Quadrilateral and Polygon

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QUADRILATERAL AND POLYGON 1.

2.

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6.

An obtuse angle made by a side of a parallelogram PQRS with other pair of parallel sides is 150°. If the perpendicular distance between these parallel sides (PQ and SR) is 20 cm, what is the length of the side RQ? [2007-I] (a) 40 cm (b) 50 cm (c) 60 cm (d) 70 cm Assertion (A) If the side of a rhombus is 10 cm. Its diagonals should have values 16 cm and 12 cm. Reason (R) The diagonals of a rhombus cut at right angles. [2007-II] (a) A and R are correct and R is correct explanation of A (b) A and R are correct but R is not correct explanation of A (c) A is correct but R is wrong (d) A is wrong but R is correct An equilateral triangle and a regular hexagon are inscribed in a given circle. If a and b are the lengths of their sides respectively, then which one of the following is correct. [2007-II] (a) a2 = 2b2 (b) b2 = 3a3 (c) b2 = 2a2 (d) a2 = 3b2 In a cricket match, the first 5 bastmen of a team scored runs : 30, 40, 50, 30 and 40. If these data represent a four sided figure with 50 as its one of the diagonals, then what does second diagonal represent? [2007-II] (a) 30 runs (b) 40 runs (c) 50 runs (d) 70 runs The incircle of a quadrilateral of perimeter 2p has radius r. What is the area of the quadrilateral? [2007-II] (a) p(r + 1) (b) 2 pr (c) pr (d) None of these

A

y

A

D

8.

8

B 6

E

C

F ABCD is a rectangle of dimensions 8 units and 6 units. AEFC is a rectangle drawn in such a way that diagonal AC of the first rectangle is one side and side opposite to it is touching the first rectangle at D as shown in the figure given above. What is the ratio of the area of rectangle ABCD to that of AEFC? [2008-II] (a) 2 (b) 3/2 (c) 1 (d) 8/9 ABCD is a square. The diagonals AC and BD meet at O. Let K, L be the points on AB such that AO = AK and BO = BL. If q = ÐLOK, then what is the value of tanq? [2008-II]

(a) 1/ 3 (b) 3 (c) 1 (d) 1/2 9. Two sides of a parallelogram are 10 cm and 15 cm. If the altitude corresponding to the side of length 15 cm is 5 cm, then what is the altitude to the side of length 10 cm? [2009-I] (a) 5 cm (b) 7.5 cm (c) 10 cm (d) 15 cm 10. Which one of the following figures has only one line of symmetry? [2009-I] (a) Rhombus (b) Rectangle (c) Isosceles trapezium (d) Parallelogram 11. A B O

D

x

E

B ABCD is a trapezium in Ðx = 120° and Ðx = 50°, (a) 50° (c) 70°

D

7.

C

F z

C which EF is parallel to BC. then what is Ðy? [2007-II] (b) 60° (d) 80°

N In the figure given above, M is the mid-point of the side CD of the parallelogram ABCD. What is ON : OB ? [2009-I] (a) 3 : 2 (b) 2 : 1 (c) 3 : 1 (d) 5 : 2

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2 12.

ABC is a triangle in which AB = AC. Let BC be produced to D. From a point E on the line AC let EF be a straight line such that EF is parallel to AB. Consider the quadrilateral ECDF thus formed. If ÐABC = 65° and ÐEFD = 80°, then what is ÐD equal to? [2009-II] (a) 43° (b) 41° (c) 37° (d) 35°

13.

D

C O

14.

15.

16.

17.

18.

19.

A X B In the figure given above, ABCD is a square in which AO = AX What is ÐXOB ? [2009-II] (a) 22.5° (b) 25° (c) 30° (d) 45° The quadrilateral formed by joining the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD is [2009-II] (a) a trapezium but not a parallelogram (b) a quadrilateral but not a trapezium (c) a parallelogram only (d) a rhombus ABCD is a square, P, Q, R and S are points on the sides AB, BC, CD and DA respectively such that AP = BQ = CR = DS. What is ÐSPQ equal to? [2010-I] (a) 30° (b) 45° (c) 60° (d) 90° The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q, respectively. If AQ and CP divide the diagonal BD into three parts BX, XY and YD, then which one of the following is correct? [2010-I] (a) BX ¹ XY ¹ YD (b) BX = YD ¹ XY (c) BX = XY = YD (d) XY = 2BX A parallelogram and a rectangle stand on the same base and on the same side of the base with the same height. If I1, I2 be the perimeters of the parallelogram and the rectangle respectively, then which one of the following is correct? [2010-I] (a) I1 < I2 (b) I1 = I2 (c) I1 > I2 but I1 ¹ 2I2 (d) I1 = 2l2 Two similar parallelograms have corresponding sides in the ratio 1 : k. What is the ratio of their areas? [2010-I] (a) 1 : 3k2 (b) 1 : 4k2 2 (c) 1 : k (d) 1 : 2k2 Consider the following statements in respect of a quadrilateral. I. The line segments joining the mid-points of the two pairs of opposite sides bisect each other at the point of intersection. II. The area of the quadrilateral formed by joining the midpoints of the four adjacent sides is half of the total area of the quadrilateral. Which of the statements given above is/are correct? [2010-I] (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II

20.

Let WXYZ be a square. If P, Q and R be the mid-points of WX, XY and ZW, respectively and K, L be the midpoint of PQ and PR, respectively. Then, what is the value area of DPKL ? of [2010-I] area of square WXYZ 1 1 (a) (b) 32 16 1 1 (c) (d) 8 64 Directions (Q. Nos. 21-23) Read the following information carefully to answer the questions that follow. Let ABCD be a quadrilateral. Let the diagonals AC and BD meet at O. Let the perpendicular drawn from A to CD, meet CD at E. Further, AO : OC = BO : OD, AB = 30 cm, CD = 40 cm and the area of the quadrilateral ABCD is 1050 sq cm. [2010-II] 21. What is BE equal to? (a) 30 cm (b) 30 2 cm 22. 23. 24.

