the curved surface area of the cone. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find it curbed surface area. [NCERT]
Sol.
Diameter of the base = 10.5 cm
Radius of the base (r) =
10.5 cm = 5.25 cm 2
Slant height ( ) = 10 cm
Ex.17
Curved surface area of the cone = r
22 × 5.25 × 10 = 165 cm2. 7
The radius of the base of a conical tent is 12 m. The tent is 9 m high. Find the cost of the canvas required to make the tent, if one square metre of canvas costs Rs. 120. (Take = 3.14)
Sol.
Here, r = 12 m and h = 9 m. Let be slant height of the conical tent. Now, 2 = r2 + h2 = (12)2 + (9)2 = 144 + 81 = 225
= 15 m
The curved surface area of the tent = r = 3.14 × 12 × 15 m2 = 3.14 × 180 m2 = 565.2 m2 Hence, the canvas required = 565.2 m2 Ex.18
Cost of the canvas at the rate of Rs. 120 per m2 = Rs. 120 × 565.2 = Rs. 67824 Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.
Sol.
(i)
Slant height ( ) = 14 cm Curved surface area = 308 cm2
r = 308
r
308 7 22 14
22 × r × 14 = 308 7
r = 7 cm.
Hence, the radius of the base is 7 cm.
[NCERT]
MANISH KUMAR (ii)
Ex.19
MATHEMATICS
Total surface area of the cone = r( r )
22 22 7 (14 7 ) 7 21 462 cm 2 7 7
Hence, the total surface area of the cone is 462 cm2. How many metres of cloth of 1.1 m width will be required to make conical tent whose vertical height is 12 m and base radius is 16 m? Find also the cost of the cloth used at the rate of Rs. 41 per metre.
Sol.
Here, h = 12 m, r = 16 m
r 2 h 2 (16) 2 (12) 2 256 144 400 20 m
Curved surface area = r
22 7040 2 16 20 m 7 7
Width of cloth 1.1 m
7040 70400 6400 Length of cloth = 7 m 1.1 77 7
Cost of the cloth used @ Rs. 14. per metre = Rs.
6400 × 14 Rs. 12800. 7
Ex.20
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres ? [NCERT]
Sol.
For conical pit Diameter = 3.5 m
Radius (r) =
3.5 m 1.75 m 2
Depth (h) = 12 m
Capacity of the conical pit
=
1 2 1 22 r h (1.75) 2 12 m 2 3 3 7
= 38.5 m3 = 38.5 × 1000 = 38.5 kl. Ex.21
The curbed surface of a right circular cone is 198 cm 2 and the radius of its base is 7 cm. Find the volume
22 of the cone. Take and 2 1.41 7 Sol.
Radius of the base of the cone = 7 cm. Let h cm be the vertical height and cm be the slant height. Here, r = 7 cm. r = 198 Now, h2 + r2 =
h=
22 7 198 7
h 2 ( 7 ) 2 ( 9) 2
9 cm
h 2 81 49 32
32 cm = 4 2 cm = 4 × 1.41 cm = 5.64 cm
Volume of the cone =
1 2 1 22 r h (7 ) 2 5.64 cm 3 3 3 7
MANISH KUMAR
MATHEMATICS = 27 × 7 × 1.88 cm3 = 289.52 cm3
Hence, the volume of the cone = 289.52 cm3.
MANISH KUMAR Ex.22
MATHEMATICS
Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm. (ii) height 12 cm, slant height 13 cm.
Sol.
(i) r = 7 cm, = 25 cm r2 + h2 = 2
(7)2 + h2 = (25)2
h2 = (25)2 - (7)2
h2 = 625 - 49
h2 = 576
h = 576
Capacity =
h = 24 cm
1 2 1 22 r h (7 ) 2 24 1232 cm 3 1.232 . 3 3 7
(ii) h= 12 cm , = 13 cm r2 + h2 = 2
r2 + (12)2 = (13)2
r2 + 144 = 169
r2 169 - 144
r2 = 25
r=
r = 5 cm
Capacity =
25
1 2 1 22 2200 2200 11 r h (5) 2 12 cm 3 . 3 3 7 7 7000 35
SPHERE Objects like football, volleyball, throw-ball etc. are said to have the shape of a sphere. In mathematical term, a sphere is a solid generated by revolving a circle about any of its diameters. Let a thin circular disc of card board with centre O and radius r revolve about its diameter AOB to describe a sphere as shown in figure.
Here, O is called the centre of the sphere and r is radius of the sphere. Also, the line segment AB is a diameter of the sphere. Formulae For a solid sphere of radius = r, we have : Surface area of the sphere = ( 4r 2 ) sq. units.
4 Volume of the sphere = r 3 cubic units. 3
SPHERICAL SHELL The solid enclosed between two concentric spheres is called a spherical shell.
MANISH KUMAR
MATHEMATICS
Formulae For a spherical shell with external radius = R and internal radius = r, we have : Thickness o shell = (R - r) units. Outer surface area = 4R 2 sq. units. Inner surface area = 4r 2 sq. units. Volume of material = 4 / 3 (R3 - r3) cubic units. HEMISPHERE When a plane through the centre of a sphere cuts it into two equal pars, then each part is called a hemisphere. For a solid sphere, the obtained hemisphere is also a solid and it has a base as shown in fig.
Formulae For a hemisphere of radius r, we have : Curved surface area = 2r 2 sq. units. Total Surface Area = ( 2r 2 r 2 ) = 3r 2 sq. units. Volume =
2 3 r cubic units. 3
HEMISPHERICAL SHELL The solid enclosed between two concentric hemispheres is called a hemispherical shell.
Formulae For a hemispherical shell of external radius = R and internal radius = r, we have : Thickness of the shell = (R - r) units. Outer curved surface area = ( 2R 2 ) sq. units. Inner curved surface area = ( 2r 2 ) sq. units. Total surface area = 2R 2 2r 2 (R 2 r 2 ) (3R 2 r 2 ) sq. units.
MANISH KUMAR
MATHEMATICS Volume of the material =
2 (R3 - r3) cubic units. 3
MANISH KUMAR Ex.23 Sol.
Ex.2 Sol.
Ex.25 Sol.
MATHEMATICS
Find the surface area of a sphere of diameter 14 m. Diameter = 14 cm
[NCERT]
14 cm = 7 cm 2
Radius (r)
Surface area = 4r 2 4
22 (7 ) 2 616 cm 2 . 7
Find the radius of sphere whose surface area is 314 cm 3. (Use = 3.14) Let r cm be the radius of the sphere whose surface area = 314 cm3
4r 2 = 314
r2
4 × 3.14 × r2 = 314
314 314 100 4 3.14 4 314
r2 = 25
r = 5 cm. Find the total surface area of a hemisphere of radius 10 cm. (Use = 3.14) [NCERT] r = 10 cm.
Total surface area of the hemisphere = 3r 2 = 3 × 3.14 × (10)2 = 942 cm2. Ex.26 A hemispherical bowl is made from a metal sheet having thickness 0.3 cm. The inner radius of the bowl is 24.7 cm. Find the cost of polishing its outer surface at the rate of Rs. 4 per 100 cm 2. (Take = 3.14) Sol. The outer radius of the bowl = (24.7 + 0.3) cm, i.e. m R = 25 cm = 2R 2 = 2 × 3.14 × (25)2 cm2 = 2 × 3.14 × 25 × 25 cm2 = 157 × 25 cm2 = 3925 cm2 Cost of polishing the outer surface at the rate of Rs. 4 per 100 cm2 The outer surface area of the bowl
4 × 3925 = Rs. 157. 100 Find the amount of water displaced by a solid spherical ball of diameter 28 cm. Diameter = 28 cm = Rs.
Ex.27 Sol.
28 cm = 14 cm 2
4 3 4 22 34496 2 r (14) 3 cm 3 11498 cm 3 . 3 3 7 3 3 There are 42 hemispherical bowls, each of radius 3.5 cm. Find the quantity of water in litres which is just
Ex.28
Radius (r) =
Amount of water displaced =
22 sufficient to fill these 42 bowls. Take 7 Sol.
[NCERT]
Here, r = 3.5 cm =
7 cm. 2
Volume of one bowl i.e., capacity of one bowl =
2 3 r 3
2 22 7 7 7 11 49 cm 3 cm 3 3 7 2 2 2 6
Capacity of 42 bowls =
11 49 42 cm 3 3773 cm 3 6
MANISH KUMAR
MATHEMATICS
Thus, the required quantity of water to fill the 42 bowls =
3773 litres = 3.773 litres. 1000
Therefore, 3.773 litres of water is just sufficient. Ex.29
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Sol.
[NCERT]
Inner radius (r) = 1 m Thickness of iron sheet = 1 cm = 0.01 m
Outer radius (R) = Inner radius (r) + Thickness of iron sheet = 1 m + 0.01 m = 1.01 m
Volume of the iron used to make the tank
2 (R 3 r 3 ) 3
2 22 {(1.01) 3 13 } 0.06348 m 3 ( Approx ). 3 7
Ex.30
22 Find the volume of a sphere whose surface area is 55.44 cm 2. Take 7
Sol.
Let r cm be the radius of the sphere. Its surface area = 55.44 cm2.
4r 2 = 55.44
r2 =
4×
55.44 7 = 4.41 = (2.1)2 4 22
Now, volume of the sphere =
22 × r2 = 55.44 7
r = 2.1 cm
4 3 1 1 r (4r 2 ) r 55.44 2.1 cm 3 3 3 3 = 55.44 × 0.7 cm3 = 38.808 cm3.
Hence, the volume of the sphere = 38.808 cm3.
THINGS TO REMEMBER 1.
Surface area of a cuboid = 2 ( b = bh = h )
2.
Surface area of a cube = 6a2
3.
Curved surface area of a cylinder = 2rh
4.
Total surface area of a cylinder = 2r(r h)
5.
Curved surface area of a cone = r
6.
Total surface area of a right circular cone = r r 2 , i.e., r ( r )
7.
Surface area of a sphere so radius r = 4r 2
8.
Curved surface area of hemisphere = 2r 2
9.
Total surface area of a hemisphere = 3r 2
10.
Volume of a cuboid = b h
11.
Volume of a cylinder = a
3
MANISH KUMAR
MATHEMATICS
12.
Volume of a cylinder = r 2 h
13.
Volume of a cone =
14.
Volume of a sphere of radius r
15.
Volume of a hemisphere =
1 2 r h 3 4 3 r 3
2 3 r 3
CBSE BASED SOME IMPORTANT QUESTIONS Q.1
Three equal cubes are placed adjacently in a row. Find the ratio of the total surface of the new cuboid to that of the sum of the surface areas of three cubes. [Hint :
Let the side of a cube be a units. Then, for the resulting cuboid, we have, Length ( ) = a + a + a = 3a units.
Q.2
Breadth (b) = a units. Height (h) = a units.] Ans. 7 : 9. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2 is Rs. 15000, find the height of the hall.
[NCERT]
[Hint : Let the height of the hall be h m. Area of 4 walls = 2( + b) h = perimeter × h Q.3
Then, 250 × h × 10 = 15000] Ans. 6 m. Ajay has built a cubical water tank in his house. The top of the water tank is covered with lid. He wants to cover the inner surface of the tank including the lid with square tiles of side 25 cm. If each inner edge of the water tank is 2 m long and the tiles cost Rs. 360 per dozen, then find the total amount required for tiles. [Hint : No. of tiles required =
Total surface area of the tank Area of a square tile
360 × No. of tiles ] Ans. Rs. 11520. 12 The length of the hall is 20 m breadth 16 m. The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls. Find the height of the hall. Total cost = Rs.
