MANISH KUMAR
MATHEMATICS AREAS OF PARALLELOGRAMS AND TRINGLES
INTRODUCTION In earlier class, we have learnt some formulae for finding the areas of different plane figures such as triangle rectangle, parallelogram, square etc. In this chapter, we will consolidate the knowledge about these formulae by studying some relationship between the area of these figures under the condition when they lie on the same base and between the same parallels.
PARALLEOGRAMS ON THE SAME BASE AND BETWEEN THE SAME PARALLELS Theorem - 1 :- Parallelograms on the same base and between the same parallels are equal in area
Given : Two parallelograms ABCD and ABEF on the same base AB and between the same parallels AB and FC To prove : ar (|| gm ABCD) = ar (|| gm ABEF). Proof : 1.
2. 3. 4. 5.
STATEMENT In BCE and ADF, we have : (i) BC = AD (ii) BCE = ADF (iii) BEC = AFD BCE ADF ar (BCE) = ar(ADF) ar (quad. ABED) + ar (BCE) = ar (quad. ABED) + ar (ADF) ar (|| gm ABCD) = ar (|| gm ABEF)
REASON Opposite sides of a || gm are equal. Corres. s are equal, as AD || BC and FC is the transversal Corres. s are equal, as BE || AF and FC is the transversal
AAS-axiom of congruence. Congruent figures are equal in area. Adding same area on both sides of 3. z ar (R1) + ar (2) = r (R1 R2)
Hence, proved. Corollary 1: In parallelogram ABCD, AB||Cd and BC||AD. If AL BC and L is the foot of the perpendicular, then or (ABCD) = BC × Al. Proof. If fig, ABCD is a parallelogram. Al BC. Now, we draw line through A and D, Be and CF Then BEFC becomes a rectangle, Here, the parallelogram ABCD and the rectangle BEFC (also BEFC is a parallelogram) have same base and both are between the same parallels, Thus, we have
MANISH KUMAR
MATHEMATICS
Area of parallelogram ABCD = Area of rectangle BEFC = BC × BE = BC × AL
( AL = BE)
Corollary 2: If a triangle and parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.
Given : AABC and ||gm BCDE on the same base BC and between the same parallels BC and AD. To Prove : ar (ABC) =
1 ar (|| gm BCDE). 2
Construction : Draw All BC and DM BC (produced). Proof : STATEMENT 1. AL = M 2.
3.
1 ar(ABC) = × BC × AL 2 1 = × BC ×DM 2 1 = ar(|| gm BCDE) 2 ar(ABC)
Ex. 1
REASON Perpendicular to the same line and between the same parallels are equal. 1 Area of a = ×Base × Height 2 AL = DM(from 1) ar (||gm BCDE) = Base × Height = BC × DM.
1 ar (||gm BCDE) 2
In fig. ABCD is a parallelogram, AL BC, AM CD, AL = 4 cm and AM = 5 cm. If BC = 6.5 cm then find CD.
Sol.
We have, BC × AL = CD × AM
[Each equal to area of the parallelogram ABCD)
6.5 × = CD × 5
CD = Ex.2
6.5 cm CD = 5.2 cm. 5
In the given fig, ABCD is a parallelogram whose diagonal intersect at O.A. line segment through meets AB at P and DC at Q. Prove that : ar (quad. APQD) =
1 ar (|| gm ABCD). 2
MANISH KUMAR Sol.
MATHEMATICS
Proof :
STATEMENT 1. 1 ar(ACD) = × ar (|| gm ABCD) 2 2.
In OAP and OCQ, we have : (i)
OA = OC
Diagonal of a ||gm bisect each other. Alt. int. s, DC || AB and CA is the transversal.
(ii) OAP = OCQ
Vert. opp s.
(iii) AOP = COQ
AAS-axiom of congruence.
3.
AOP OQ
4.
ar (OAP) = ar (OCQ)
5.
