Lesson: Areas of Parallelograms and Triangles

EXERCISE 9.1

Question: 1

The median of a triangle divides it into two

(a) Triangles of equal area

(b) Congruent triangles

(c) Right triangles

(d) Isosceles triangles

Solution:

a

Median of a triangle divides it into two triangles of

equal area.

Question: 2

In Fig. 9.3, which one of the figures has two polygons

on the same base and between the same parallels?

Solution:

d

In fig (D) two polygons PARQ and BSRQ lie on same

base QR and between same parallel lines PS and QR.

Question: 3

The figure obtained by joining the mid-points of the

adjacent sides of a rectangle of sides 8 cm and 6 cm

is:

(a) A rectangle of area 48 cm

2

.

(b) A square of area 48 cm

2

.

(c) A trapezium of area 24 cm

2

.

(d) A rhombus of area 24 cm

2

.

Solution:

d

Let PQRS be a rectangle with PQ=8cm and RQ=6cm.

Let ABCD be the quadrilateral formed by joining the

midpoints of sides PQ, QR, RS and SP, respectively.

Quadrilateral formed by joining the mid-points of the

adjacent sides of a rectangle is a rhombus.

Thus, ABCD is a rhombus.

Then, diagonal of the rhombus, AC= RQ=6cm and

BD=PQ=8cm.

∴ Area of rhombus ABCD=

Product of diagonals

cm

2

8 6

24

2 2

Question: 4

In Fig. 9.4, the area of parallelogram ABCD is:

(a)

AB BM

(b)

BC BN

(c)

DC DL

(d)

AD DL

Fig. 9.4

Solution:

c

Area of a parallelogram= base × corresponding

altitude.

Hence, for base AB, the height is DL.

So, area=AB×DL

Similarly, for base AD the height is BM.

So, area=AD×BM,

For base DC ,the height is DL.

So, area=DC×DL

And for base BC, the height is not given.

Hence, option (c) is correct.

Question: 5

In Fig. 9.5, if parallelogram ABCD and rectangle ABEM

are of equal area, then:

(a) Perimeter of ABCD = Perimeter of ABEM

(b) Perimeter of ABCD < Perimeter of ABEM

(c) Perimeter of ABCD > Perimeter of ABEM

(d) Perimeter of ABCD =

1

2

(Perimeter of ABEM)

Solution:

c

It is given that ABCD is a parallelogram,

∴AB=CD ……….. (i)

ABEM is a rectangle

∴

EM=AB ……… (ii)

Adding eq. (i) and eq. (ii), we get

AB+EM=CD+AB ….…… (iii)

In ∆AMD,

AD>AM …..(iv) (Hypotenuse is the longest side in a

right angled triangle)

Similarly, in ∆BEC

vBC BE ........

(Hypotenuse is the longest side in a

right angled triangle)

Adding Eq. (v) and Eq. (vi), we get

AD BC AM BE

Adding AB+CD on both sides,

Using Eq.

AD BC AB CD AM BE AB CD

AD BC AB CD AM BE AB EM iii

⇒ Perimeter of the parallelogram ABCD > Perimeter

of the rectangle ABEM.

Question: 6

The mid-point of the sides of a triangle along with any

of the vertices as the fourth point make a

parallelogram of area equal to

(a)

ar

1

( )

2

ABC

(b)

ar ( )

1

3

ABC

(c)

ar ( )

1

4

ABC

(d)

ar ( )ABC

Solution:

a

Let ABC is a triangle and D, E and F are the midpoints

of BC, AC and AB, respectively.

Then, Area (∆AFE) =Area (∆BFD) =Area (∆DEF) = Area

(∆EDC)

∴Area (∆DEF) =

1

4

Area (∆ABC)

Let BDEF be the parallelogram with vertices B as the

fourth point.

Then area of parallelogram BDEF

= Area (∆BFD) + Area (∆DEF)

=Area (∆DEF) + Area (∆DEF)

=2 Area (∆DEF)

=

Area ( )= Area ( )ΔABC ΔABC

1 1

2

4 2

Question: 7

Two parallelograms are on equal bases and between

the same parallels. The ratio of their areas is

(a) 1 ∶ 2

(b) 1∶ 1

(c) 2 ∶ 1

(d) 3 ∶ 1

Solution:

b

If two parallelograms are on equal bases and

between the same parallels then they are equal in

area.

Question: 8

ABCD is a quadrilateral whose diagonal AC divides it

into two parts, equal in area, then ABCD

(a) Is a rectangle

(b) Is always a rhombus

(c) Is a parallelogram

(d) Need not be any of (a), (b) or (c)

Solution:

d

Two triangles with equal areas doesn’t imply that

these triangles are also congruent.