Q.4
[Hint : ( × b) + ( × b) = 2 ( + b) h] Q.5.
Ans. 8.88 m (approx).
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind. [NCERT]
Sol.
Surface area of one box of size 25 cm × 20 cm × 5 cm = 2[25 × 20 + 20 × 5 + 5 × 25) cm 2 = 1450 cm2 Surface area of 250 such boxes = 250 × 1450 m2 = 362500 cm2
MANISH KUMAR
MATHEMATICS
Surface are of one box of size 15 cm × 12 m × 5 cm = 2[15 × 12 + 12 × 5 × 5 × 15) cm 2 = 630 cm2 Surface area of 250 such boxes = 250 × 630 m2 = 157500 cm2 Total surface area of the boxes of two types = 362500 m2 + 157500 cm2 = 520000 cm2 Area of sheet required for making 250 boxes of each including extra required area of 5% for overlaps etc.
5 520000 520000 cm 2 546000 cm 2 100 Total cost of sheet at the rate of Rs. 4 for 1000 cm2 = Rs.
4 × 546000 = Rs. 2184 1000
MANISH KUMAR Q.6
MATHEMATICS
The sum of the length, breadth and height of a cuboid is 21 cm and the length of its diagonal is 13 cm. Find the surface area of the cuboid. Also find the cost of painting the surface at the rate of Rs. 1.40 per cm2. [Hind : + n + h = 21 &
2 b 2 h 2 = 13 or 2 + b2 + h2 = 169
( + b + h)2 = 2 + b2 + h2 + 2( b + bh + h )
Q.7
2( b + bh + h ) = ( + b + h)2 - ( 2 + b2 + h2)]
Ans. 272 cm2 ; Rs. 380.80
The cost of papering the four walls of a room at 70 paise per square metre is Rs. 157.50. The height of the room is 5 metres. Find the length and the breadth of the room if they are in the ratio 4 : 1. [Hint : Let length be 4x m and breadth be x m. 2(4x + x) (5) ×
Q.8.
Q.9
70 = 157.50] 100
Ans. 18 m ; 4.5 m
The capacity of a cubical tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m. [NCERT] [Hint : 50,000 litres = 50 m3] Ans. 2 m. A rectangular water reservoir is 10.8 m by 3.75 m at the base. Water flows into it at the rate of 18 m per second through a pipe having cross section 7.5 cm × 4.5 cm. Find the height to which the level of water reach in 15 minutes.
Sol.
The base of the reservoir is 10.8 m × 3.75 m. Let the height of water reached in 15 minutes be h metres. The cross section of the pipe is 7.5 cm × 4.5 cm Speed of water = 18 m/sec. = 18 × 60 m/min = 1080 m/min. Cross section of the pipe is 7.5 cm × 4.5 cm, i.e., Volume of water that flows in 15 minutes
7.5 4.5 75 45 m m i.e., m m 100 100 1000 1000
75 45 1080 15 m 3 1000 1000
Now, we have volume of water in reservoir equal the volume of water that flows into it through the pipe in 15 minutes. Thus, we have 10.8 × 3.75 × h =
Q.10
75 45 × 1080 × 15 1000 1000
108 375 75 45 1080 15 h 10 100 1000 1000
h
75 45 1080 15 75 45 15 135 m m m 1.35 m. 1000 108 375 100 375 100
Hence, the height of the water level in the reservoir in 1.35 m. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube ? Also, find the ratio between their surface areas.
Q.11
[NCERT]
[Hint : Let the side of new cube be a cm. Volume of bigger cube = 8 × volume of a smaller cube] Ans. 6 cm ; 4 : 1 The areas of three adjacent faces of a cuboid are p,q and r. If its volume is v, prove that v 2 = pqr.
MANISH KUMAR [Hint
MATHEMATICS p = h × h ; q = × b ; r = xh pqr = 2 b 2 h 2 ]
MANISH KUMAR Q.12
Q.13
In figure, you see the frame of a lampshade. It is to be covered with decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.
[Hint : r = 10 cm, h = (30 + 2.5 + 2.5) cm = 35 cm] Ans. 2200 cm 2. If the radius of the base of a right circular cylinder is halved, keeping the height same , what is the ratio of the volume of the reduced cylinder to that of the original one ? [Hint : let r be the radius, h be the height & v1, be the volume of the original cylinder. The, for reduced cylinder, we
r , height = h and volume = v2] Ans. 1 : 4. 2 If costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m depp. If the cost of painting is at the rate of Rs. 20 per m2, find (i) Inner curved surface area of the vessel, (ii) Radius of the base, (iii) Capacity of the vessel. [NCERT] [Hint : Inner curved surface area × Rs. 20 = Rs. 2200] Ans. (i) 110 m2 ; (ii) 1.75 m ; (iii) 96.25 m3 (or 96.25 kl). Water is supplied to a city population for general use (not for drinking) from a river through a cylindrical pipe. The radius of the cross-section of the pipe is 20 cm. The speed of water through the pipe is 18 km per hour. Find the quantity of water in litres which is supplied to the city in two hours. (Take = 3.14 and 1 m3 = 1000 litres.) [Hint : 18 km/hr = 18000 m /hr distance covered by water in 2 h = 36000 m 1 Radius of cross section = 20 cm = m 5 1 Volume of water which flows in two hours = 3.14 × × 18000 × 2 m3 25 315 180 2 = × 1000 litres] Ans. 4521600 litres. 25 The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it ? [NCERT] [Hint : Let the radius of the vessel be r m. 22 Volume of vessel = 15.4 = 0.154 m3 × r2 × h = 0.0154] Ans. 0.47 m2. 7 A corn cob, shaped like a right circular cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 cm2 of the surface of the cob carries an average of four grains, find the number of grains of the entire cob. have radius =
Q.14
Q.15
Q.16.
Q.17
MATHEMATICS
MANISH KUMAR
MATHEMATICS
[Hint: Number of grains on the entire cob =4 curved surface area.] Q.18
Q.19
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use = 3.14) [NCERT] [Hind : Area of Tarpaulin required = Curved surface of the conical tent i.e., × b = r ] Ans. 63 m. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained. [NCERT]
[Hint :
Q.20
Radius, r = 5 cm ; height, h = 12 cm & slant height, = 13 cm ]
Ans. 100 cm3
If the triangle ABC in the question 19 above is revolved about the side 5 cm, than find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 19 and 20. [NCERT]
[Hint :
Q.21
Radius, r = 12 cm ; height, h = 5 cm & slant height , = 13 cm] Ans. 240 cm2 ; 5 : 12. The base radii of the two right circular cones of the same height are in the ratio 3 : 5. Find the ratio of their volumes. [Hint : Let r1 and r2 be the radii of two cones ; v1 and v2 be their respective volumes and h be their height.
1 2 2 r1 h r1 r2 r 3 v1 Then, ; 2 12 1 ] 1 2 r2 5 v2 r2 r2 h r2 3 Q.22
Ans. 9 : 25
If h, c and v be the height, curved surface and volume of a cone, show that 3vh 3 c 2 h 2 9v 2 0. [Hint : h = height of cone ; c = curved surface of cone = r; ; v = volume of cone =
Q.23
1 2 r h 3
Substitute the values of LHS to obtain RHS] The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas o the balloon in the two cases. [NCERT] [Hint : r1 = 7 cm & r2 = 14 cm and let S1 and S2 be the surface areas of respective spheres.
MANISH KUMAR
MATHEMATICS 2
Q.24
S1 4r12 r12 r1 Ans. 1 : 4 ] S2 4r22 r22 r2 The diameter of the moon is a approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. [NCERT] [Hint : Let d1 and d2 be the diameters of the moon and the earth respectively and S1 and S2 be their respective surface areas. d1
Q.25
d 2r r 1 1 1 1 d2 1 1 1 ] 4 d2 4 2r 2 4 r2 4
Ans. 1 : 16.
A right circular cylinder just encloses a sphere of radius r. Find (i) Surface area of the sphere, (ii) Curbed surface area of the cylinder, (iii)
Ratio of the areas obtained in (i) and (ii).
[NCERT]
[Hint : Radius of cylinder = radius of sphere = r Height of cylinder = 2 × radius of sphere = 2r] Q.26
Ans. (i) 4 r 2 (ii) 4 r 2 (iii) 1 : 1.
A shot-put is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm 3, find the mass
22 of the shot-out. Take . 7
Q.27
[Hint : Mass of 1 m3 of metal = 7.8 g Mass of the shot put = volume of shot-put × 7.8 g] Ans. 3.845 kg (approx). The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon ? [NCERT] [Hint : Let d1 and d2 be the diameters of the moon and the earth respectively. Then, d1 =
4 3 3 r1 r r1 Volume of moon 1 ; 3 1 ] 4 3 r2 r2 4 Volume of earth r 2 3 Q.28
Ans.
1 d 4 1
1 64
If the number of square centimetres of the surface of a sphere is equal to the number of cubic centimetres in its volume what is the diameter of the sphere ? [Hint : 4r 2
4 3 r r 1 ] 3 3
Q.29
A cone and hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Sol.
Let the radius of base of hemisphere and cone, each be r cm. Let the height of the cone be h cm. Volume of the cone
1 2 r h cm 3 3
Volume of the hemisphere = According to the question,
2 3 r cm 3 3
1 2 2 r h r 3 3 3
h = 2r
Height of the cone = 2r cm.
MANISH KUMAR
MATHEMATICS
Height of the hemisphere = r cm. Q.30
Ratio of their heights = 2r : = 2 : 1 Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the (i) radius r’ of the new sphere, (ii) ratio of S and S’. [NCERT] [Hint : Volume of 27 solid iron sphere each of radius r = volume of new sphere of radius R.
R = 3r
S = 4r
2
S’ = 4(3r ) 2 ]
Ans. 3r ; 1 : 9
MANISH KUMAR
MATHEMATICS
EXERCISE SUBEJCTIVE TYPE QUESTIONS. (A)
VERY SHORT ANSWER TYPE QUESTIONS :
1.
Write the lateral surface area of a cuboid having length units, breadth b units and height h. units.
2.
Write the total surface area of a cuboid having three edges of length as 10 cm, 5 cm and 3 cm.
3.
Write the curved surface area of a right circular cylinder whose radius is 3 cm and height is 5 cm.
4.
The volume of right cylinder having base radius 10 cm is 600 cm3 . Find the height of the cylinder.
5.
22 Write the curved surface area of a right circular cone having radius 7 cm and slant height 10 cm. Take 7
6.
Write the total surface area of a right circular solid cone having radius 10 cm and slant height 25
22 cm. Take 7 7.
Find the vertical height of a right circular cone whose radius is 6 cm and slant height is 10 cm.
8.
22 Find the volume of a right circular cylinder having radius 8 cm and height 10.5 cm. Take 7
9. 10.
Write the volume of a right circular cone having radius r and height h. Find the quantity of water in litres in a hemispherical bowl of radius 21 cm. The bowl is completely filled with water.
22 Take 7 11. 12. 13. (B)
The volume of a cuboid is 440 cm3 and the area of its base is 88 m2. Find its height. The volume of cube is 1000 cm. Find its total surface area. How many 3 metre cubes can be cut from a cuboid measuring 18 m × 12 m × 19 m ? SHORT ANSWR TPE QUESTIONS : QUESTIONS ON CUBOID & CUBE
1. 2.