(c) 30 3 cm (d) None of these What is the area of the DADC equal to? (a) 300 cm2 (b) 450 cm2 2 (c) 600 cm (d) None of these What is ÐAEB equal to? (a) 30° (b) 45° (c) 60° (d) None of these In the given figure, ABCD is a quadrilateral with AB parallel to DC and AD parallel to BC, ADC is a right angle. If the perimeter of the DABE is 6 units. What is the area of the quadrilateral ? [2010-II] A

D

60°

60°

B

C

E

(a)

25.

(b) 4 sq units 2 3 sq units (c) 3 sq units (d) 4 3 sq units In the figure given below, ABCD is a parallelogram. P is a point in BC such that PB : PC = 1 : 2. DP produced meets AB produced at Q. If the area of the DBPQ is 20 sq units, what is the area of the DDCP? [2010-II] A B Q P

26.

27.

D C (a) 20 sq units (b) 30 sq units (c) 40 sq units (d) None of these The sides of a parallelogram are 12 cm and 8 cm long and one of the diagonals is 10 cm long. If d is the length of other diagonal, then which one of the following is correct? [2012-I] (a) d < 8 cm (b) 8 cm < d < 10 cm (c) 10 cm < d < 12 cm (d) d > 12 cm ABCD is a rhombus with diagonals AC and BD. Then, which one among the following is correct? [2012-I]


3

28.

29.

30.

31.

32.

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34.

35.

(a) AC and BD bisect each other but not necessarily perpendicular to each other (b) AC and BD are perpendicular to each other but not necessarily bisect each other (c) AC and BD bisect each other and perpendicular to each other (d) AC and BD neither bisect each other nor perpendicular to each other. Let ABCD be a parallelogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2 mn, BD = m2 – n2 and AB = (m2 + n2)/2. Statement I AC > BD Statement II ABCD is rhombus Which one of the following is correct in respect of the above statements? [2012-I] (a) Both Statements I and II are true and Statement II is the correct explanation of Statement I (b) Both Statement I and II are true but Statement II is not the correct explanation of Statement I (c) Statement I is true but Statement II is false (d) Statement II is true but Statement I is false ABCD is a rectangle. Let E be a point on AB and F be a point on CD, such that DE is parallel to BF. If AE = 3 cm and if the area of DBFC = 6 sq cm. Consider the following statements I. Area of rectangle ABCD can be of the form pq2 sq cm, where p and q are distinct primes. II. Area of the figure EBFD is of the form r2 sq cm, where r is rational but not an integer. Which of the above statements is/are correct? [2012-I] (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II Let LMNP be a parallelogram and NR be perpendicular to LP. If the area of the parallelogram is six times the area of D RNP and RP = 6 cm what is LR equal to? [2012-I] (a) 15 cm (b) 12 cm (c) 9 cm (d) 8 cm If the diagonals of a quadrilateral are equal and bisect each other at right angles, then the quadrilateral is a [2012-II] (a) rectangle (b) square (c) rhombus (d) trapezium If two parallel lines are cut by two distinct transversals, then the quadrilateral formed by the four lines is always a [2012-II] (a) square (b) parallelogram (c) rhombus (d) trapezium ABCD is a parallelogram. If the bisectors of the ÐA and ÐC meet the diagonal BD at point P and Q respectively, then which one of the following is correct? [2012-II] (a) PCQA is a straight line (b) DAPQ is similar to DPCQ (c) AP = CP (d) AP = AQ Let X be any point within a square ABCD. On AX a square AXYZ is described such that D is within it. Which one of the following is correct ? [2012-II] (a) AX = DZ (b) ÐADZ = ÐBAX (c) AD = DZ (d) BX = DZ The locus of a point in rhombus ABCD which is equidistant from A and C is [2012-II]

(a) a fixed point on diagonal BD (b) diagonal BD (c) diagonal AC (d) None of the above 36. ABCD is a trapezium with parallel sides AB = 2 cm and DC = 3 cm. E and F are the mid-points of the non-parallel sides. The ratio of area of ABFE to area of EFCD is [2013-I] (a) 9 : 10 (b) 8 : 9 (c) 9 : 11 (d) 11 : 9 37. In the figure given below, PQRS is a parallelogram. If AP, AQ, CR and CS are bisectors of ÐP, ÐQ, ÐR and ÐS respectively, then ABCD is a [2013-I] P Q C D

B A

S

R (a) square (b) rhombus (c) rectangle (d) None of these 38. In the figure given above, ABCD is a trapezium. EF is parallel to AD and BG. Ðy is equal to [2013-I] A D x y

120°

E

B

z 90°

F

50°

C (a) 30° (b) 45° (c) 60° (d) 65° 39. A quadrilateral ABCD is inscribed in a circle. If AB is parallel to CD and AC = BD, then the quadrilateral must be a [2013-II] (a) parallelogram (b) rhombus (c) trapezium (d) None of these 40. ABCD is a quadrilateral such that BC = BA and CD > AD. Which one of the following is correct? [2013-II] (a) ÐBAD = ÐBCD (b) ÐBAD < ÐBCD (c) ÐBAD > ÐBCD (d) None of these 41. Two light rods AB = a + b, CD = a – b symmetrically lying on a horizontal AB. There are kept intact by two strings AC and BD. The perpendicular distance between rods is a. The length of AC is given by [2014-I] (a) a (b) b (c)

(d) a2 - b2 a 2 + b2 42. If PQRS be a rectangle such PQ = 3 QR. Then, what is ÐPRS equal to? [2014-I] (a) 60° (b) 45° (c) 30° (d) 15° 43. In a trapezium, the two non-parallel sides are equal in length, each being of 5 cm. The parallel sides are at a distance of 3 cm apart. If the smaller side of the parallel sides is of length 2 cm, then the sum of the diagonals of the trapezium is [2014-I] (a)

10 5 cm

(b)

(c)

5 5 cm

(d) 3 5 cm

6 5 cm


4 44.