REASON Diagonal AC divides ||gm ABCD into two s of equal area
Adding ar (quad. AOQD) on both sides of 4.
ar (OAP) + ar (quad. AOQD)
From 1 to 6
= ar (OCQ) + ar (quad. AOQD) ar (quad. APQD) = ar (ACD) 6. 7.
ar (quad. APQD) =
1 × ar (||gm ABCD) 2
Hence, proved.
Ex.3
Prove that of all parallelograms of which the sides are given the parallelogram which is rectangle has the greatest area.
Sol.
Let ABCD be parallelogram in which AB = a and AD = b. Let h be the altitude corresponding to the base AE then,
ar (||
gm
ABCD) = AB × h = ah.
Since the sides a and b are give. Therefore, with the same sides a and b we can construct infinitely many parallelograms with different heights. Now, ar (||
gm
ar (||
gm
ABCD) = ah
ABCD) is maximum or greatest when h is maximum. [ a is given i.e., a is constant]
But, the maximum value which h can attain is AD = b and this is possible when AD is perpendicular to AB i.e. the ||
gm
ABCD becomes a rectangle.
MANISH KUMAR gm
Thus, (ar ||
MATHEMATICS ABCD) is greatest when AD AB i.e., when (||
gm
ABCD) is a rectangle.
MANISH KUMAR Ex.4
MATHEMATICS
In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that : ar (OAB) + ar (OCD) = ar (OB) + ar (OAQ).
Sol.
Given : A || gm ABCD in which O is a point. To prove : ar (OAB) + (OCD) = ar (OBC) + ar (OAD) Construction : Draw EOF || AB and GOH || AD.
Proof :
STATEMENT
REASON
1.
AE || BF (given), EF||AB (By construction) AOB and || gm ABFE being on same base and between the same parallels.
2.
ABFE is a || gm 1 ar (OAB) = a (||gm ABFE) 2
3.
EFCD is a || gm
4. 5.
6. 7.
8.
1 ar (||gm EFCD) 2 1 ar (OAD) = ar (||gm AGHD) 2 1 ar (BOC) = ar (||gm GBCH) 2 ar (OAB) + ar (OCD) 1 1 = ar (||gm ABFE) + ar (|| gm 2 2 EFCD) 1 = × [ar || gm ABFE) + ar ( || gm 2 EFCD) 1 = × ar (|| gm ABCD) 2 ar (OAD) + ar (BOC) 1 1 = ar (|| gm AGHD) + ar (|| 2 2 gm GBCH) 1 = × [ar || gm AGHD) + ar (|| 2 gm GBCH) 1 = × ar (|| gm ABCD) 2 ar (OCD) =
EF || AB || DC (by const.) & FC || ED (given), OCD and || gm EFCD being on same base and between the same parallels. Similarly Similarly. Adding 2 and 4.
Adding 5 and 6.
Each = ar (|| gm ABCD), from 7 and 8
MANISH KUMAR 9.
Ex.5
ar (OAB) + ar (OCD) = ar (OAD) + ar (BOC)
In fig, ABCD is parallelogram and EFCD is a rectangle. Also AL DC. Prove that (i) ar (ABCD) = (EFCD)
Sol.
MATHEMATICS
(ii) ar (ABCD) = CD × AL
(i) We know that a rectangle is also a parallelogram.
Thus, parallelogram ABCD and rectangle EFCD are on the same base CD and between the same parallels CD and BE.
ar (||gm ABCD) = ar (EFCD)
(ii) From (i), we have ar (ABCD) ar (EFCD)
ar (ABCD) = CD × FC
[ Area of a rectangle = Base + Height]
ar (ABCD) = CD × AL
[ AL = FC as ALCF is a rectangle]
ar (ABCD) = DC × AL
TRIANGLES ON THE SAME BASE AND BETWEEN THE SAME PARALLELS Theorem -2: Triangles on the same base and between the same parallels are equal in area. Given : Two s ABCD and DBC on the same base BC and between the same parallels BC and AD. To prove : ar(ABC) = ar (DBC). Construction : Draw BE||CA, meeting DA produced at E and draw CF||BC, meeting AD produced at F. Proof : STATEMENT
REASON
MANISH KUMAR
MATHEMATICS BC || EA and BE || CA (By construction)
1.