The dimensions of cuboid are in the ratio of 1 : 2 : 3 and its total surface are is 88 m 2. Find the dimensions. Three cubes each of side 5 m are joined end to end. Find the surface are of the resulting cuboid.
3.
A swimming pool is 20 m in length, 15 m in breadth, and 4 m in depth. find the cost of cementing its floor and walls
4. 5.
at the rate of Rs. 12 per square metre. A cuboid has total surface area of 40 m2 and its lateral surface area is 26 m2. Find the area of its base. The length of a cold storage is double its breadth. Its height is 3 metres. The area of its four walls (including
6.
doors) in 108 m2. find its volume. The sum of length, breadth and depth of a cuboid is 19 cm and the length of its diagonal is 11 cm. Find the
7.
surface area of the cuboid. An open box is made of wood 3 cm thick. It external length, breadth and height are 1.48 m, 1.16 m and 8.3 dm.
8.
Find the cost of painting the inner surface at Rs. 50 per sq. metre. A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the
9.
base are 15 cm and 12 cm. Find the rise in water level in the vessel. A solid cube is cut into two cuboids of equal volumes. Find the ratio of the total surface area of the given cube and that of one of the cuboids.
MANISH KUMAR 10.
MATHEMATICS
Three metal cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find its edge. Also, find the surface area of the new cube.
MANISH KUMAR
MATHEMATICS
QUESTIONS OF RIGHT CURCULAR CYLINDER. 11.
The area of the base of right circular cylinder is 154 cm 2 and its height is 15 cm. Find the volume of the cylinder.
12.
The thickness of a hollow wooden cylinder is 2 cm. It is 35 cm long and its inner radius is 12 cm. Find the volume of the wood required to make the cylinder, assuming it is open at either end.
13.
22 The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm 3. Find its radius. Use 7
14.
The volume of metallic cylindrical pipe is 748 cm3. Its length is 14 cm and its external radius is 9 cm. Find its thickness. QUESTIONS OF RIGHT CIRULAR CONE
15.
The diameter of a cone is 14 cm and its slant height is 9 cm. Find the area of its curved surface.
16.
Find the total surface area of a cone, if its slant height is 9 m and the radius of its base is 12 m.
17.
The radius of a cone is 3 cm and vertical height is 4 cm. Find the area of the curved surface.
18.
The radius and slant height of a cone are in the ratio 4 : 7. It its curved surface are is 792 m 2, find its radius.
22 Use 7 19.
The lateral surface of a cylinder is equal to the curved surface of a cone. If the radius be the same, find the ratio of the height of the cylinder and slant height of the cone.
20.
Find the volume of a right circular cone 1.02 m high, if the radius of its base is 28 cm.
21.
The diameter of a right circular cone is 8 cm and its volume is 48 cm 3 . What it its height ?
22.
A right circular cone is 3.6 cm high and radius of its base is 1.6 cm. It is melted and recast into a right circular cone with radius of its base as 1.2 cm. Find its height.
23.
A conical vessel whose internal radius is 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which the water rises.
24.
A cone a cylinder are having the same base. Find the ratio of their heights if their volumes are equal. QUESTIONS ON SPHERE
25.
Find the surface are and total surface area of a hemisphere of radius 21 cm.
26.
A sphere, a cylinder and a cone are of the same radius and same height. Find the ratio of their curved surface.
27.
Sow that the surface area of a sphere is the same as that of the lateral surface of a right circular cylinder that just encloses the sphere.
28.
The internal and external diameters of a hollow hemi-spherical vessel are 24 cm and 25 cm respectively. The cost of paint one sq. cm of the surface is 7 paise. Find the total cost to paint the vessel all over. (ignore the area of edge).
29.
Find the volume of a sphere whose surface area is 154 square cm.
30.
A solid sphere of radius 3 cm is melted and then cast into small spherical balls each of diameter 0.6 cm. Find the number of balls thus obtained.
MANISH KUMAR 31.
MATHEMATICS
How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter.
32.
A solid lead ball of radius 7 cm was melted and then drawn into a wire of diameter 0.2 cm. Find the length of the wire.
(C)
LONG ANSWER TYPE QUESTIONS : QUESTION ON CUBE & CUBOID
1.
Length of class-room is two times its height and breadth is 1
2.
walls at the rate of Rs. 1.60 per m2 is Rs. 179.20. Find the cost of tilling the floor at the rate of Rs. 6.75 per m2. The dimensions of a rectangular box are in the ratio 2 : 3 : 4 and the difference between the cost of covering it
3.
with sheet of appear at the rate of Rs. 4 and Rs. 4.50 per square metre is Rs. 416. Find the dimensions of the box. Find the number of bricks, each measuring 25 cm × 12.5 cm × 7.5 cm required to construct a wall 6 m long, 5 m
4.
high and 0.5 m thick, while the cement and sand mixture occupies 1/20 of the volume of the wall. A class room is 7 m long, 6.5 m wide and 4 m high. It has one door 3 m × 1.4 m and three windows, each
1 times its height. The cost of white-washing the 2
measuring 2 m × 1 m. The interior walls are to be colour washed. The contractor charges Rs. 525 per sq. m. Find 5.
the cost of colour washing. A room is half as long again as it is broad. The cost of carpeting the room at Rs. 3.25 per m 2 is Rs. 175.50 and the
6.
cost of papering the walls at Rs. 1.40 per m2 is Rs. 240.80. If 1 door and 2 windows occupy 8 m2, find the dimensions of the room. A wooden bookshelf has external dimensions as follows : Height = 110 cm, Depth = 25 cm, Breadth = 85 cm. The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm 2 and the rate of painting is 10 paise per cm2. Find the total expenses required for polishing and painting the surface of the bookshelf.
7.
In fig. the shape of a solid csopper piece (made of two pieces with dimensions as shown in the figure) is shown. The face ABCDEFA is the uniform cross section. Assume that the angle at A, B, C, D, E and F and are right angles. Calculate the volume of the piece.
MANISH KUMAR
MATHEMATICS
8.
A plot of land in the form of a rectangle has dimension 240 m × 180 m. A drain let 10 m wide is dug all around it
9.
(on the outside) and the earth dug out is evenly spread over the plot, increasing its surface level by 25 cm. Find the depth of the drain let. A metallic sheet is of the rectangular shape with dimensions 48 cm × 36 cm. From each one of its corners, a
10.
square of 8 cm is cutoff. An open box is made of the remaining sheet. Find the volume of the box. Water in a canal, 30 dm wide and 12 dm depp, is flowing with a velocity of 20 km per hour. How much area will it irrigate in 30 min, if 9 cm of standing water is desired ? QUESTIONS ON RIGHT CIRCULAR CYLINDER
11.
A cylindrical road roller made of iron is 1 m wide. Its inner diameter is 54 cm and thickness of the iron sheet rolled
12.
into the road roller is 9 cm. Find the weight of the roller it 1 c.c. of iron weights 8 g. A solid cylinder has total surface area of 462 square cm. Its curved surface area is one-third of its total surface area. Find the volume of the cylinder (Take = 22/7)
13.
A well with 10 m inside diameter is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of embankment.
14.
QUESTIONS ON RIGHT CIRCULAR CONE A tent is of the shape of right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum h eight of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at rate of Rs. 2 per square metre, if the radius of the base is 14 metres.
15. 16.
A solid cube of side 7 cm is melted to make a cone of height 5 cm, find the radius of the base of the cone. From a right circular cylinder with height 10 cm and radius of base 6 cm, a right circular cone of the same height
17.
and base is removed. Find the volume of the remaining solid. QUESTIONS ON SPHERE The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. The cost
22 to paint 1 cm2 surface is Rs. 0.05. Find the total cost to paint the vessel all over. Use 7
MANISH KUMAR 18.
MATHEMATICS
A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the cost of painting the toy at the rate of Rs. 5 per 1000 cm 2.
MANISH KUMAR 19.
MATHEMATICS
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small
supports as whose in fig. Eight such spheres are used for this purpose, and are to be painted silver. Each support is cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
20.
A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is four times
(D) 1.
the radius of its base, find the radius of the ice-cream cone. NCERT QUESTIONS : A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine : (i)
The area of the sheet required for making the box.
2.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs. 20. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the
3.
walls of the room and the ceiling at the rate of Rs. 7.50 per m2. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high. (i)
4.
Which ox has the greater lateral surface area and by how much ?
(ii) Which box has the smaller total surface area and by how much ? A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
5.
(i) What is the area of the glass ? (ii) How much of tape is needed for all the 12 edges ? It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How
6.
many square metres of the sheet are required for the same ? A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its :
7.
(i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area. A cylindrical pillar is 50 cm is diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar
8.
at the rate of Rs. 12.50 per m2. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its
9.
height. The inner diameter so a circular well is 3.5 m. It is 10 m deep. Find (i)
10.
its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs 40 per m2. Find the total surface area of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required t make 10 such caps.
MANISH KUMAR 12.
MATHEMATICS
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard, Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2, what will be the cost of painting all these cones ? (Use = 3.14 and take
13.
Find the surface are of a sphere of radius : (i) 10.5 cm
14.
1.04 = 1.02)
(ii) 5.6 cm
(iii) 14 cm
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2.
15.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold ? (1 m3 = 1000 )
16.
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m 3.
17.
A godown measures 40 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
18.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute ?
19.
A soft drink is available in two packs - (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much ?
20.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find (i) radius of its base
21.
(ii) its volume. (Use = 3.14)
A lead pencil consist of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
22.
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?
23.
Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
24.
The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use = 3.14)
25.
If the volume of a right circular cone of height 9 cm is 48 cm3, find the diameter of its base.
26.
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find (i) height of the cone
27.
(ii) slant height of the cone
(iii) curbed surface area of the cone
A heap a wheat is in the form of a cone whose diameter is 1 0.5 m and height is 3 m. Find its volume. The heap its to e covered by canvas to protect it from rain. Find the area of the canvas required.
28.
Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m
MANISH KUMAR
MATHEMATICS
29.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ?
30.
Find the volume of a sphere whose surface area is 154 cm2.
31.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm 3) is needed to fill this capsule ?
MANISH KUMAR
MATHEMATICS
OBJECTIVE TYPE QUESTIONS 1.
A rectangular sand box is 5 m wide and 2 m long. How many cubic metres of sand are needed to fill the box upto depth of 10 cm ? (a) 1
2.
(b) 10
(c) 100
(d) 1000
A beam 9 m long, 40 cm wide and 20 cm deep is made up of iron which weight 50 kg per cubic mer. The weight of the beam is : (a) 27 kg
3.
(c) 48 kg
(d) 56 kg
The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm × 6 cm × 2 cm is: (A) 2 13 cm
4.
(b) 36 kg
(B) 2 14 m
(C 2 26 cm
(D) 10 2 cm
How many bricks, each measuring 25 cm × 12.5 cm × 7.5 cm will be needed to construct a wall 15 m long 1.8 m high and 37.5 cm thick ? (a) 4400
5.
(b) 4660
(c) 4320
(d) 4575
A wooden box of dimensions 8 m × 7 m × 6 m is to carry rectangular boxes of dimensions 8 cm × 7 cm × 6 cm. The maximum number of boxes that can be carried in the wooden box, is : (a) 9800000
6.
(d) 19683 cm3
(b) 1000
(c) 800
(d) 160
(b) 256 cm3
(c) 484 cm3
(d) 512 cm3
(b) 10 2 cm
(c) 10 3 cm
(d) 20 cm
If the length of diagonal of a cube is 4 3 cm, then the length of its edge is : (a) 2 cm
11.