45.

46.

47.

48.

49.

The area of a rectangle lies between 40 cm2 and 45 cm2. If one of the sides is 5 cm, then its diagonal lies between [2014-I] (a) 8 cm and 10 cm (b) 9 cm and 11 cm (c) 10 cm and 12 cm (d) 11 cm and 13 cm Let ABCD be a parallelogram. Let P, Q, R and S be the mid-points of sides AB, BC, CD and, DA, respectively. Consider the following statements. I. Area of triangle APS < Area of triangle DSR, if BD < AC. II. Area of triangle ABC = 4 [Area of triangle BPQ]. Select the correct answer using the codes given below. [2014-I] (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II Consider the following statements I. Let ABCD be a parallelogram which is not a rectangle. then, 2(AB2 + BC2) ¹ AC2 + BD2 II. If ABCD is a rhombus with AB = 4cm, then AC2 + BD2 = n3 for some positive integer n. Which of the above statements is/are correct? [2014-I] (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II ABCD is a parallelogram, E is a point on BC such that BE : EC = m : n. If AE and DB intersect in F, then what is the ratio of the area of DFEB to the area of DAFD? [2014-I] (a) m/n (b) (m/n)2 (c) (n/ m2) (d) [m/(m + n)]2 Let ABCD be a parallelogram. Let X and Y be the mid– points of the sides BC and AD, respectively. Let M and N be the mid–points of the sides AB and CD, respectively. Consider the following statements: [2014-II] 1. The straight line MX cannot be parallel to YN. 2. The straight lines AC, BD, XY and MN meet at a point. Which of the above statements is/are correct ? (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) Neither 1 nor 2 What is the maximum distance between two points of a cube of side 2 cm ? [2014-II] (a)

3 cm

(b)

2 3 cm

(c) 4 3 cm (d) 2 2 cm Directions (Q. Nos. 50–52) Read the following information carefully and answer the given questions that follow. A piece of land is in the form of a parallelogram and the perimeter of the land is 86m. The length of one side exeeds the other by 13 m and one of the diagonals is 41m. [2014-II] 50. What is the area of the parallelogram ? (a) 63 m2 (b) 96 m2 2 (c) 126 m (d) 252 m2 51. What is the shorter height of the parallelogram ? (a) 9.0 m (b) 7.5 m (c) 5.5 m (d) 4.5 m 52. Consider the following statements : 1. The difference between the diagonals of the parallelogram is more than 20 m. 2. The difference between the heights of the parallelogram is more than 10 m. Which of the above statements is / are correct ? (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) Neither 1 nor 2

53.

What is the number of pairs of perpendicular planes in a cuboid ? [2014-II] (a) 4 (b) 8 (c) 12 (d) None of these 54. How many equilateral triangles can be formed by joining any three vertices of a cube ? [2014-II] (a) 0 (b) 4 (c) 8 (d) None of these Directions (Q. Nos. 55-56) Read the following information carefully and answer the given questions that follow. ABCD is a trapezium, in which AB is parallel to CD. Let M be the mid-point of BC. [2014-II] 55. Consider the following statements : 1. ‘Area of DADM + Area of DDCM’ is equal to threefourth of the area of trapezium ABCD, if AB = CD. 2. ‘Area of DDCM + Area of DABM’ is always greater than half of the area of trapezium ABCD. Which of the above statements is/are correct? (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) Neither 1 nor 2 56. Consider the following statements : 1. ‘Area of DADM – Area of DABM’ is always equal to area of DDCM, if AB = CD. 2. Half of area of DABM is equal to one-eight of area of trapezium ABCD, if AB = CD. Which of the above statements is /are correct? (a) Only 1 (b) Only 2 (c) Both 1 and 2 (d) Neither 1 nor 2 57. ABCD is a parallelogram. P and R are the mid-points of DC and BC, respectively. The line PR intersects the diagonal AC at Q. The distance CQ will be [2014-II] (a) AC/4 (b) BD/3 (c) BD/4 (d) AC/3 58. Bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect each other at a point P. Which one of the following is correct ? [2014-II] (a) 2ÐAPB = ÐC + ÐD (b) ÐAPB = ÐC + ÐD (c) ÐAPB =180° – (ÐC + ÐB) (d) ÐAPB =180° – (ÐC + ÐD) 59. If each interior angle of a regular polygon is 135°, then the number of diagonals of the polygon is equal to [2015-I] (a) 54 (b) 48 (c) 20 (d) 18 60. AD is the diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is [2015-I] (a) 17 cm (b) 15 cm (c) 13 cm (d) 8 cm 61. If a star figure is formed by elongating the sides of a regular pentagon, then the measure of each angle at the angular points of the star figure is [2015-I] (a) 36° (b) 35° (c) 32° (d) 30° 62. The area of a rhombus with side 13cm and one diagonal 10 cm will be [2015-I] (a) 140 square cm (b) 130 square cm (c) 120 square cm (d) 110 square cm 63. The diagonals of a trapezium are at right angles, and the slant sides, if produced, form an equilateral triangle with the greater of the two parallel sides. If the area of the trapezium is 16 square cm, then the distance between the parallel sides is [2015-I] (a) 2 cm (b) 4 cm (c) 8 cm (d) Cannot be determined due to insufficient data


5

HINTS & SOLUTIONS 1. (a)

Þ

Given that, ÐSPQ = 150° and PM = 20 cm In parallelogram PQRS, ÐRSP + SPQ = 180° (interior angles) ÐRSP = 180° – 150° = 30° ÐRSP = q = 30° M R S q

20 cm 150° P

In DODE, cos 60°= Þ \ 4. (c) Þ 5. (c)

Q

In DPSM, sin q = sin 30° =

1 20 = Þ SP = 40 cm 2 SP RQ = SP = 40 cm.