2.
BCAE is a || gm. ar (ABC) =
Diagonal BA divides || gm BCAE into two s of equal areas.
1 ar (|| gm BCAE) 2
BC || DF and BD || CF (By construction) Diagonal CD divides || gm BCFD into two s of equal
3.
BCFD is a || gm
areas. || gmson same base BC and between the same
4.
ar (DBC) =
1 ar (|| gm BCFD) 2
5.
ar (|| gm BCAE) = ar (|| gm BCEF)
6.
ar (ABC) = ar (DBC)
parallels BC and EF are equal in area From 2 ,4 and 5.
BC || EA and BE || CA (By construction) Diagonal BA divides || gm BCAE into two s of equal areas. BC || DF and BD || CF (By construction) Diagonal CD divides || gm BCFD into two s of equal areas. || gm on same base BC and between the same parallels BC and EF are equal in area From 2 ,4 and 5. Corollary : Area of a triangle =
1 × Base × Height. 2
Given : A ABC with base BC and height AL. To prove : ar(ABC) =
1 × BC × AL 2
Construction : Draw CD||BA and AD||BC, intersecting each other at D.
Proof :
STATEMENT
REASON
MANISH KUMAR 1.
ABCD is a || gm
3.
ar (ABC) =
=
Hence,
MATHEMATICS BC || AD and BA||CD (By construction) Diagonal CA divides || gm ABCD into two
1 ar (|| gm ABCD) 2
s of equal areas.
1 × BC × AL 2
Areal of triangle =
ar (|| gm ABCD) = BC × AL
1 × Base × Height 2
Ex.6
Show that a median of a triangle divides it into two triangles of equal area.
Sol.
Given : ABC in which AD is a median. To prove : ar(ABC) = ar (ADC). Construction : Draw AL BC. Proof : Since AD is the median ABC.
MANISH KUMAR
MATHEMATICS
Therefore, D is the mid-point of BC.
BD = DC BD × AL = DC × AL [Multiplying both sides by AL]
1 1 1 (BD × AL) = (DC × AL) [Multiplying both sides by ] 2 2 2
ar (ABD) = ar (ADC) Hence, proved. Ex.7
In the given figure, ABCD is a quadrilateral in which M is the mid-point of diagonal AC. Prove that : ar (quad. ABMD) = ar(quad. DMBC).
Sol.
Given : ABCD is quadrilateral in which M is the mid-point of diagonal AC. To prove : ar(quad. ABMD) = ar (quad. DMBC) Proof : 1.
STATEMENT ar (ABM) ar (CBM)
2.
ar (AMD) = ar (DCM)
3.
ar (ABM) + ar (AMD)
REASON Median BM divides ABC into two triangles of equal area. Median DM divides DAC into two triangles of equal are.
= ar (CBM) + ar (DCM) 4.
ar (quad. ABMD) = ar (quad.
Adding 1 and 2.
DMBC)
Hence, proved. Ex.8
In the given figure, PQRS and PXYZ are two parallelograms of equal area. Prove that YR||QZ.
Sol.
We have : ar (|| gm PQRS) = (|| gm PXYZ)
ar (|| gm PQRS) - ar (|| gm PQOZ) = ar (|| gm PXYZ) - ar (|| PQOZ)
ar (|| gm (ZORS) = ar (|| gm QXYO)
1 1 ar (|| gm ZORS) = ar (|| gm QXYO) 2 2
ar (ZOR) = ar (OQY) ar (ZOR) + ar (OYR) = ar (OQY) + ar (OYR) ar (ZYR) = ar (QYR)
MANISH KUMAR
MATHEMATICS
YR||QZ. [ZYR & QYR are equal in area & have the same base. So, they must be between the same parallels]
Ex.9
Prove that the area of the quadrilateral formed by joining the mid-points of the adjacent sides of a quadrilateral is half the area of the given quadrilateral.
Sol.
Given : A quadrilateral ABCD, and PQRS is the quadrilateral formed by joining mid-points of the sides AB, BC, CD and DA respectively.