(c) 4374 cm 2
The length of the longest rod that can fit in a cubical vessel of side 10 cm, is (a) 10 cm
10.
(b) 729 cm2
The volume of a cube with surface area 384 sq. cm, is : (a) 216 cm3
9.
(d) 1200000
The perimeter of one face of a cube is 40 cm. The volume of the cube (in cm3) is : (a) 1600
8.
(c) 1000000
The surface area of a cube of side 27 cm is : (a) 2916 cm2
7.
(b) 7500000
(b) 2 cm
(c) 4 cm
(d) 6 cm
If the diameter of the base of a cylindrical pillar is 4 m and its height is 21 m, then the cost of construction of the pillar at Rs. 1.50 per cubic metre is :
MANISH KUMAR (a) Rs. 396
MATHEMATICS (b) Rs. 400
(c) Rs. 410
(d) Rs. 420
MANISH KUMAR 12.
The volume of the cylinder whose height is 14 cm and diameter of base 4 cm, is : (a) 176 cm3
13.
MATHEMATICS (b) 196 cm3
(c) 276 cm3
(d) 2 cm2
Given that 1 cm3 of a metal weighs 5 gms, the weight of a cylindrical metal container with base radius 10.5 cm and height 60 cm, is : (a) 97.65 kg
14.
(b) 2992
(c) 2292
(d) 2229
(b) 15 cm
(c) 20 cm
(d) 40 cm
The volume (in cm3) of a right circular cone of height 12 cm and base radius 6 cm, is : (a) 12
17.
(d) 102.45 kg
If the curved surface area of a cylinder is 1760 sq, m and its base radius is 14 cm, then its height is : (a) 10 cm
16.
(c) 103.95 kg
If the diameter of a cylinder is 28 cm and its height is 20 cm, then total surface area (in cm 2) is : (a) 2993
15.
(b) 48.75 kg
(b) 36
(c) 72
(d) 144
If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is : (a)
18.
1 3 r 3
6 7
(b) 52
6 7
(c) 31
3 7
(d) 15
5 7
(b) 4.5 m
(c) 5 m
(d) 6 m
(b) 525 m
(c) 490 m
(d) 860 m
The volume of a sphere is 38808 cu. cm. The curved surface area of the sphere (in cm 2) is : (a) 5544
22.
(d) 9r 3
The length of canvas 1.1 m wide required to build a conical tent of height 14 m and the floor area 346.5 m 2, is (a) 655 m
21.
(c) 3r 3
A cone of height 8 m has a curved surface area 188.4 m2. The radius of the base is : (Take = 3.14): (a) 4m
20.
2 3 r 3
The lateral surface area (in cm2) of a cone with height 3 cm and radius 4 cm, is : (a) 62
19.
(b)
(b) 1386
(c) 8316
(d) 4158
The volume of a spherical shell whose internal and external diameters are 8 cm and 10 cm respectively (in cubic cm) is : (a)
23.
122 3
(c) 212
(d) 257
(B) 1 L 4
(C) 1 : 8
(D) 1 : 16
If a hemi-spherical dome has an inner diameter of 28 m, then its volume (in m 3) is : (a) 6186.60
25.
244 3
If the ratio of volumes of two spheres is 1 : 8, then the ratio of their surface areas is : (A) 1 : 2
24.
(b)
(b) 5749.33
(c) 7099.33
(d) 7459.33
Three solid spherical beads of radii 3 cm, 4 cm 5 cm are melted into a spherical bead. Its radius is :
MANISH KUMAR (A) 6 cm
MATHEMATICS (B) 7 cm
(C) 8 cm
(D) 9 cm
MANISH KUMAR
MATHEMATICS ANSWER KEY
SURFACE AREAS & VOLUMES (A)
(B)
VERY SHORT ANSWER TYPE QUESTIONS : 1. 2 ( + b)h
2. 190 cm2
8. 2112 cm3
9.
3. 30 cm 2
1 2 r h 3
4. 6 cm
10. 19.404 litres 11. 5 cm
5. 220 cm2
6. 1100 cm2
12. 600 cm2
13. 72
4. 7 m2
5. 216 m3
7. 8 cm
SHORT ANSWER TYPE QUESTIONS : 1. 2 m, 4 m and 6 m
2. 350 cm2
3. Rs. 6960
7. Rs. 279.70
8. 4.05 cm
9. 3 : 2
10. 6 cm, 216 cm2
11. 2310 cm3
12. 5720 cm3
13. 5 cm
14. 1 cm
15. 198 cm2
17. 62.85 cm2
18. 12 cm
19. 1 : 2
21. 9 cm
22. 6.4 cm
23. 2 cm
24. 3 : 1
26. 4 : 4 : 5
28. Rs. 132.11
29. 179.66 cm3
20. 83776 cm
3
25. 2772 cm2 ; 4158 cm2 31. 2541 (C)
EXERCISE
6. 240 cm2
30. 1000
32. 457.33 m
LONG ANSWER TYPE QUESTIONS : 1. Rs. 324
2. 8m, 12 m and 16 m
3. 6080
5. length = 9 m, breadth = 6m and height = 6 m 3
10. 4,00,000 m
15. 8.09 cm
3
9. 5120 cm
16. 754.28 cm
2
4. Rs. 513.45 6. Rs. 6275 3
7. 880 cm3
8. 1.227 m 14. Rs. 2068
11. 1424.304 kg
12. 539 cm
13. 4.66 m
17. Rs. 96.28
18. 51 paise
19. Rs. 382.80 20. 3 cm
(D) NCERT QUESTIONS : 1. (i) 5.45 m2
(ii) Rs. 109
3. (i) Lateral surface area of cubical box is greater by 40 cm2
2. Rs. 555
(ii) Total surface are of cuboidal box is greater by 10 cm2 5. 7.48 m2
4. (i) 450 cm2 (ii) 320 cm
6. (i) 968 cm2 (ii) 1064.8 cm2 (iii) 2038.08 cm2 2
9. (i) 110 m (ii) Rs. 4400
10. 1244.57 cm
2
13. (i) 1386 cm2 (ii) 394.24 cm2 (iii) 2464 cm2 16. Rs 4320
17. 16000
20. (i) 3 cm (ii) 141.3 cm3
18. 4000 m3
8. 1 m
11. 5500 cm
12. Rs. 384.34 (approx)
14. Rs. 27.72
15. 135000 litres
19. The cylinder has the greater capacity by 85 cm3
21. Volume of wood = 5.28 cm3, Volume of graphite = 0.11 cm3
22. 38500 cm3 or 38.5 litre of the soup. 26. (i) 48 cm (ii) 50 cm (iii) 2200 cm2 29. 0.303 litre (approx)
7. Rs. 68.75 2
23. (i) 264 cm3 (ii) 154 cm3 27. 86.625 m3 ; 99.825 m2
30. 179 179
2 cm3 3
24. 10 cm
28. (i) 11498
25. 8 cm
2 cm3 (ii) 0.004851 m3 3
31. 22.46 mm3 (approx)
ANSWER KEY ue.
1
3
4
5
6
7
8
9
10
11
Ane.
A
B
C
C
C
C
B
D
C
C
Que.
11
12
13
14
15
16
17
18
19
20
Ans.
A
A
C
B
C
D
D
A
D
B
MANISH KUMAR
MATHEMATICS
Que
21
22
23
24
25
Ans.
A
B
B
B
A
TRIANGLES
INTRODUCTION You have leant about triangles and some of their properties in your previous classes. In this chapter, we will study about the congruency of triangles and some more properties. We many had some idea about these properties in lower classes but here, we will study these properties in greater details.
CONGRUENCE OF TRIANGLES Congruent figures : Two geometrical figures, having exactly the same shape and size are known as congruent fig. For congruence, we use the symbol . Thus, two line segments are congruent, if they have the same length. Two angles are congruent, if they have the same measure. Congruent Triangles : Two triangles are congruent, if and only if one of them can be made to superpose on the other, so as to cover it exactly. Thus, Thus, congruent triangles are exactly identical, i.e., their corresponding three sides and the three angles are equal.
If ABC is congruent of DEF, we write ABD DEF. This happens when AB = De, BC = EF, AC = DF and A = D, B = E, C = F. In this case, we say that the sides corresponding to AB, BC and AC are DE, EF and DF respectively. And, the angles corresponding to A, B and C are D, E and F respectively. Thus, the corresponding parts of two congruent triangles are equal. We show it by the abbreviation C.P.C.T., which means corresponding parts of congruent triangles. Congruence Relation in the Set of All Triangles : From the definition of congruence of two triangles, we obtain the following results:
(i)
Every triangle is congruent to itself i.e. ABC ABC.
(ii)
If ABC ADEF, then DEF ABC.
(iii)
if ABC DEF, and DEF PQR, then ABC PQR.
CRITERIA FRO CONFRUENCE OF TRIANGLES In earlier classes, we have leant some criteria from congruence of triangles. Here, in this class we will learn the truth of these either experimentally or by deductive proof. In the previous section we have studied that two triangles are congruent if and only if there exists a correspondence between their vertices such that the corresponding sides and the corresponding angles of two
MANISH KUMAR
MATHEMATICS
triangles are equal i.e. six equalities hold good, three of the corresponding sides and three of the corresponding angles. In this section, we shall prove that if three properly chooses conditions out of the six conditions are satisfied, then the other three are automatically satisfied. Let us now discuss those conditions which ensure the congruence of two angles.
MANISH KUMAR I.
MATHEMATICS
SIDE-ANGLES(SAS)CONGRUENCE CRITERION AXIOM : Two triantgles are congruent if two sides and included angle of one triangle are equal to the sides and the included angle of the other triangle. In the given figure, in ABC and DEF, we have : AB = DE, AC = DF and A D.
ABC DEF
[By SAS-criteria]
REMARK : SAS-Congruence rule holds but ASS or SSA or SSA rule does not hold.
II.
ANGLE (ASA) CONGRUENCE CRITERION THEOREM-1 : Two triangles are congruent, if two angles and the included side of the one triagle are equal to two angles and the included side of the other triangle. Given : Two triangles ABC and DEF such that B E , C F and BC = EF. To prove : ABC DEF . Proof : To prove ABC DEF . we need to consider three possible situations : Case 1. Let AB = DE.
STATEMENT 1.
2.
In ABC and DEF (i) AB = DE (ii) B E (iii) BC = EF
ABC DEF
REASON
Supposed Given Given By SAS criteria
Case2. Let AB > DE.
Construction : Take a point H on AB such that HB = DE.
MANISH KUMAR
MATHEMATICS STATEMENT
In HBC and DEF (i) HB = DE (ii) B E (iii) BC = EF
1.
HBC DEF HCB DFE ACB DFE HCB ACB
2. 3. 4. 5.
REASON By Construction Given Given By SAS criteria C.P.C.T Given From (3) and (4)
It can be possible only if H coincides with A. In other words, AB = DE. Case 3. Let AB < DE.
Contruction : Take a point M on DE such that ME = AB. By repeating the same arguments as in Case 2, we prove that AB = DE, and so ABC DEF . Hence. proved
III. ANGLE-ANGLE-SIDE (AAS) CONGRUENCE CRITERION THEOREM-2 : Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. Given : Two s ABC and DEF such that A D, B E , BC EF . To prove : ABC DEF
Proof : STATEMENT 1. 2. 3.
4.
5.