Þ \ 2. (a) \ Þ Þ

6. (b) \

PM SP

\ \ 7. (c)

H2 = P2+ B2 AB2 + BC2 + CD2 + AD2 = AC2 + BD2 (10)2 + (10)2 + (10)2 + (10)2 = (16)2 + (12)2 400 = 400

D 8

8 cm 90°

E

cm

3 a. 2

Þ

C Þ Þ

O

D Length of OC=

3 2 a a´ = = radius 2 3 3

Now, AC = 82 + 62 = 10 2 In DAED, AE + ED2 = AD2 AE2 = AD2 – x2 = 36 – x2 ...(i) and in DCFD, CF2 + DF2 = CD2 CF2 = (8)2 – (10 – x)2 ... (ii) From eqs. (i) and (ii), we get 36 – x2 = 64 – (10 – x)2 (Q AE = FC) 36 – x2 = 64 – (100 + x2 – 20x) (because AECE is rectangle) 18 5

20x = 72 Þ x = æ 18 ö From eq. (i) AE = 36 - ç ÷ è 5ø

B

b 2

x

2

2

F

DF= b Þ DE =

C

F

We know altitude of equilateral DABC is

60°

10 –

m

B 10 cm Hence, both A and R are true and R is the correct explanation of A.

A

6

8 D

6c

B 10

6 x

A

Also,

8

A

M

\

1 3b = Þ a = 3b 2 2a a2 = 3b2 2 Here we see (50) = (30)2 + (40)2 2500 = 900 + 1600 Þ 2500 = 2500 It means given scores are the sides of a rectangle. So, other diagonal should be 50 runs. We know that, if r be the radius of incircle and 2p be the perimeter of a quadrilateral, then Area of quadrilateral = pr ABCD is a trapezium. AD||BC and EF||BC (given) Hence, EF||AD Ðx + Ðy = 180°(interior angles) Ðy= 180° – 120° = 60° Let ED = x and area of rectangle ABCD = AB × BC = 8 × 6 = 48 units

C 6 cm

3. (d)

DE b/2 = OD a / 3

AE2 = 36 Þ

AE2 =

576 25

Þ

AE =

24 5

324 900 - 324 = 25 25


6 \ 8. (c)

10. (c) Isosceles trapezium has only line of symmetry.

Area of rectangle ABCD = Area of rectangle AEFC

8´6 =1 24 10 ´ 5 Let sides of the square be a.

D

C 11. (b) In DDMN and DBMC, DM = MC

O a/ 2

A

q q 2 L M K a/ 2

A

B

D

AC = a 2 and AO = OC =

Here,

a AM = 2

a

In DOML,

tan

q = 2

–

2 a

a 2

-

2 a 2

a

Þ 9. (b)

Ð1 = Ð2 (vertically opposite angle) Ð3 = Ð4 + Ð9(alternate interior angle) Since, BC||AD and intersects by CD. DDMN @ DBMC DN = BC = AD (ASA)

a a and OM = 2 2

=

2 -1 2 = 1 2

2( 2 - 1)

=

AN = 2 BC Þ

So, 2 -1

q 2( 2 - 1) 2 = tan q = q 1 - (2 + 1 - 2 2) 1 - tan 2 2

=

3

N

2

2 tan

\

7 O 6 9 5 2 4 C 1 M

8

Then,

LM =

B 10

a/ 2

\

2( 2 - 1)

1- 3+ 2 2 2 2-2 tan q = 1 Area of parallelogram = Base × Height = 15 × 5 = 75 sq cm

\

ON 2 = OB 1 12. (d) Here, ÐB = ÐC = 65° Here, GF||AB, which is intersects.

Þ

A

[from eq. (i)]

F 80°

10 c m

5 cm

E

N

65°

\

... (i)

AN ON = BC OB

D

Þ

AN 2 = BC 1

In DOAN an DOBC, Ð5 = Ð6 (vertically opposite angle) Ð7 = Ð8 (alternate interior angle) Ð9 = Ð10 (rest angle) DOAN ~ DOBC So, the sides will be in same ratio

C

A

(mid-point) (given)

M

15 cm

B

B

Now, BG

Area of parallelogram = Base × Height = 10 × DN 10 × DN = 75 75 = 7.5 cm DN = 10

Þ

1 65° G

C

D

Ð1 = ÐB = 65° (corresponding angles)

In DFGD, Ð1 + ÐF + ÐD = 180° 65° + 80° + ÐD = 180° Þ ÐD = 35°


7 13. (a)