To prove : ar (quad. PQRS) =
1 × ar(quad. ABCD) 2
Construction : Join AC and AR.
STATEMENT 1.
2.
3.
ar (ARD) = ar (SRD) = ar (SRD) =
1 2 1 2 1 4 1 4
× ar (ACD)
REASON Median of a triangle divides it into two triangles of equal area.
× ar (ARD)
Same as in 1.
× ar (ACD)
From 1 and 2.
4.
ar (PBQ) =
× ar (ABC)
5.
ar (SRD) + ar (PBQ) =
1 [ar (ACD) + ar(ABC) 4
1 × (quad. ABCD) 4 1 ar (APS) + ar (QCR) = × ar(quad. ABCD) 4 ar (APS) + ar (PBQ) + ar (QCR) + ar (SRD) 1 = × ar (quad. ABCD) 2 ar (APS) + ar (PBQ) + ar (QCR) + ar (SRD) + ar (quad. PQRS) = ar (quad. ABCD) 1 ar (quad. PQRS) = × ar (quad. ABCD) 2 =
6. 7.
8.
9.
Ex.10
As in 3. Adding 3 and 4.
As in 5. Adding 5 and 6.
Subtracting 7 from 8.
If the diagonals of quadrilateral separate it into four triangle of equal area, show that it is a parallelogram.
Sol.
Given : A quad. ABCD whose diagonals intersect at O such that : ar (AOD) = ar (AOB) = ar (BOC) = ar (COD) To prove : ABCD is a parallelogram Proof : ar (AOD) = ar (BOC)
ar (AOD) + ar (AOB) = ar (BOC) + ar (AOB)
MANISH KUMAR
MATHEMATICS
ar (ABD) = ar (ABC)
DC || AB
[Equal s on same base must be between the same parallels]
Similarly, AD||BC. Hence ABCD is || gm.
EXERCIESE SUBEJCTIVE TIYE QUESTIONS (A)
SHORT ANSWERS TYPE QUESTIONS :
1.
In the adjoining, BD is a diagonal of quad. ABCD. Show that ABCD is a parallelogram and calculate area of || gm ABCD.
2.
In a || gm ABCD, it is given that AB = 16 cm and the altitudes corresponding to the sides AB and AD are and 8 cm respectively. Find the length of AD.
3.
Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram. divides if two equal parallelograms.
4.
In the given figure, the area of II gm ABCD is 90 cm2. State giving reasons : (i) ar (|| gm ABEF)
(ii) ar (ABD)
(iii) ar (BEF).
MANISH KUMAR 5.
MATHEMATICS
In the given figure, the area of ABC is 64 cm2. State giving reasons : (i) ar (|| gm ABCD)
(ii) ar (rect. ABEF)
6.
In the given figure. ABCD is a quadrilateral. A line through D, parallel to AC, Meets BC Produced in P, Prove that: ar (∆ ABC) = ar (quad. ABCD).
7.
Answer the following questions as per exact requirement: (i) ABCD is a parallelogram in which AB || CD and AB = CD = 10 cm. If the perpendicular distance between AB and CD be 8 cm, find the area of the parallelogram ABCD. 2 (ii) ABCD is a parallelogram having area 240 cm , BC = AD = 20 cm and BC || AD, Find the distance between the parallel sides BC and AD. 2 (iii) ABCD is a parallelogram having area 160 cm , BC || AD and the perpendicular distance between BC and AD is 10 cm. Find the length of the side BC. 2 (iv) ABCD is a parallelogram having area 200 cm . If AB || CD pints P and Q divide AB and DC respectively , find the area of the parallelogram APQD . (v) ABCD is a parallelogaram having area 450 cm2 . If AB || CD points P and Q divide AB and DC respectively in the ratio 1: 2 find the area of the parallelogram APQD and parallelogram PBCQ.
8.
In fig.
9.
In fig ABCD is a trapezium in which AB= 7 cm, AD = BC = 5 cm. DC = x cm, and distance between AB and DC is 4 cm.Find the value of x and area of trapezium ABCD.