Question: 9

If a triangle and a parallelogram are on the same base

and between same parallels, then the ratio of the

area of the triangle to the area of parallelogram is

(a) 1 ∶ 3

(b) 1 ∶2

(c) 3 ∶ 1

(d) 1 ∶4

Solution:

b

If a triangle and a parallelogram are on the same base

and between same parallels then

area of the triangle =

1

2

area of the parallelogram

Area of triangle

Area of paralleogram

1

2

⇒ Area of triangle∶ Area of parallelogram=1: 2

Question: 10

ABCD is a trapezium with parallel sides AB = a cm and

DC = b cm (Fig. 9.6). E and F are the mid-points of the

non-parallel sides. The ratio of Area (ABFE) and Area

(EFCD) is

(a) a ∶b

(b) (3a + b) ∶ (a + 3b)

(c) (a + b) ∶ (3a + b)

(d) (2a + b) ∶ (3a + b)

Solution:

b

Let us join DB which intersects EF at O.

Since E and F are the mid-points of AD and CB

respectively,

therefore,

.EF AB CD

So, the distance between AB, EF and EF, CD will be

same say h.

Now, in

ABD

,

E is the midpoint of AD and EO

∥

AB,

1 1

2 2

EO AB ia

and O is the midpoint BD.

Similarly, in

BCD

,

O and F are the midpoints of BD and BC respectively.

1 1

2 2

OF CD iib

Adding (i) and (ii),

1 1 1

2 2 2

EO OF a b a b iii

Now,

Area of a trapezium=

1

2

(Sum of the parallel lines)×(distance between

parallel lines)

So,

1 1

2 2

( ) ( )Area Trapezium ABFE a a b h

(Using (iii))

=

1

(3 )

4

a b h

And

ar Trapezium using(

1 1

2 2

)EFCD b a b h iii

=

1

(3 )

4

b a h

Hence,

ar(trapeziumABFE)

ar(trapeziumEFCD)

1

4

1

4

(3 )

(3 )

(3 ) (3 )

a b h

a b

b a h b a

EXERCISE 9.2

Question: 1

ABCD is a parallelogram and X is the mid-point of AB.

If Area (AXCD) =24 cm

2

, then Area (ABC) = 24 cm

2

.

Solution:

False,

Since X is the midpoint of AB,

= a ar BXC r AXC ar ABC i

1

2

Now, ∆ABC and parallelogram ABCD lie on same base

AB and between same parallel lines AB and CD,

a

a

1

2

1

2

[ ( ) ]

[ ]

1 1

24

2 2

ar ABC ar ABCD

r ABC r AXCD ar BXC

r ABC ar ABC

a

[∵

ar cm

2

( ) 24AXCD

is given and using (i)]

a

a

2

1

12

4

1

12

4

3

12

4

12 4

16

3

]r ABC ar ABC

r ABC ar ABC

ar ABC

ar ABC cm

Question: 2

PQRS is a rectangle inscribed in a quadrant of a circle

of radius 13 cm. A is any point on PQ. If PS = 5 cm,

then Area (PAS) = 30 cm

2

.

Solution:

PQRS is a rectangle inscribed in a quadrant of a circle.

Radius of circle= 13cm and PS=5cm

In right angled triangle ∆PQS,

cm

2 2 2 2

( ) ( ) (13) (5) 12PQ SQ QR

Now, Area (∆PQS)

2

1

2

1

2

1

5 12 30

2

base height

PS PQ

cm

So, the given statement is true, if A coincides with Q.

Question: 3

PQRS is a parallelogram whose area is 180 cm

2

and A

is any point on the diagonal QS. The area of

2

90 .ASR cm

Solution:

False,

It is given that

2

180ar gm PQRS cm

.

We know that diagonal of a parallelogram divides it

into two triangles of equal area.

r

2

( ) 90a QSR cm

Also,

Area of ∆ASR<Area of ∆QRS

2

90Area of ASR cm

Question: 4

ABC and BDE are two equilateral triangles such that

D is the mid-point of BC.

Then,

1

( ( )

4

)ar BDE ar ABC

.

Solution:

True,

It is given that D is the midpoint of BC.

So,

1

2

BD BC

Area of an equilateral trinagle= side)

2

3

( .