Let ÐXOB = q

D

C

O q A

B

X

In DOXB, ÐXOB + ÐOBX + ÐOXB = 180° Þ q + 45° + ÐOXB = 180° Þ ÐOXB = 180° – 45° – q Þ ÐOXB = 135° – q Here, ÐOXA + ÐOXB = 180° Þ ÐOXA + 135° – q = 180° Þ ÐOXA = 45° + q In DOAX, AO = AX \ ÐOXA = ÐAOX = 45° + q Since, ÐAOX+ ÐXOB = 90° Þ 45 + q + q = 90° Þ 2q = 45° \ q = 22.5° 14. (c) The quadrilateral formed by joining the mid-points of the sides is a parallelogram. 15. (d) In DAPS and DPBQ,

D

C

R

\ DY = YX ...(i) Similarly YX = XB ...(ii) From equations (i) and (ii) DY = YX = XB 17. (c) If a parallelogram and a rectangle stand on the same base and on the same side of the base with the same height, then perimeter of parallelogram is greater than perimeter of rectangle. \ I1 > I2 18. (c) Let the sides of a parallelograms are x, y and xk, yk. C

T

E

y A

S

yk B

D

Q R xk Since, sides of two parallelogram are in 1 : k.

\ \

x

P

AC BC BC y 1 = Þ = = PT QT QT yk k Let BC = z and QT = zk Ratio of areas of two similar parallelograms

DABC ~ DPQT Þ

=

D

19. (c)

S

R

1 x´ z = 2 xk ´ zk k

C

Q

S Q

A A

B P AS = PB (given) AP = BQ (given) and ÐA = ÐB = 90° (since, ABCD is square) Therefore, DAPS @ DBQP are congruent. \ SP = PQ ÐSPA = ÐBQP and ÐASP = ÐBPQ \ ÐSPQ = 90° (by RHS rule) 16. (c) Since ABCD is a parallelogram and P, Q are the midpoints of AB, CD respectively. \ AP = QC and AP || QC Þ APCQ is a parallelogram. Now in DDXC, Q is the mid point of CD and QY || CX, therefore Y will be the mid point of DX. Q C D

Y

A

P

B

AP = PB CQ = QB RC = DR SD = AS According to above ABCD is a parallelogram. Then, the diagonals PR and SQ bisect each other. Now,

ar (RSQ) =

1 ar (SQCD) 2

1 ar(ABQS) 2 From addition of both (i) and (ii), we get

and

ar (PQS) =

... (i) ... (ii)

1 ar(ABCD) 2 Thus, both statements are correct.

ar (PQRS) =

20. (b)

X

P

ar(PRQ) = =

1 ar(WXQR) 2

1æ1 ö 1 ç ar(WXYZ ) ÷ = ar(WXYZ ) 2è2 ø 4

... (i)

B


8 Z

24. (a) AB||DC and AD||BC

Y

R

A

W

B

60°

X

P

ar( PRQ) (2 LP ) 2 = ar( PLK ) LP 2 ar(PRQ) = 4 ar(PLK)

40 cm E

Þ Þ

Þ Þ

C

ÐEAB = ÐABE = 60° ÐAEB = 60° D ABE is an equilateral triangle. Now, AB = BE = EA perimeter of DABE = 6 AB + BE + EA = 6 AB = 2 units AB = BE = EA = 2 and in DADE, AE2 = AD2 + ED2 4 = AD2 + 1 (since, E is mid-point of CD) AD = 3 units Therefore, area of quadrilateral ABCD = AB × AD

= 2 × 3 = 2 3 sq units. 25. (d) We know that, ratio of the areas of two similar triangles are equal to the ratios of squares of their corresponding sides. \

C

E

In DABE,

[from eq. (i)]

ar( PLK ) 1 1 Þ ar(WXYZ) = ar(PLK) Þ = ar(WXYZ ) 16 16 Explanations (Q. Nos. 21-23) : Given, AO : OC = BO : OD and AB = 30 cm and CD = 40 cm

D

D

Þ Þ

1 ar(WXYZ) = 4 ar(PLK) 4

Þ

60°

K

ar( PRQ) RP 2 = ar( PLK ) LP 2 (by properties of similar triangle)

Þ

60°

Q L

Þ

... (Given)

ar(DBPQ ) PB 2 = ar(DDPC ) PC 2

20 1 = ar(DDPC ) 4 Þ ar (D DPC) = 80 sq units 26. (d) In parallelogram, d2 + d22 = 2 (l2 + b2)

Þ

O A

\ \

30 cm

D

B

OA AB 3 = = OC CD 4 ÐOAB = ÐOCD and ÐOBA = ÐODC It means DC || AB. So, it is a trapezium.

DAOB ~ ACOD Þ

1 Area of quadrilateral ABCD = ( AB + CD) ´ AE 2

Þ Þ Also, 21. (b) In right DEAB, EB =

1 1050 = (30 + 40) ´ AE 2 AE = 30 cm ÐBAE = 90°

AE 2 + AB 2 Þ EB =

C d1

\ Þ Þ Þ 27. (c) \

b=12cm cm 0 1 d 1= A B I=8cm d2 + (10)2 = 2 (64 + 144) d2 = 2 × 208 – 100 d2 = 416 – 100 = 316 Þ d = d = 17.76 cm d > 12 ABCD is a rhombus. AB = BC = CD = DA A

316

D

302 + 302 = 30 2 cm

1 1 ´ CD ´ AE = ´ 40 ´ 30 = 600 cm2 2 2 23. (b) Also ÐBAE = 90°, AE = AB = 30 cm \ ÐAEB = ÐABE = 45°

22. (c) Area of DADC =

B C and diagonals bisect each other at right angles


9 Area of parallelogram = 6 × Area of D NPR

28. (b) In parallelogram ABCD. R A

D

S

Q

B

C

P

AC = 2 mn, BD = m2 – n2 and AB =

Þ

m2 + n2 2

We know that, (AC2 + BD2) = 2 (AB2 + BC2) 2 2 (m + n + m4 + n4 – 2m2n2)