AOB 90 0 , AC = BC, OA = 12 cm and OC = 6.5 cm, Find the area of ∆AOB .
MANISH KUMAR 10.
MATHEMATICS
In fig. OCDE is a rectangle inscribed in a quadrant of circle of radius 10 cm. If OE = the rectangle.
2 5 , find the area of
MANISH KUMAR
MATHEMATICS
11.
In fig ABCD is a trapezium in which AB || DC. Prove that ar ( ∆AOD) = ar (∆BOC)
12.
In fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar(∆ADE) = AR (∆BCF).
(B)
Long Answer type QUESTINS:
1.
ABCD is a quadrilateral If AL BD , prove that : ar(quad, ABCD) =
2.
In the given figure, D is the mid-point of BC and E is any point on AD. Prove that: (i) ar (∆EBD) = ar(∆EBC). (ii) ar(∆ABE) = ar (∆ACE).
3.
In the given figure, D is the mid-point of BC and E is the, mid-point of AD. Prove that : ar (ABE) =
4.
1 BD ( AL CM ) 2
1 ar (ABC ). 4
In the given figure. a point D is taken on side BC of ∆ABC and AD is produced to E. making DE= AD, Show That : ar(∆BEC) = ar (∆ABC).
MANISH KUMAR
MATHEMATICS
5.
If the medians of a ABC intersect at G, show that : ar (AGB) = ar (AGC) = ar (BGG) =
6.
D is a point on base BC of a ABC such that 2BD = DC. Prove that : ar (ABD) =
7.
In the given figure, AD is a median of ABC and P is a point on AC such that :
1 ar (ABC.) 2
1 ar (ABC). 3
ar (ADP) : ar (ABD) = 2 : 3. Find (i) AP : PC (ii) ar (PDC) : ar (ABC).
8.
In the given figure, P is a point on side BC of ABC such that BP : PC = 1 : 2 and Q is a point on AP such the PQ : QA = 2 : 3. Show that ar (AQC) : ar (ABC) = 2 : 5
9.
In the adjoining figure, ABCD is a parallelogram. P and Q are any two points on the sides AB and BC respective Prove that : ar (CPD) = ar (AQD)
10.
In the adjoining figure, DE || BC. Prove that : (i) ar (ABE) = ar (ACD). (ii) ar (OBD) = ar (OCE).
MANISH KUMAR 11.
MATHEMATICS
In the given figure, ABCD is a parallelogram and P is a point on BC. Prove that : ar (ABP) + ar (DPC) = ar (APD)
12.
In the adjoining figure, ABCDE is a pentagon. BP drawn parallel to AC meets DC produced at P and EQ draw parallel to AD meets CD produced at Q. Prove that : ar (Pentagon ABCDE) = ar (APQ).
13.
I the adjoining figure, two parallelograms ABCD and AEFB are drawn on opposite sides of AB. Prove that : ar (|| gm ABCD) + ar (|| gm AEFB) = ar (|| gm EFCD).
14.
In the adjoining figure, ABCD is a parallelogram and O is any point on its diagonal AC. Show that : ar (AOB) = ar (AOD).
15.
In the given figure, XY||BC, BE||CA and FC||AB. Prove that: ar (ABE) = ar (ACF).
16.
In the given figure, the side AB of || gm ABCD is produced to a point P. A line through A drawn parallel to CD meets CB produced in Q and the parallelogram PBQR is completed. Prove that: ar (|| gm ABCD) = ar (|| gm BPRQ).
MANISH KUMAR 17.
MATHEMATICS
In the adjoining figure, CE is drawn parallel to DB to meet AB produced at E. Prove that : ar (quad. ABCD) = a (DAE).
MANISH KUMAR 18.
MATHEMATICS
In the adjoining figure, ABCD is parallelogram. B is produced to a point P and DP intersects BC at Q. Prove that : ar (APD) = ar (quad. BPCD).
9.
In the adjoining figure, ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q. Prove that: ar (BPC) = ar (DPQ).
20.