A D B E A B D E 180 0 C 180 0 F
REASON Given Given Addint (1) and (2)
A B C 180 0 A B 180 0 C 0 Similarly. D E 180 F
C F In ABC and DEF (i) B E (ii) BC = EF (iii) C F
ABC DEF
Given Given From(3) By ASA criteria
MANISH KUMAR IV
MATHEMATICS
SIDE-SIDE-SIDE (SSS) CONGRUENCE CRITERION Two triangles are congruent, if the three sides of one triangle are equal to the corresponding three sides of the other triangle. If the given figure, in ABC and DEF, we have : AB = DE, BC = EF and AC = DF.
V
ABC DEF [By SSS-criteria]
RIGHT ANGLE-HYPOTENUSE-SIDE (RHS) CONGRUENCE CRITERION Two right angles are congruent, if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle. In the given figure, ABC and DEF are right - angled triangles in which Hyp. AC = Hyp. DF and BC = EF.
Ex.1
ABC DEF [By RHS-criteria]
In fig. OA = OB and OD = OC. Show that
[NCERT]
(i) AOD BOC and (ii) AD || BC. Sol.
Given : In fig. OA = OB and OD = OC. To prove : (i) AOD BOC and (ii) AD || BC. Proof :
STATEMENT 1.
2.
In AOD and BOC (i) (ii)
OA = OB OD = OC
(ii)
AOD = BOC
AOD BOC
3.
OAD = OBC
4.
OAD = OBC AD || BC.
Hence, proved. Ex.2
IN ABC, AB = AC. If P is a point of AB and Q is a point on AC such that AP = AQ. Prove that (i) APC AQB
Sol.
(ii) BPC CQB.
Given : In ABC, AB = AC. P is a point on AB and Q is a point on AC such that AP = AQ.
REASON Given Given Vertically opposite angles By SAS criteria C.P.C.T.
OAD and OBC from a pair of alternate angles and are equal.
MANISH KUMAR
MATHEMATICS
To prove : (i) APC AQB (ii) BPC CQB. Proof :
STATEMEN 1.
In APC and AQB (i) AP = AQ (ii) AC = AB (iii) CAP = BAQ
2. 3.
REASON
APC AQB AB = AC and AP = AQ
Given Given Common By SAS criteria Given
(AB - AP) = (AC - AQ) BP = CQ 4.
In BPC and CQB
From (3) C.P.C.T. APC and AQB
(i) BP = CQ (ii) PC = QB
Common By SSS criteria
(iii) BC = CB 5.
BPC CQB
Hence, proved. Ex.3
In figure, AC = AE, AB = AD and BAD = EAC, prove that BC = DE.
Sol.
Given : In figure, AC = AE, AB = AD and BAD = EAC.
To prove : BE = DE. Construction : Joint DE.
[NCERT]
MANISH KUMAR
MATHEMATICS STATEMEN
1.
REASON
In ABC and ADE (i) AB = AD
Given
(ii) AC = AE
Given
(iii) BAD = EAC
Given
BAD + DAC = DAC + EAC
Adding DAC to both sides
BAC DAE 3.
ABC ADE
By SAS criteria
3.
BC = DE
C.P.C.T.
Hence, proved. Ex.4
In figure, diagonal AC of a quadrilateral ABCD bisects the angles A and C. Prove that AB = AD and CB =
Sol.
Given : In figure, diagonal AC of a quadrilateral ABCD
CD/ bisect the angles A and C. To prove : AB = AD and CB = CD
STATEMEN 1.
REASON
In ABC and ADC (i) BAC = DAC (ii) ACB = ACD (iii) AC = AC
2.
ABC ADC
3.
AB = AD BC = CD
Given Given Common ASA Axiom C.P.C.T. C.P.C.T.
Hence, proved. Ex.5
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD =
ABE and EPA = DPB. Show that (i) DAP EBP (ii) AD = BE. Sol.
[NCERT]
Given : AB is a line segment and P is its mid-point. D and E are points of the same side of AB such that BAD =
ABE and EPA = DPB.
MANISH KUMAR
MATHEMATICS
To prove : (i) DAP EBP (ii) AD = BE Proof :
STATEMEN 1.
In DAP and EBP (i) AP = BP
P is the mid point of AB.
(ii) DAP = EBP
Given
(iii) EPA = DPB
Given
EPA + EPD APD = APE 2. 3.
REASON
DAP EBP
Adding EPD to both sides. By ASA criteria C.P.C.T.
AD = BE
Hence, proved Ex.6
AB is a line-segment. As and By are two equal line - segments drawn on opposite sides on line AB such that AX || BY. if AB and XY intersect each other at P. Prove that : (i) APX BPY
Sol.
(ii) AB and XY bisect each other at P.
Given : AB is a line-segment. AX and BY are to equal line-segments drawn on opposite sides of line AB such that AX || BY. To prove : (i) APX BPY
(ii) AB and XY bisect each other at P.
MANISH KUMAR
MATHEMATICS STATEMEN
1.
2.
REASON
AX || BY PAX = PBY
Alternate Interior Angles
and PXA = PYB
Alternate Interior Angles
In APX and BPY (i) PAX = PBY
From (1)
(ii) PXA = PYB
From (1)
(iii) AX = BY 3. 4.
APX BPY PX = PY and PA = PB. AB and XY bisect each other at P.
Given By ASA criteria C.P.C.T.
Hence, proved. Ex.7
In fig., AD is a median and BL, CM are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BL = CM.
MANISH KUMAR Sol.
MATHEMATICS
Given : In fig., AD is a median and BL, CM are perpendiculars drawn from B and C respectively on AD and AD produced. To prove : BL = CM
STATEMEN
REASON
1. 2.
In ABDL and CDM Each equal to 900
3.
(i) BLD = CMD (ii) BDL = CDM
Vertically Opposite Angles D is the mid point of BC By AAS criteria C.P.C.T.
(iii) BD = DC
BDL CDM BL = CM
Ex.8
Sol.
Hence, proved. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Given : AD and BC are equal perpendiculars to a line segment AB. To Prove : CD bisects AB.
STATEMEN 1.
REASON
In OAD and OBC (i) AD = BC (ii) OAD = OBC (iii) AOD = BOC
2. 3.
OAD OBC
Given Each = 900 Vertically Opposite Angles By AAS criteria C.P.C.T.
OA = OB CD bisects AB
Hence, proved. Ex.9
If fig. BCD = ADC and ACB = BDA. Prove that AD = BC and A = B/
[ NCERT]
MANISH KUMAR Sol.
MATHEMATICS
Given : In fig BCD = ADC and ACB = BDA To prove : (i) AD = BC (ii) A = B
STATEMENT 1.
In APC and BPC (i) ACP = BDP (ii) APC = BPD (iii) Third A = Third B
2.
In ACD and BDC (i) CD = CD (ii) A = B (iii) BCD = ADC
3. 4.
Ex.10
Sol.
REASON
ACB = BDA ACD = BDA ACD = BDC ADC BDC
Given Vertically Opposite Angles The sum of the three angles of a is 1800
Common From (1) Given Given By AAS criteria C.P.C.T.
AD = BC Hence, proved. In the fig. ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that : (i) AC bisects each of the angles A and C. (ii) BE = ED. (iii) ABC = ADC. Can we say that AE = EC ? Given : In the fig. ABCD is a quadrilateral in which AB = AD and BC = DC. To prove : (i) AC bisects each of the angles A and C. (ii) BE = ED. (iii) ABC = ADC. Can we say that AE = EC ?
MANISH KUMAR
MATHEMATICS STATEMENT
1.
In ABC and ADC (i) AB AD (ii) BC = DC (iii) AC = AC
2. 3.
ABC ADC
4.
(i) AB = AD
1 = 2 and 4 =4 AC bisects each of the angle A and C In ABE and ADE (ii) 1 = 2 (iii) AE = AE
ABE ADE 5. 6. 7.
Ex.11
REASON Given Given Common By SSS criteria C.P.C.T.
Given From (3) Common By SAS criteria C.P.C.T.
BE = ED
ABC ADC
C.P.C.T.
ABC = ADC
No. AE EC. From AE to be equal to EC, it is necessary that AED CED for which 2 must be equal to 4. Hence, proved. AB is a line segment, P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is the perpendicular bisector of AB.
Sol.
[NCERT]
Given : AB is line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. To prove : PQ is the perpendicular bisector of AB.
MANISH KUMAR
MATHEMATICS
STATEMENT
REASON
In PAQ and PBQ
Given
(i) AP = BP
Given
(ii) AQ = BQ
Common
(iii) PQ = PQ
By SSS criteria
2.
PAQ PBQ
C.P.C.T.
3.
APQ = BPQ
1.
4.
in PAC and PBC (i) AP = BP (ii) APC = BPC
APQ = BPQ, from (3) Common By SAS criteria
(iii) PC = PC
C.P.C.T.
PAC PBC
Linear pair of angles
5.
AC = BC and ACP = BCP
6.
ACP + BCP = 1800
7.
Given
2ACP = 1800 ACP = 900
From (6) and (7) we conclude that PQ is the perpendicular bisector of AB. Hence, proved.
MANISH KUMAR Ex.12
MATHEMATICS
ABC and PBC are two isosceles triangles on the same base BC and vertices A and P are on the same side of BC. A and P are joined, show that : (i)
Sol.
ABP ACPand (ii) AP bisects A of ABC.
Given : ABC and PBC are two isosceles triangles on the same base BC and vertices A and P are on the same side of BC. To prove :
(i) ABP ACP and (ii) AP bisects A of ABC.
STATEMENT 1.
2.
Ex.13
3.
REASON
In ABP and ACP
ABC is isosceles
(i) AB = AC (ii) PB = PC (iii) AP = AP
PBC is isosceles
ABP ACP
Common By SSS criteria C.P.C.T.
BAP = CAP AP bisect A.
Hence, proved In fig. P is a point equidistant from the lines and m intersecting at point A. Show that the line n (along AP) bisects the angle between and m.
Sol.
Given : In fig. P is a point equidistant from the lines and m intersecting at point A. To prove : line n (along AP) bisects the angle between and m.
[NCERT]
MANISH KUMAR
MATHEMATICS STATEMENT
1.
REASON
In right PAB and PAC (i) PB = PC (ii) Hyp PA = Hyp PA
Given Common By RHS criteria
PAB PAC 2. 3.
BAP = CAP Line n bisects angle between and m.
C.P.C.T.
Hence, proved. Ex.14
AD, BE and CF, the altitudes of ABC are equal. Prove that ABC is an equilateral triangle.
Sol.
Given : AD, BE and CF are the altitudes of ABC and are equal. To prove : ABC is an equilateral triangle
STATEMENT 1.
In right BCE and CBF (i) Hyp. BC = Hyp. CB (ii) BE = CF
2. 3.
BCE CBF
B = C AC = AB
4.
Similarly, ABD BAE
B = A
REASON Common Given By RHS criteria C.P.C.T. Sides opposite to equal angles are equal C.P.C.T. Sides opposite to equal angles are equal From (3) and (4)
AC = BC 5.
AB = BC = AC
Hence ABC is an equilateral triangle. Ex.15
In a ABC, the internal bisectors of B and C meet at O. Prove that OA is the internal bisector of A.
MANISH KUMAR
Sol.
Given : In a ABC, the internal bisectors of B and C meet at O. To prove : OA bisects A. Construction : Draw OD BC, OE, CA and OF AB.
MATHEMATICS
MANISH KUMAR
MATHEMATICS
Proof :
1.
STATEMENT In ODC and OEC (i) OCD = OCE (ii) ODC = OEC (iii) OC = OC
2. 3. 4. 5. 6.