ì1 2 2 2 2ü = 2 í (m + n ) + BC ý î4 þ Þ

(m2 + n2)2 =

Þ

2BC2 = BC2 =

Þ

NR × PL = 6 ×

A

D

1 2 (m + n 2 )2 + 2 BC 2 2

B

1 2 (m + n2 )2 2 2

C

Case II If two distinct transversals are parallel, then always (Trapezium + Parallelogram)

2 2

(m + n ) 4 2

A

D

2

m +n 2 Therefore, ABCD is a rhombus. Let AC > BD Þ 2mn > m2 – n2 Þ (m + n)2 > 2m2 which is always true for every positive integers m and n, where n < m < 2n. 29. (a) Statement I Let side BC = x cm. Þ

1 ´ NR ´ PR 2 Þ PL = 3PR (here, PL = PR + RL) Þ PR + RL = 3PR Þ RL = 2PL = 2 × 6 = 12 cm 31. (b) We know that in a square diagonals are equal and bisect each other at 90°. Hence, the required quadrilateral is a square. 32. (d) If two parallel lines are cut by two distinct transversals, the quadrillateral formed by the four lines is always a ‘Trapezium’. Case I If two distinct transversals (are not parallel), then always ® (Trapezium)

\

BC =

4

A

D

C

33. (b) Since, line segment AP and CQ bisects the ÐA and ÐC, respectively. Then, AP||CQ Now in DAPQ and DCQP Q AP || QC D C P

5

3

F

5

E

Q 3

B

4 Given Area of DBFC = 6 cm2

N

A

B ÐAPQ = ÐPQC PQ = ÐPQ Also, PC || AQ \ ÐCPQ = ÐPQA \ DAPQ ~ DCQP Thus, DAPQ is similar to DPCQ. 34. (d) In D ABX and DADZ. AB = AD (side of and AX = AZ (side Let ÐBAX = q \ ÐXAD = 90° – q Q

C

1 ´ 3 ´ x = 6 Þ x = 4 cm 2 In DBFC, BF2 = x2 + 9 = 16 + 9 Þ BF2 = 25 \ BF = 5 \ area of rectangle ABCD, pq2 = p (2)2 cm2 Statement II which is of the form pq2. While the area of EBFD cannot be the form of r2 cm2.

30. (b)

B

A

(alternate angle) (common) (alternate angle) (by ASA) a square ABCD) of square AXYZ)

B

M X Z D

P

6 cm R

L

C

Y Also, AXYZ is a square,


10 \ Þ

ÐZAX = 90° Þ ÐZAD + ÐXAD = 90° ÐZAD = 90° – (90° – q) = q i.e., ÐBAX = ÐZAD \ DABX @ DADZ \ BX = DZ (by CPCT) 35. (b) Diagonals of a rhombus are perpendicular bisector of each other. Hence each point on the diagonal BD is equidistant from vertices A and C. Therefore required locus is the diagonal BD. 36. (c) Join AC. In DACD, EG||DC and E and G are mid-points of AD and AC, respectively. A B

\ \

q + z° = 180° (linear pair) q = 180° – 50° = 130° In quadrilateral AQFD, x° + y° + 120° + q = 360° 50° + y° + 120° + 130° = 360° y = 360° – 300° = 60° 39. (c) The quadrilateral must be a trapezium because a quadrilateral where only one pair of opposite sides are parallel (in this case AB || CD) is a trapezium. 40. (c) Constraction : In quadrilateral ABCD, form A to C. Now, in DABC A B

h E

F

G

h D

D

C

\

EG = Similarly, in DABC GF =

1 3 DC = 2 2

In DADC,

Area of trapezium =

1 AB = 1 2 3 5 = 2 2

1 (Sum of parallel side × Height) 2

Area of ABFE Now, the ratio = Area of EFCD

ÐQ ÐR 180° + = 2 2 2 In D PDS, ÐD = 90° (Because ÐP + ÐS = 180° and ÐP ÐS 180 + = = 90° 2 2 2 \ ABCD is a rectangle. 38. (c) From figure.

E

\

CD > AD ÐDAC > ÐDCA (since in a triangle, angle opposite to greater side is bigger than the angle opposite to smaller side) On adding eqs. (i) and (ii), we get ÐBAD > ÐBCD 41. (d) Since, they are symmetrically lying on horizontal plane. A x C E

a–b

1æ 5ö ç2 + ÷´h 9 2è 2ø = = 1æ 5ö 11 ç3+ ÷´ h 2è 2ø 37. (c) In parallelogram PQRS, AP, AQ, CR and CS are bisector of ÐP, ÐQ, ÐR and ÐS. In D RBQ ÐB = 90° (Because, ÐQ + ÐR = 180° and

A

C AB = BC ...(Given) ÐBAC = ÐBCA (angles opposite to equal side)

Q \

EF = EG + GF = 1 + \

\

Q

F \ \ \

x B

Now, i.e., Now in DACE,

a–b D AC AE AB a+b x

= BD = BF = x = (a – b) + 2x = a – b + 2x Þ 2b = 2x =b

x2+a2 = AC2

Þ AC2 = b2 + a2 Þ AC = 42. (c) In rectangle PQRS, S R

b2 + a 2

D

x°

y° q Q

120° z°

P

F

90° 50° B C BC || EF || AD x° = z° = 50° (corresponding interior angle

\

PQ || RS

3 QR

Q

ÐRPQ = ÐPRS (Alternate interior angles) ... (i) Now in DPQR, tan ÐQPR =

RQ PQ


11 tan ÐQPR = Þ \ 43. (b) In DBCF,

QR

3 3QR ÐQPR = 30° ÐPRS = 30°

D 5

D

1

=

O [from eq. (i)]

A 4 E

5

3 F

2

B

4

Pythagoras theorem, (5)2 = (3)2 + (BF)2 Þ BF = 4cm AB = 2 + 4 + 4 = 10 cm Now in DACF, AC2 = CF2 + FA2 Þ AC2 = 32 + 62

Similarly, \ 44.