In the adjoining figure, ABCD is a parallelogram, P is a point on BC such that BP : PC = 1 : 2 DT produced meets AB produced at Q. Given ar (CPQ) = 20 cm2. Calculate: (i) ar (CDP) (ii) ar (|| gm ABCD).
21.
In the adjoining figure, ABCD is a parallelogram, P is a point on DC such that ar (APD) = 25 cm2 and ar (BPC) = 15 cm2. Calculate : (i) ar (|| gm ABCD) (ii) DP : PC
22.
In the given figure, AB || DC || EF, AD || BE and DE || AF. Prove that : ar (|| gm DEFH) = ar (|| gm ABCD).
23.
In the given figure, squares ABDE and AFGC are drawn on the side AB and hypotenuse AC of right triangle ABC and HB FG. Prove that: (i) EAC BAF. (ii) ar (sq. ABDE) = ar (rect. ARHF).
MANISH KUMAR 24.
MATHEMATICS
If fig, ABC is a right triangle right angles at A, BCED, ACFG and ABMN are squares on the sides BC, CA an AB respectively. Line segment AX DE meets BC at Y. Show that (i) MBC ABD
(ii) ar (BYZD) = 2 ar (MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) FCB ACE
(v) ar (CYXE) = 2 ar (FCB)
(vi) ar (CYXE) = ar (ACFG)
(vi) ar (BCED) = ar (ABMN) + ar.(ACFG) (C)
NCEFT QUESTIONS
1.
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.
2.
In fig. ABCD is parallelogram, AE DC and CF AD. if AB = 16 cm, AE = 8 cm and CF = 10 cm, find AI
3.
If E.F.G. and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = ar (ABCD).
4.
P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that (APB) = ar (BQC).
5.
In fig, P is a point in the interior of parallelogram ABCD. Show that (i)
ar (APB) + ar (PCD) =
1 ar (ABCD) 2
(ii) ar (APD) + ar (PB) = ar (APB) + ar (PCD) 6.
In fig. PQRS and ABRS are parallelograms and X is any point on side BR. Show that : (i)
ar (PQRS) = ar (ABRS)
MANISH KUMAR (ii) ar (AXS) =
7.
MATHEMATICS 1 ar (PQRS) 2
A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divides? what are the shapes of these parts ? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it ?
8.
In fig. E is any point on median AD and a ABC. Show that ar (ABE) = ar (ACE)
9.
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) =
10.
11. 12.
1 ar(ABC). 4 Show that the diagonals of a parallelogram divide it into four triangles of equal area.
In fig, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O. show that ar(ABC) = ar (ABD). D, E and F are respectively the mid-points of the sides BC, CA and AB of a ABC. Show that
1 1 ar (ABC) (iii) ar (BDEF) = ar(ABC) 2 4 In fig, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that : (i) ar (DOC) = ar (AOB) (ii) ar (DCB) = ar (ACB) (iii) DA || CB or ABCD is a parallelogram. (i) BDEF is parallelogram.
13.
14. 15. 16.
(ii) ar (DEF) =
D and E are points on sides AB and AC respectively of ABC such that ar(DBC) = ar(EBC). Prove that DE||BC. XY is a line parallel to side BC of a triangle ABC. If BE|AC and CF||AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF) The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed Show that ar (ABCD) = ar (PBQR)
MANISH KUMAR
MATHEMATICS
17.
Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at O. Prove that ar (AOD) = ar
18.
(BOC). In fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar (ACB) = ar (ACF) (ii) ar (AEDF) = ar (ABCDE)
19.
A villager Itwaari has a plot of land of the shape of quadrilateral. The Gram Pandhayat of the village decide to take over some portion of his plot from on the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining plot so as to form a triangular plot. Explain how this proposal will be implemented.
20.
ABCD is a trapezium with AB||DC. A line parallel to AC intersects AB at X and BC at Y. prove that ar (AD) = ar (ACY)
21.
In fig. AP||BQ||CR. Prove that ar (AQC) = ar (PBR).
22.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove the ABCD is a trapezium.
23.