7. 8.
ODC OEC OD = OE OD = OF OE = OF In right OEA and OFA (i) OE = OF (ii) Hyp. OA = Hyp, OA
OEA OFA
OAE = OAF OA bisects A.
REASON OC bisects C. Each = 900 (by construction). Common By AAS criteria C.P.C.T. Similarly, (ODB OFB). From (3) and (4) From (5) Common By RHS criteria C.P.C.T .
Hence, proved.
PROPERTIES OF AN ISOSCELES TRAINGLE In this section, we will learn some properties related to a triangle whose two sides are equal. We know that a triangle whose two sides are equal is called an isosceles triangle. Here, we will apply SAS congruence criteria and ASA (or (AAS) congruence criteria to study some properties of an isosceles triangle.
THEORM - 3 : Angles opposite to equal sides of an isosceles triangle are equal.
Given : ABC is an isosceles triangle and AB = AC. To prove : B = C. Construction : Draw AD the bisector of A. AD meets BC at D. Proof :
MANISH KUMAR
MATHEMATICS STATEMENT
1.
2. 3.
In BAD and CAD (i) AB = AC
REASON Given Construction Common By SAS criteria C.P.C.T.
(ii) BAD = CAD (iii) AD = AD
BAD CAD
B = C
Hence, proved.
COROLLARY each angle of an equilateral triangle is of 600.
Given : ABC is an equilateral triangle. To prove : A = B = C = 600. Proof :
STATEMENT 1. 2. 3. 4. 5.
REASON
AB = AC = BC
ABC is an equilateral triangle.
B = C A = C A = B = C A + B + C = 1800
AB = AC AB = BC From (2) and (3) Angle sum property.
A = B = C =
1 × 1800 = 600 3
Hence, proved. THEOREM-4 : The sides opposite to equal angles of a triangle equal.
Given : ABC in which B = C.
MANISH KUMAR
MATHEMATICS
To prove : AB = AC. Construction : Draw AD BC. AD meets BC in D. Proof :
STATEMENT 1.
REASON
In ABD and ACD Given Each = 900 Common By AAS criteria C.P.C.T.
(i) ABD = ACD (ii) ADB = ADC (ii) AD = AD 2. 3.
ABD ACD AB = AC
Ex.16
Hence, proved. In the adjoining fig, find the value of x.
Sol.
We have,
CAD + ADC + DCA = 1800
[Angle sum property]
CAD + ADC + 640 = 1800
CAD + ADC = (1800 - 640) = 1160
But CD = CA CAD = ADC So,
CAD = ADC =
[s opposite to equal sides of a are equal]
1160 = 580 2
Now, ADC = ABD + DAB
[Ext. of a = sum of int. opp. s]
But, AD = BD ABD = DAB. So,
ADC = 2DAB DAB =
x0
1 580 290 2
Hence, x = 29
1 ADC 2
MANISH KUMAR Ex.17
MATHEMATICS
ABC is a isosceles triangle is which AB = AC. Side BA is produced to D such that AD = AB. Show that
BCD is a right angle. Sol.
Given : ABC is an isosceles triangle is which AB = AC. Side BA is produced to D such that AD = AB To prove : BCD = 900
Proof :
STATEMENT 1.
REASON
AB = AC
ABC = ACB 2. 3.
AB = AD AC = AD
Given
CDA = ACD
Given From (1) & (2)
s opposite to equal sides of a are equal
or CDB = ACD 4.
ABC + CDB = ACB + ACD ABC + CDB = BCD
5.
In BCD
BCD + DBC + CDB = 180
0
s opposite to equal sides of a are equal Adding (1) & (3) Angle sum property Using (4)
BCD + ABC + CDB = 1800 BCD + BCD = 1800 2 BCD = 1800 BCD = 900 BCD is a right angle. Ex.18 Sol.
Hence, proved. In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side. Given : A ABC in which B = 900 and ACB = 2 CAB. To prove : AC = 2BC. Construction : Produce CB to D such that BD = BC. Join AD.
MANISH KUMAR
MATHEMATICS
Proof :
1.
STATEMENT In ABD and ABC (i) BD = BC
2. 3. 4. 5. 6.
(ii) ABD = ABC (iii) AB = AB
ABD ABC AD = AC, DAB = CAB = x0
DAC = ACB AD = DC AC = DC AC = 2BC
REASON By construction Each = 900 Common By SAS criteria C.P.C.T. Side opposite to equal s are equal. AD = AC, from (3) DC = BD + BC = 2BC, as BD = BC.
Hence, proved.
INEQUALITIES IN A TRIANGLE THEOREM-5 : If two sides of a triangle are unequal, then the greater side has greater angle opposite to it. Given : A ABC in which AC > AB. To prove : ABC > ACB. Construction : Mark a point D on AC such that AD = AB. Join BD. Proof :
STATEMENT 1. 2. 3. 4. 5. 6. 7.
AB = AD ABD = BDA BDA > DCB ABD > DCB ABC > ABD ABC > DCB ABC > ACB
Hence, proved.
REASON By construction s opposite to equal sides of a are equal (Ext. of BCD) > (Each of its int. Opp. s) Using (2) ABD is a part of ABC. Using (5) DCB = ACB.
MANISH KUMAR
MATHEMATICS
THEOREM-6 (Converse of Theorem-5) : If two angles of a triangles are unequal, then the greater angle has greater side opposite to it. Given : A ABC in which ABC > ACB. To prove : AC >AB. Proof :
STATEMENT 1.
We may have three possibilities only (i) AC = B (ii) AC < AB (iii) AC >AB Out of these exactly one must be true. Case-I. AC = AB. ABC = ACB. This is contrary to what is given. AC = AB is not true Case -II AC < AB AB > AC ACB > ABC This is contrary to what is given AC < AB is not true. Thus, we are left with the only possibility AC > AB which must be true.
REASON
s opposite to equal sides of a are equal
Greater side has greater angle opp. to it.
Hence. proved.
THEOREM-7 : The sum of any two sides of a triangle is greater than its third side. Given : A ABC. To prove :
(i) AB + AC > BC (ii) AB + BC > AC (iii) BC + AC > AB.
Construction : Produce BA to D such that AD = AC. Join CD. Proof :
MANISH KUMAR
MATHEMATICS STATEMENT
1.
AD = AC
REASON By construction
ACD = ADC BCD > ACD 3. BCD > ADC BD > BC BA + AD > BC BA + AC > BC
s opposite to equal sides of a are equal
2.
4.
Using (1) & (2) Greater angle has greater side opp. to it. BAD is a straight line, BD = BA + AD. AD = AC, by construction.
or AB + AC > BC. Similarly, AB + BC > AC and BC + AC > AB
Hence, proved. REMARK :
(i) The largest side of a triangle has the greatest angle opposite to it and converse is also true (ii) The smallest side of triangle has the smallest angle opposite to it and converse in also true
Ex.19
In fig, show that : (i) AB > AC (ii) AB > BC and (iii) BC > AC.
Sol.
Given : A ABC in which B = 400 and ACD = 1000. To prove :
(i) AB > AC (ii) AB > BC (iii) BC > AC.
Proof :
1.
2. 3.
4. 5.
STATEMENT A + B = 1000 A + 400 = 1000 A = 600 C + 100.0 = 1800 C = 800 C > B AB > AC C > A AB > BC A > B BC > AC
REASON Ext. = sum of int. opt. s Linear pair of angles.
C = 800 and B = 400 Greater angle has greater side opp. to it.
C = 800 and A = 600 Greater angle has greater side opp. to it.
A = 600 and B = 400 Greater angle has greater side opp. to it.
Hence, proved. Ex.20
Show that of all the line segments that can be drawn to a given straight line from a given point outside it, the perpendicular is the shortest.
Sol.
Given : A straight line AB and a point P outside it; PM AB and N is any other point of b. To prove : PM < PN.
[NCERT]
MANISH KUMAR
MATHEMATICS
Proof :
STATEMENT 1. 2.
REASON
PMN = 900
PM AB
PNM < 900
In a right , one angel measures 900 and each one of the remaining two is acute.
3. 4.
From (1) and (2) Side opp. the smaller angle is smaller.
PNM < PMN PM < PN
PM is the shortest of the line segments from P to AB.
Hence, proved. Ex.21
Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle.
Sol.
Given : A AC in which AD, BE and CF are its altitudes. To prove : AD + BE + CF < AB + BC + CA
Proof :
STATEMENT 1. 2. 3. 4.
In ABD , ADB = 90
REASON
0
ADB > ABD AB > AD
Given Side opp greater angle is longer.
Similarly , BC > BE and CA > CF AB + CB + CA > AD + BE + CF or AD + BE + CF < AB + BC + CA Hence, Proved.
From (2) and (3)
MANISH KUMAR
MATHEMATICS THINGS TO REMEBER
1. 2. 3.
Two figures are congruent , if they are of the same shape and of the same size. Two circles of the same radii are congruent. Two squares of the same sides are congruent.
4.
If two triangle ABC and PQR are congruent under the correspondence A P, B Q and C R, then symbolically, it is expressed as ABC PQR.
5.
If two sides and the included angle of one triangle re equal to two sides and the included angle of the other triangle, then the two triangle are congruent (SAS congruence rule).
6.
If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent (ASA congruence rule). If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent (AAS Congruence rule)
7. 8. 9.
Angle opposite to equal sides of a triangle are equal. Sides opposite to equal angles of a triangle are equal.
10. 11.
Each angle of an equilateral triangle is of 600. If three sides of one triangle are equal to three sides of the other triangle, then the two triangles are congruent (SSS congruence rule).
12.
If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent (RHS congruence rule).
13. 14.
In a triangle, angle opposite to the longer side is larger (greater). In a triangle, side opposite to the larger (greater) angle is longer.
15.
Sum of any two sides of a triangle is greater than the third side.
MANISH KUMAR
MATHEMATICS CBSE BASED SOME IMPORTANT QUESTIONS
Q.1
In fig, DP = BQ, DPB = BQD and ADP = CBQ, Show that ADP CBQ.
[Hint : DPB = BQD
1800 - DPB = 1800 - BQD APD = CQB] Q.2
and m are two parallel lines intersected by another pair of parallel lines p and q. Show that
ABC CDA.
[Hint:
Q.3
Q.4.
[NCERT]
(i) BAC = DCA
(ii) ACB = CAD (iii) AC = CA] Ram wishes to determine the distance between two objects A and B, but there is an obstacle between these two objects as shown in fig, which prevents his from making a direct measurement. He devises an ingenious way to overcome this difficultly. First, he fixes a pole at a convenient point O so that from O, both A and B are visible. Then he fixes another pole at the point D on the line AO (produced) such that AO = DO. In a similar way, he fixes a third pole at the point C on the line BO (produced) such that BO = CO. Then he measures CD and finds that CD = 170 m. Prove that the distance between the object A and B is also 170 m.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that : (i) (ii) (iii)
AMC BMD DBC is a right angle. DBC ACB
MANISH KUMAR (iv)
CM =
MATHEMATICS 1 AB. 2
[NCERT]
[Hint : (i) By SAS congruence rule, AMC BMD (ii) ACM = BDM
(CPCT)
CA || BD BCA + DBC = 1800 DBC = 900 (iii) By SAS congruence rule, DBC ACB (iv) DBC ACB CD = AB
AMC BMD CM = DM CM = CM = Q.5
1 CD 2
1 AB] 2
In fig AB||QR, BC||PR and AC = PQ. Prove that ABC QRP.