A B So it is not true. II. ABCD is a rhombus and diagonals AC and BD bisect each other. \ AO = OC and OB = OD In DAOB, AB2 = AO2 + OB2

C

2 3

C

AC =

45 cm

BD =

45 cm

Sum of diagonal = 2 ×

2

45 = 2 × 3 5

= 6 5 cm (b) Here, Area of Rectangle lies between 40cm2 and 45cm2 Given that one sides = 5cm. Area of Rectangle = 5 × second sides Now, If Area = 40cm2 then, 40 = 5 × second sides Q second sides = 8cm. Again, If Area = 45cm2 45 = 5 × second sides. Q Second sides = 9cm. it means that second sides varies. between 8 cm to 9 cm. Let diagonal = d

2

æ AC ö æ BD ö (4) = ç ÷ +ç ÷ è 2 ø è 2 ø 2 2 \ AC + BD = 64 = (4)3 i.e., n3 So only II is true. 47. (d) In DAFD and DBFE, ÐAFD = ÐBFE (vertically opposite angles) D C 2

nx E mx

F A

B ÐADF = ÐFBE DAFD ~ DEFB

and \ So,

ar(DEFB ) EB 2 = ar(DAFD) AD 2

(mx )2

48.

(alternate angles) (By AA)

m2

2

é m ù = = = ê ú (mx + nx )2 ( m + n) 2 ë ( m + n) û (b) From Statement 1. Given, ABCD is a parallelogram. X and Y are mid–points of BC and AD, respectively. M and N are the mid–points of AB and CD, respectively.

Þ 82 + 52 < d < 52 + 92

D

N

C

Þ 64 + 25 < d < 25 + 81 Y

Þ 87 < d < 106

X

Þ 81 < 89 < d < 106 < 121 B M From statement 2. Here join point A and C. In DABC, M and X are mid–points of AB and BC. A

Þ 81 < d < 121 Þ 9cm < d < 11cm. 45. (b) Area of DAPS = Area of DDSR

D S A

Q \ 46. (b)

R

C

MX || AC and MX =

\

YN || AC and YN =

Q

B P AS = SD and AP = DR ar (DABC) = 4ar (DBPQ) I. ABCD is a parallelogram, then AC2 + BD2 = 2(AB2 + BC2)

1 AC ...(i) 2 In DADC, Y and N are mid–points of AD and CD.

\

1 AC ...(ii) 2 From equations (i) and (ii), we get MX || YN From statement 2 So, Statement 1 is not correct. Clearly, straight lines AC, BD, XY and MN meet at a point, So Statements 2 is correct.


12 49.

50.

(b) Side of cube = 2 cm \ Maximum distance between two points of a cube = Length of diagonal = 3 × side = 2 3 cm (d) Perimeter of parallelogram land = 86 m and diagonal = 41m Suppose one side of parallelogram be x m other side = (x + 13) m \ Perimeter = 2(x + x +13) = 86 Þ 2x + 13 = \

54.

(c) In a cuboid, 4 perpendicular face pairs in bottom surface, 4 perpendicular face pairs in top surface and 4 perpendicular face pairs in vertical surface. Total perpendicular pairs are 12. (c) In a cube, there are six faces. Let the sides of a cube be a. \

Z C (0, 0, a)

30 = 15 2 one side of parallelogram = 15 m other side = 15 + 13 = 28 m

x=

15 m N D

28 m

M

=

F (0,0, a) B

41 m

area of DABD =

Y In DABC,

15 m

28 m

C

s( s - a)(s - b)( s - c)

42(42 - 15)(42 - 28)(42 - 41)

55.

252 = 16.8 m 15 Difference between the heights = 16.8 – 9 = 7.8 which is not more than 10. Statement 2 is not correct.

\

O (0, 0, 0) D (a, a, 0) B (0, a, 0)

Area of DAMD = A

1 area of ABCD ...(i) 2 B

M C D Because, both are between same parallels and in same base. By according to question

252 =9m 28 Therefore, shorter height of the parallelogram is 9 m.

Þ 15 × CN = 252 Þ CN =

(a, a, a) A (a, 0, 0) X

and AC = a 2 + a 2 = a 2 DABC, is an equilateral triangle. Similarly, In DABE, DODG, DODE, DCEB, DCEA, DFGO and DFGD Eight equilateral triangles are possible. (c) 1. Given, ABCD is a trapezium. If AB = CD, then it becomes a parallelogram. M is the mid-point of BC.

15 + 41 + 28 84 é ù = = 42ú êQ s = 2 2 ë û 2 = 42 ´ 27 ´14 ´ 1 = 126 m Required area of parallelogram = 2 × Area of DABD = 2 × 126 = 252 m2 (a) From the above figure, shorter height of parallelogrm = AM From question 30 Area of parallelogram = Base Height = 252m2

(a) 1. Let second diagonal be x. Now, x2 + 412 = 2(152 + 282) Þ x2 = 337 = 18.36 (approx) \ Difference between the diagonals = 41 – 18.36 = 22.64 Which is more than 20 Therefore, Statement 1 is correct. 2. Q Second height of parallelogram ABCD = CN \ Base × Height = Area

E

AB = a 2 + a 2 = a 2, BC = a 2 + a 2 = a 2

28 × AM = 252 Þ AM =

52.