In fig. ar (DRC) = ar (DPC) and ar (BD) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR a trapeziums.
OBJECTIVE TYPE QUESTIONS 1.
In fig. ABCD is a parallelogram, Al CD and AM BC. If AB = 12 cm, AD = 8 cm and AL = 8 cm, then AM
(A) 15 cm 2.
(B) 9 cm
(C) 10 cm
(D) None of these
In fig. ABCD is parallelogram and P is mid-point of AB. If ar (APCD) = 36 cm2 then ar (ABC) =
MANISH KUMAR (A) 36 cm2 3.
(B) 48 cm2
(C) 24 cm2
(D) None of these
(C) 16 cm2
(D) None of these
2
In fig. if (ABC) = 28 cm , then ar (AEDF) =
(A) 21 cm2 4.
MATHEMATICS
(B) 18 cm2
In fig. ABCD is a quadrilateral. BE||AC. BE meets DC (produced) at E. AE and BC intersect at O. Which one is the correct answer from the following ?
5.
(A) ABEC is parallelogram
(B) ar (AOC) = ar (BOE)
(C) ar(OAB) = ar(OCE)
(D) ar (ABE) = ar (ACE)
In fig, D and E are the mid-points of the sides AC and BC respectively and ABC. If (BED) = 12 cm2. then are (ABED) =
6.
(A) 36 cm2
(B) 48 cm2
(C) 24 cm2
(D) None of these
Two parallelograms stand on equal bases and between the same parallels. The ratio fo their areas is (A) 1 : 2
7.
(C) 1 : 1
(D) 1 : 3
If a rectangle and parallelogram are equal in area and have the same base and are situated on the same side, then the quotient :
8.
(B) 2 : 1
Perimeter of rec tan gle is Perimeter of || gm
(A) Equal to 1
(B) Greater than 1
(C) Less than 1
(D) Indeterminate
If ABCD is a rectangle, E, F are the mid points of BC and AD respectively and G is any point on EF, then GAB equals. (A)
1 (ABCD) 2
(B)
1 (ABCD) 3
(C)
1 (ABCD) 4
(D)
1 (ABCD) 6
MANISH KUMAR
MATHEMATICS
E, F, F are mid points of the sides BC, CA & AB respectively of ABC, then area of || gm BDEF is equal to
9.
10.
(A)
1 ar (ABC) 2
(B)
1 ar (ABC) 4
(C)
1 ar (ABC) 3
(D)
1 ar (ABC) 6
ABCD is a quadrilateral P,Q,R and S are the mid-points of AB, BC, CD and DA respectively, then PQRS is a (A) Square
(B) Parallelogram
(C) Trapezium
(D) Kite
AREAS OF PARALLELOGRAMS AND TRAINGLES
(A)
ANSWER KEY
EXERCISE
SHORT ANSWERS TYPE QUESTIONS : 1.
48 cm2
2. 2
12 cm
4.
(i) 90 cm
1
(iii) 45 cm2
7.
(i) 80 cm2, (ii) 12 cm,
(iii) 16 cm,
8.
30 cm2
(ii) 45 cm
5.
(i) 128 cm2,
(ii) 128 cm2
(iv) 100 cm2,
(v) 150 cm2, 300 cm2
x = 13 cm, 40 cm2
9.
10. 40 cm2
(B)
LONG ANSWERS TYPE QUESTIONS : 7.
(i) 2 : 1
(ii) 1 : 6
(i) 40 cm2,
20.
(ii) 120 cm2
21. (i) 80 cm2, (ii) 5 : 3
(C)
NCERT BASED QUESTIONS : 1.
(i) Base DC, parallels DC and AB,
(iii) Base QR, parallels QR & PS.
(v) Base AD, parallels AD and BQ. 2.
12.8 cm
7.
Wheat in APQ and pulses in other two triangles or pulses in APQ and wheat in other two triangles.
(OBJECTIVE)
ANSWER KEY
EXERCISE
Qu.
1
2
3
4
5
7
8
8
9
10
Ans.
B
C
D
C
A
C
C
C
A
B