[Hint : BAC = RQP (alternate interior angles) AC = PQ
BCA = RPQ (alternate exterior angles)] Q.6
E and F are respectively the mid-points of equal sides AB and AC of ABC. Show that BF = CE. [NCERT]
[Hint : To show BF = CE, prove ABF ACE.] Q.7
In fig, BL AC, AM LN, AL = CN and BL = CM. Prove that ABC NML.
MANISH KUMAR
MATHEMATICS
[Hint : By SAS congruence rules, ALB ANCM
AB = NM and LAB = CNM In ABC and NML, AB = NM, CAB = LNM, AL = CN AL + LC =CN + LC AC = NL.] Q.8
In an isosceles triangle ABC with AB = AC, D and E are points of BC such that BE = CD. Show that AD = AE.
[NCERT]
[Hint : AB = AC
Q.9
....(i)
B = C
....(ii)
BE = CE BE - DE = CD - DE BD = CE
....(iii)
If two isosceles triangle have a common base, the line joining their vertices bisects them at right angles. [Hint : There can be two possible situations : By SSS congruence rule, ABD ACD 1 = 2
By SSS congruence rule, BAE CAE 3 = 4 & Be = EC.] Q.10
In an isosceles triangle ABC, with AB = AC, the bisectors of B and C interest each other at O. Join A to O. Show that : (i)
OB = OC
(ii) AO bisects A
[Hint : (i) AB = AC B = C
[NCERT]
1 1 B = C OBC = OCB 2 2
Ob = OC (ii) By SAS congruence rule, AOB AOC
OAB = OAC] Q.11
Prove that any two sides of a triangle are together greater than twive the median drawn to the third side. [Hint :
MANISH KUMAR
MATHEMATICS
In ABC, AD is the median. Produced AD to E such that AD = DE. Join EC.
ADB EDC AB = EC In AEC, AC + EC > AE AC + AB > 2AD] Q.12
ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that : (i)
ABD ACD
(ii)
ABP ACP
(iii)
AP bisects A as well as D.
(iv)
AP is the perpendicular bisector of BC.
[NCERT]
[Hint : (i) By SSS congruence rule, ABD ACD (ii) BAP = CAP In ABP & ACP, AB = AC, AP = AP, BAP = CAP (iii) To show AP bisects D, prove DBP CDP] Q.13
A point O is taken inside an equilateral four sides figure ABCD such that its distances from the angular points D and B are equal. Show that AO and OC are is one such the same straight line. [Hint :
AOD AOD 1 = 2 DOC BOC 3 = 4
1 + 2 + 3 + 4 = 3600 Q.14
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR. Show that : (i)
ABM PQN
(ii)
ABC PQR
[NCERT]
MANISH KUMAR
MATHEMATICS
[Hint : (i) By SSS congruence rule, ABM PQN (ii) By SSS congruence rule, ABC PQR] Q.15
In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side.
Sol.
Given : A right angled ABC with ABC = 900, BAC = x and BCA = 2x. To prove : CA = 2BC. Construction : Produce CB to point P such that BP = BC. Joint PA.
STATEMENT 1.
REASON
In ABP and ABC (i) BP = BC (ii) ABP = ABC
Each = 900
(iii) AB = AB 2.
By construction
Common
ABP ABC
By SAS criteria
PA = CA 3. PAB = AC = 4. PAC = x + x = 2x
C.P.C.T. C.P.C.T.
5.
PA = PC
PAC = PCA = 2x
6.
PA = 2BC
BP = BC
CA = 2BC
From (3) and (5)
Hence, proved. Q.16.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that : (i) ABE ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
[NCERT]
[Hint : A = A, AED = AFC, BE = CF] Q.17
In fig. ABC is right angled at C, PQR is right angled at R. If AB = PQ and BC = PR, prove that ACP
QRB.
MANISH KUMAR
[Hint : By RHS congruence rule, ACB ARP
AC = QR Now, BC = RP
BC + CR = CR + RP BR = CP]
MATHEMATICS
MANISH KUMAR Q.18
MATHEMATICS
In fig. AB = AC, AD BC, BE = DE and CF = DF. Prove that : (i) ABE ACF (ii) BAE = CAF [Hint : By RHS congruence rule, ADB ADC
BD = DC and B = C Now, BD = DC
1 1 BD = BC BE = FC] 2 2
Q.19 In fig, AOB = POQ = 900, OB = OQ and AB = PQ. Prove that : (i) OAB OPQ. (ii) 1 = 2 Q.20
Show that in a right angled triangle, the hypotenuse is the longest side.
[NCERT]
[Hint : In a right angled ABC with C = 900, A + B = 900
A < 900, B < 900 AB > BC, AB > AC] Q.21
Show that the difference of any two sides of a triangle is less than the third side.
Sol.
Given : A ABC To prove :
(i)
AC - AB < BC
(ii)
BC - AC < AB
(iii)
BC - AB < AC
Construction : Take a point D on AC such that AD = AB. Join BD. Proof :
STATEMENT 1. 2. 3.
3 < 1
Ext. of a is greater than each of int. opp. s.
2 > 4
Ext. of a is greater than each of int. opp. s.
2 = 1
By construction AB = AD From (1) (2) and (3)
3 > 1 = 2 > 4 3 > 4
4.
REASON
BC > CD
Side opp to greater angle is larger
or CD < BC 5.
AC - AD < BC AC - AB < BC
By construction AB = AD
MANISH KUMAR Similarly BC - AC < AB and BC - AB < AC. Hence, proved.
MATHEMATICS
MANISH KUMAR Q.22
MATHEMATICS
In fig. B < A and C < D. Show that AD < BC. [Hint: B < A OA < OB
...(i)
C < D OD < OC
...(ii)
[NCERT]
Add (i) & (ii)] Q.23
In fig. AP and PR > PQ. Show that AR > AQ.
Sol.
Given : In fig. AP and PR > PQ. To prove : AR > AQ. Construction : Mark a point S on PR such that PS = PQ. Join AS. Proof :
STATEMENT 1.
In APQ and APS
Common
(i) AP = AP
Each = 900 By construction By SAS criteria C.P.C.T.
(ii) APQ = APS (iii) PQ = PS 2. 3. 4. 5.
REASON
APQ APS
2 = 3 3 > 2
Ext. of a is greater than each of int. opp. s From (3) and (4) Side opp. to greater angles is larger.
1 > 2 AR > AQ
Hence, proved Q.24
In fig. PR > PQ and PS bisects QPR. Prove that PSR > PSQ. [Hint : PR > PQ Q > R and PS = RPS
Q + QPS > R + RPS PSR > PSQ] Q.25
In fig. AB > AC, PB and PC are bisectors of B and C respectively. Show that PB > PC.
[NCERT]
MANISH KUMAR
MATHEMATICS
Q.26
In fig. O is an interior point of ABC. BO meets AC at D. Show that OB + OC < AB + AC.
Sol.
Given : O is an interior point of ABC. BO meets AC at D. To prove : OB + OC < AB + AC Proof :
STATEMENT 1. 2. 3.
REASON
In ABD, AB + AD > BD
Sum of two sides of a is greater than the third
AB + AD > BO + OD
side.
In COD, OD + DC > OC AB + AD + OD + DC > BO + OD + OC
AB + AD = DC > BO + OC AB + (AD + DC) > OB + OC AB + AC > OB + OC
Sum of two sides of a is greater than the third side. Adding (1) and (2).
or OB + OC < AB + AC Hence, proved Q.27
In fig, ABCD is a quadrilateral in which diagonals AC and BD intersect at O. Show that 2(AC + BD) > AB + BC + CD + DA.
[Hint : OA + OB < AB OB + OC > BC OC + OD > CD OD + OA > DA]
MANISH KUMAR
MATHEMATICS
EXERCISE SUBEJCTIVE TYPE QUESTINS (A) 1.
2.
VERY SHORT ANSWER TYPE QUESTIONS : Whish of the following pairs of triangles are congruent ? (a)
ABC and DEF in which : BC = EF, AC = DF and C = F.
(b)
ABC and PQR in which : AB = PQ BC = QR and C = R.
(c)
ABC and LMN in which : A = L = 900, AB = LM, C = 400 and M = 500.
(d)
ABC and DEF in which : B = E = 900 and AC = DF.
Answer the following as per the exact requirement : (a)
In s ABC and PQR, AB = PQ, AC = PR and BAC = QPR. Here, ABC PQR. Justify the statement by writing the congruence criteria applicable in this situation.
(b)
In fig.
BAC = QRP.
Justify that ABC RQP. 3.
In ABC, AB = AC. OB and OC are bisectors of B and C respectively. Show that OB = OC.
4.
In fig. 1 > 2. Show that AB > AC.
5.
In ABC, we have, A > B > C, then determine the shortest and the longest side of the triangle.
6.
If ABC PQR, B = 400 and C = 950, find P.
7.
In ABC, AB = BC = 5 cm and A = 550, find B.
8.
State the angle angle-side congruence criteria for triangles.
9.
In fig, AB = AC and ACD = 1150. Find A.
MANISH KUMAR
MATHEMATICS
10.
In ABC, BC = AC and B = 640, find C.
11.
In PQR, P = 500 and R = 700, Name (i) the shortest side (ii) the longest side of the triangle.
(B)
SHORT ANSWER TYPE QUESTIONS :
1.
In the given fig, the line segments AB and CD intersect at a point M in such a way that AM = MD and CM = MB. Prove that, AC = BD but AC many not be parallel to BD.
2.
In the given fig. AY ZY and BY XY such that AY = ZY and BY = XY. Prove that AB = ZX.
3.
If the bisector of the exterior vertical angles of a triangle is parallel to the base, show that the triangle is isosceles.
4.
In each of the following figures, find the value of x :
5.
In each of the following figures, find the value of x:
MANISH KUMAR
MATHEMATICS
MANISH KUMAR
MATHEMATICS
6.
In the given fig, BD || CE; AC = BC, ABD = 200 and ECF = 700. Find GAC.
7.
In the given figure, AB || CD and CA = CE. Find the value of x, y and z.
8.
In the given figure, AB = AD; CD; A = 420 and C = 1080, find ABC.
9.
In the given figure, side BA of ABC has been produced to D such that CD = CA and side CB has been produced to E. If BAC = 1060 and ABE = 1280, find BCD.
10.
In the given figure, AB = BC and AC = CD. Show that BAD : ADB = 3 : 1.
11.
In the given figure, AD is the internal bisector of A and CE || DA. If CE meets BA produces at E, prove that
CAE is isosceles.
MANISH KUMAR
MATHEMATICS
MANISH KUMAR 12.
In the given figure, AD bisects A. Arrange AB, BD and DC in ascending order.
13.
In the given fig. AB = AC. Prove that : AF > AE.
14.
In the given figure, side AB of ABC is produced to D such that BD = BC.
MATHEMATICS
IF A = 600 and B = 500 prove that :
15.
(i)
AD > CD
(ii)
AD > AC
In the given figure, AD bisect A. If B = 600, = 400, then arrange AB, BD and DC in ascending order of their lengths.
(B)
LONG ANSWER TYPE QUESTIONS :
1.
In the given fig. ABCD is a square and PAB is an equilateral triangle.
(i) Prove that APD BPC. (ii) Show that DPC = 150.
2.
In the given fig. in ABC, B = 900. if ABPQ and ACRS are squares, prove that :
MANISH KUMAR
MATHEMATICS
(i) ACQ ABS. (ii) CQ = BS.
3.