Diagonal of face = a 2 + a 2 = a 2 Hence, there is no equilateral triangle will be formed in faces.

86 = 43 Þ 2x = 43 – 13 = 30 2

A

51.

53.

Q

Area of DABM + Area of DDCM =

1 area of ABCD 2

Area of DABM = Area of DDCM 1 Þ 2 Area of DDCM = Area of ABCD 2 1 Þ Area of DDCM = Area of ABCD ...(ii) 4 On adding equations (i) and (ii), we get Area of DAMD + Area of DDCM Q

=

1 1 Area of ABCD + Area of ABCD 2 4

3 Area of ABCD 4 Therefore, Statement 1 is correct.

=


13 2.

ABCD is a trapezium.

\

B

A

\ M C D \ Area of trapezium = Area of DDCM + Area of DABM + Area of DAMD Þ Area of DDCM + Area of DABM = Area of trapezium ABCD – Area of DAMD If AD = BC, then using eq. (i), we get Area of DDCM + Area of DABM

1 1æ1 ö 1 OC = ç AC ÷ = AC 2 2è2 ø 4 (a) A quadrilateral ABCD, AP and BP are bisectors of ÐA and ÐB, respectively. \

58.

1 AC 2 In DCBD, P and R are mid-points of DC and BC. PR || BD or PQ || DO and RQ || BO Again in DOCD, PQ || OD So, Q is mid-point of OC. OC =

CQ =

D

C P

1 Area of trapezium ABCD 2 If AD = BD, then it is true otherwise are of DDCM and area of DABM is greater than half of the area of trapezium ABCD. Therefore Statement 2 is also correct. (a) 1. If AB = CD, then ABCD is a parallelogram =

56.

Area of DADM = A

A

1 æ1 ö ÐAPB = 180° – ç Ð A + ÐB ÷ 2 è2 ø We know that sum of all angles of a quadrilateral = 360° Þ ÐA + ÐB + ÐC + ÐD = 360° \

1 Area of ABCD 2 B

M

Þ

2. Þ Þ 57.

(a)

C D [since, both are in same base and between same parallels] Area of DADM – Area of DABM 1 = Area of ABCD 2 1 – Area of ABM = Area of DDMC 4 Statement 1 is correct. 1 Area of DABM + area of DDCM = Area of ABCD 2 1 2 Area of DABM = Area of ABCD 2 [Q area D ABM = area of D DCM] 1 Area of D ABM = Area of ABCD 4 Statement 2 is not correct. Given, ABCD is a parallelogram. Join AC and BD which intersect each other at O. B A

O

D

R Q P

C

B

\

1 1 1 1 360° ÐA + ÐB + ÐC + ÐD = 2 2 2 2 2

Þ

1 1 1 ö æ1 ÐC + ÐD = 180° – ç ÐA + B ÷ 2 2 2 2 ø è

1 (ÐC + ÐD) = ÐAPB 2 Þ ÐC + ÐD = 2ÐAPB

Þ

59.

[from eq. (i)]

(c) Sum of angle of regular polygon =

( n - 2 )180° n

Þ 135 n = 180n – 360 Þ 45 n = 360 360 =8 45 Number of diagonals = 8C2 – 8 n=

8´ 7 – 8 = 20 2 (d) Let AD is diameter of circle of centre O. Find OP = q =

60.

A

17 cm O 15 P

D 17 cm

B AD = 34 cm, AO = OD = 17 cm


14 AB = 30 cm, AP = OP =

30 = 15 cm 2

E 63.

(17)2 – (15)2

(b)

D

= 64 = 8cm. 61.

y G

A

(a) 72° G

B

H

O x

72° F

E

108°

Pentagon

J I

C

D

Sum of Interior angles = (number of sides –2) × 180° = (5 – 2) × 180° = 540° Interior angle of regular pentagon =

C

A

x B

F

D EAB is equilateral D EDC is also equilateral Area of trapezium ABCD

æ1 ö 1 = ç ´ DB ´ OA ÷ + ( DB ´ OC ) è2 ø 2 540 = 108° 5

1 = ´ DB ´ AC 2 Let AO = OB = x and DO = OC = y

Area (ABCD) =

72° 108°

1 ( x + y ) 2 = 16 (given) 2

Þ x+y= 4 2 ... (i) D AOB is a right angled isosceles triangle. So, AB = x 2 + x 2 = 2 x Similarly, DC =

Now suppliment angle = 180° – 108° = 72° Now, we can find angle at the top point of the star by adding the two equal base angles. = 180° – (72° + 72°) = 180° – 144° = 36° So each point of the stars = 36° 62.

B

A

(c)

Now, FG = EF – EG Þ FG = AB sin 60° – DC sin 60°

3 6 ( AB – DC ) = ( x – y ) ... (ii) 2 2 Area of trapezium = Area DEAB – Area DEDC =

=

3 ( AB 2 – DC 2 ) 4

=

3é ( x 2) 2 – ( y 2)2 ù û 4 ë

O 5cm

D

90°

13cm

C

OC =

169 – 25 = 144 = 12 cm Þ AC = 2OC = 24 cm 1 Area of Rhombus = ´ d1 ´ d 2 2

=

1 ´ 10 ´ 24 = 120 cm2 2

2y

Þ Area =

Now,

3 (x + y) (x – y) 2

3 ( x + y )( x – y) = 16 2

Þ x–y=

Height =

32 3( x + y )

Þ x– y =

8 6

(Q x + y = 4 2 )

6 6 8 ( x – y) = ´ = 4 cm 2 2 6


Quadrilateral and Polygon

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