Squares ABPQ and ADRS are drawn on the sides AB and AD of a parallelogram ABCD. Prove that : (i) SAQ = ABC (ii) SQ = AC.
4.
In the given fig, ABCD is a square and P, Q, R are points on AB, BC and CD respectively such that AP = B = CR and PQR = 900. Prove that : (i) PB = QC, (ii) PQ = QR (iii) QPR = 450.
5.
6.
In the given fig, ABCD is a square, EF||BD and R is the mid-point of EF. Prove that : (i)
BE = DF
(ii)
AR bisects BAD
(iii)
If AR is produced, it will pass through C.
In a ABC, AB = AC and BC is produced to D. From D, DE is drawn perpendicular to BA produced and DF is drawn perpendicular to AC produced. Prove that BD bisects EDF.
7.
Prove that the perimeter of a triangle is greater than the sum of its three medians.
MANISH KUMAR 8.
In the adjoining figure, prove that : (i)
AB + BC + CD > DA
(ii)
AB + BC + CD + DA > 2AC
(iii)
AB + BC + CD + DA > 2BD
(iv)
AB + BC + CD + DA > AC + BD
MATHEMATICS
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MATHEMATICS
9.
In the adjoining figure, O is the centre of a circle, XY is a diameter and XZ is a chord. Prove that XY > XZ.
10.
In the given figure, AD = AB and AE bisects A. Prove that : (i)
BE = ED
(ii)
ABD > BCA.
(C)
NCERT QUESTIONS :
1.
In quadrilateral ACBD, AC = AD and AB bisects A. Show that ABC ABD. What can you say about BC and BD ?
2.
3.
ABCD is a quadrilateral in which AD = BC and DAB = CBA. Prove that (i)
ABD BAC
(ii)
BD = AC
(iii)
ABD = BAC.
Line is the bisector of an angle A and B is any point of . BP and BQ are perpendicular from B to the arms of
A. Show that :
MANISH KUMAR (i)
APB AQB
(ii)
BP = BQ or B is equidistant from the arms of A.
MATHEMATICS
4.
In ABC, AD is the perpendicular bisector of BC. Show that ABC is an isosceles triangle in which AB = AC.
5.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.
6.
ABC is a right angled triangle in which A = 900 and AB = AC. Find B and C.
7.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that : (i) AD bisects BC (i) AD bisects A.
8.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
9.
ABC is an isosceles triangle with AB = AC. Draw AP BC to show that B = C.
10.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that A > C and B > D.
11.
D is a point on side BC and ABC such that AD = AC. Show that AB > AD.
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MATHEMATICS
WHICH OF THE FOLLOWING STATEMENT AE TRUE (T) AND WHICH ARE FALSE (F) : (a) (b) (c) (d) (e)
(E)
Angles opposite the equal sides of a triangle are equal. Each angle of an equilateral triangle is 600 Side opposite the equal angles of a triangle may be unequal. The bisectors of the two equal angles of a triangle are equal. Two right triangles are congruent, if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle. (f) If any two sides of right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent. (g) The two altitudes corresponding to two equal sides of a triangle need not be equal. (h) If two sides and an angle of a triangle are respectively equal to two sides and an angle of other triangle, the two triangles are congruent. (i) If two sides and the included angle of a triangle are respectively equal to two sides and the included angle of the other triangle, the two triangles are congruent. (j) If two angles and the included side of one triangle are respectively equal to the two angles and the included side of the other triangle, the two triangles are congruent. (k) If ABC PRQ, then AB = PQ. (l) If two angles and a side of a triangles are respectively equal to two angles and a side of the other triangle, then the two triangles are congruent. (m) If DEF RPQ, then D = Q. (n) If PQR CAB, then PQ = CA. (o) If two sides of a triangle are unequal, then larger side has the smaller angle opposite to it. (p) If two angles a triangle are unequal, then the grater angle has the larger side opposite to it. (q) Sum of any two sides of a triangle is greater than the third side. (r) Difference of any two sides of a triangle is equal to the third side. (s) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest. (t) Sum of the three sides of a triangle is less than the sum of its three altitudes. FILL IN THE BLANKS : (a) (b) (c) (d) (e)
Sides opposite the equal angles of a triangles are _____. Angles Opposite the equal sides of a triangles are ___. In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ____ CE. If altitudes CE and BF of a triangle ABC are equal, then AB = ____. In right triangles PQR and DEF, if hypotenuse PQ = hypotenuse EF and side PR = DE, then PQR ____.
(f)
In a triangle ABC, if BC = AB and C = 800, then B = ____.
(g) (h)
In a triangle PQR, if P = R, then PQ = ____. If two sides and the ___ angle of one triangle are respectively equal to two sides and the included angle of the other triangle, then the triangles are congruent. If ____ sides of a triangle are respectively equal to the three sides of the other triangle, then the triangles are congruent.
(i) (j)
If in two triangles ABC and PQR, AB = QR, A Q and B = R, then ABC ____.
(k) (l) (m) (n) (o) (p)
If in two triangles ABC and DEF, AB = DF, BC = DE and B = D, then ABC ____. If in two triangles PQR and DEF, PR = EF, QR = DE and PQ = FD, then PQR _____. Sum of any two sides of a triangle is ____ than the third side. If two angles of a triangle are unequal. then the smaller angle has the ___ side opposite to it. Of all the line segments drawn from a point to a line not containing it, the ___ line segment in the shortest. Difference of any two sides of a triangle is ____ than the third side.
MANISH KUMAR (q) (r) (s) (t)
MATHEMATICS
If any two sides of a triangle are unequal, then the larger side has the ___ angle opposite to it. The sum of the three altitudes of a triangle is ____ than its perimeter. In a right triangle, the hypotenuse is the ___ side. The perimeter of a triangle is ___ than the sum of its medians.
OBJECTIVE TYPE QUESTIONS 1.
For a triangle ABC, the true statement is : (A) AC2 = AB2 + BC2
2.
(C) AC > AB + BC
(D) AC < AB + BC
The internal bisectors of the angles B and C of a triangle ABC meet at O. The, BOC is equal to : (A) 900 + A
3.
(B) AC = AB + BC
(B) 2A
(C) 900 +
1 A 2
(D) 1800 - A
If the sides of triangle are produced, then the sum of the exterior angles i.e. a + b + c is equal to : (A) 1800 (B) 3600 (C) 900 (D) 2700
4.
In a ABC, the sides AB and AC have been produced to D and E. Bisectors of CBD and BCE meet at O. If
A = 640, then BOC is : (A) 520 5.
(B) 580
(C) 260
(D) 1120
D is the mid point of the base BC of a triangle ABC, DM and DN are perpendiculars on AB and AC respectively. If DM = DN, the triangle is : (A) Isosceles
6.
(B) Equilateral
(C) Right angled
(D) Scalene
The distance between the top of two trees 20 m and 28 m high is 17 m. The horizontal distance between the trees is : (A) 11 m
7.
(B) 31 m
(C) 15 m
(D) 9 m
Consider the following statements : (i)
If three sides of a triangle are equal to three sides of another triangle, then the triangles are congruent.
(i)
If three angles of a triangle are equal to three angles of another triangle respectively, then the two triangles are congruent.
Of these statements,
8.
(A) (i) is correct and (ii) is false.
(B) Both (i) and (ii) are false.
(C) Both (i) and (ii) are correct.
(D) (i) is false and (ii) is correct.
In a triangle, the perpendicular from vertex to the base bisects the base. The triangle is : (A) Isosceles
9.
(B) Right angled
In the adjoining figure, AB = AC and AP BC. Then, (A) AB = AP (B) AB < AP (C) AB > AP
(C) Equilateral
(D) Scalene
MANISH KUMAR
MATHEMATICS
(D) AB AP 10.
The measures of three angles of a triangle are in the ratio 1 : 2 : 3 Then the triangle is : (A) Right angled
11.
(B) Equilateral
(C) Isosceles
(D) Obtuse angled
The hypotenuse of a right angled triangle is 25 cms. The other two sides are such that one is 5 cm longer than the other. Their lengths (in cm) are : (A) 10 , 15
12.
13.
(C) 15, 20
(D) 25, 30
In the given figure, AM BC and AN is the bisector of A. Then MAN is : (A) 32
10 2
(B) 16
10 2
(C) 16
0
(D) 32
0
For a triangle ABC, which of the following is true ? (A) BC2 - AB2 = AC2
14.
(B) 20, 25
(B) AB - AC = BC
(C) (AB - AC) > BC
(D) (AB - AC) < BC
ABC is a triangle in which AB = AC. The base BC is produced to D and ACD = 1300. Then, A equals : (A) 800 (B) 600 (C) 500 (D) 400
15.
If ABCD is a square and DCE is an equilateral triangle in the given figure, then DAE is equal to : (A) 450 (B) 300 (C) 150 (D) 22
16.
10 2
In the adjoining figure, AD = BD = AC ; CAE = 750 and ACD = x0. Then, the value of x is : (A) 450 (B) 500 (C) 600 (D) 37
17.
10 2
O is any point on the bisector of the acute angle XYZ. The line OP is parallel to ZY. Then, YPO is : (A) Scalene (B) Isosceles but not right angled (C) Equilateral (D) Right d & isosceles.
MANISH KUMAR 18.
MATHEMATICS
One side other than the hypotenuse of a right angled isosceles triangle is 4 cm. The length of the perpendicular on the hypotenuse from the opposite vertex is : (A) 8 cm
19.
(B) 4 2 cm (B) Greater than 7
(C) Less than 7
In a ABC, A = 90 , AB = 5 cm and AC = 12 cm. If AD BC, then length of AD is :
13 cm 2
(B)
2 15 cm 13
(C)
EXERCISE
2. (a) SAS congruence criteria
5. Shortest side is AB and the longest side is BC.
6. 450
7. 700
10. 520
9. 500
11. (i) QR, (ii) PQ
SHORT ANSWER TYPE QUESTIONS : 5. (i) 22, (ii) 40
0
9. 54
0
6. 1300
7. x = 36, y = 68, z = 44
12. BD < AB < DC
15. BD = DC < AB
(g) F
(j) T
NCERT QUESTIONS : 6. Each of 450
1. They are equal
(F)
13 cm 60
1. (a), (c)
8. 105
(E)
(D)
VERY SHORT ANSWR TYPE QUESTIONS :
4. (i) 110, (ii) 55
(D)
60 cm 13
ANSWER KEY
TRIANGLES
(B)
(D) None of these
0
(A)
(A)
(D) 2 2 cm
ABC is a triangle such that AB = 10 and AC = 3. The side BC is : (A) Equal to 7
20.
(C) 4 cm
TRUE & FALSE : (a) T
(b) T
(c) F
(d) T
(e) T
(f) F
(o) F
(p) T
(q) T
(r) F
(s) R
(t) F
(h) F
(i) T
(k) F
(l) F
(m) F
(n) T
FILL IN THE BLANKS : (a) Equal
(b) Equal
(c) Equal to
(d) AC
(e) EFC
(f) 20
(g) RQ
(h) Included
(i) Three
(j) QRP
(k) FDE
(l) EFE
(m) Greater
(n) Smaller
(o) Perpendicular
(p) Less
(q) Greater
(r) Less
(s) Largest
(t) Greater
ANSWER KEY Que.
1
3
4
5
6
7
8
9
10
11
Ans.
D
C
B
B
A
C
A
A
C
A
Que.
11
12
13
14
15
16
17
18
19
20
Ane.
C
C
D
A
C
B
B
D
B
C
MANISH KUMAR
MATHEMATICS
Important Notes