4

So,

ar BDE

ar ABC

2

2

2 2

2 2 2

2

3( )

( )

( ) ( ) 1

4 2

( ) ( ) 4( ) 4

3( )

4

BD BC

BD BC

BC BC BC

BC

Question: 5

In Fig. 9.8, ABCD and EFGD are two parallelograms

and G is the mid-point of CD. Then

1

)

2

( ) (ar DPC ar EFGD

.

Solution:

False

Let us join PG.

Since G is the midpoint of CD, in ∆DPC, PG is a

median.

So,

1

. 1

2

ar DPG ar GPC ar DPC

Now,

∆DPG and parallelogram EFGD lie on same base DG

and between same parallel lines EF and DG.

1

.. 2

2

ar DPG ar gm EFGD

From Eq. (1) and (2),

1 1

2 2

ar DPC ar gmEFGD

ar DPC ar gmEFGD

EXERCISE 9.3

Question: 1

In Fig.9.11, PSDA is a parallelogram. Points Q and R

are taken on PS such that

PQ = QR = RS and PA || QB || RC.

Prove that Area (PQE) = Area (CFD).

Solution:

It is given that PSDA is a parallelogram,

PS AD PQ AB

Also,

||PA QB

(given)

So, PABQ is a ∥gm.

PQ AB

….. (i) (Opposite sides of ∥gm)

Similarly, QRCB is a ∥gm.

∴QR=BC….. (ii) (Opposite sides of ∥gm)

And RSDC is a ∥gm.

∴RS=CD….. (iii) (Opposite sides of ∥gm)

Now, it is given that PQ = QR = RS…… (iv)

From Eq. (i), (ii), (iii) and (iv), we have

PQ QR RS AB BC CD

……… (v)

Now, in ∆PRF

and ||PQ QR QE RF

(Given)

So, by the midpoint theorem

PE EF

………. (vi)

Similarly, in

DEB

BC CD

(As proved above) and

||CF EB

(Given)

So, by the midpoint theorem

DF EF

……… (vii)

From (vi) and (vii)

PE DF

…………… (viii)

Now,

in and PQE DFC

PQ DC

(Using (v))

QPE FDC

(Alternate interior angle)

PE FD

(Using (viii))

So, by SAS congruence rule

PQE DCF

∴Area (∆PQE) = Area (∆DFC) (∵Congruent figures have

equal area)

Question: 2

X and Y are points on the side LN of the triangle LMN

such that LX = XY =YN. Through X, a line is drawn

parallel to LM to meet MN at Z (See Fig. 9.12). Prove

that Area (LZY) = Area (MZYX).

Solution:

Since ∆LXZ and ∆MXZ lie on same base XZ and

between same parallel lines LM and XZ.

Area (∆LXZ) = Area (∆XMZ)

On adding Area (∆XYZ) to both sides, we get

Area (∆LXZ) + Area (∆XYZ) = Area (∆XMZ) + Area

(∆XYZ)

Area (∆LYZ) = Area (MZYX)

Hence it is proved.

Question: 3

The area of the parallelogram ABCD is 90 cm

2

(see

Fig.9.13). Find

(i) Area (ABEF)

(ii) Area (

∆

ABD)

(iii)Area (

∆

BEF)

Solution:

(i) Parallelogram ABCD and parallelogram ABEF lie on

the same base AB and between the same parallel

lines AB and FC.

So, Area (ABEF) = Area (ABCD) = 90cm

2

(Parallelograms that lie on the same base and

between same parallel lines have equal area.)

(ii) Parallelogram ABCD and ∆ABD lie on the same

base AB and between the same parallel lines AB and

DC.

So, Area (∆ABD) =

1

2

Area (ABCD) =

1

2

×90cm

2

=45 cm

2

(If a triangle and a parallelogram lie on the same base

and between the same parallel lines then the area of

the triangle is equal to half of the area of the

parallelogram).

(iii) Parallelogram ABEF and ∆BEF lie on the same

base BE and between the same parallel lines BE and

FA.

So, Area (∆BEF) =

1

2

Area (ABEF) =

1

2

×90cm

2

=45

cm

2

(If a triangle and a parallelogram lie on the same base

and between the same parallel lines then the area of

the triangle is equal to half of the area of the

parallelogram).

Question: 4

In

ABC

, D is the mid-point of AB and P is any point

on BC. If

||CQ PD

meets AB in Q (Fig. 9.14), then

prove that

Area (

∆

BPQ) =

1

2

Area (ABC).

Fig. 9.14

Solution:

Let us join CD.

In

ABC

, CD is the median (Since D is the mid-point

of AB).

So, it will divide

ABC

into two triangles of equal

area.

.

..

ar BCD ar ABC

ar BDP ar DPC ar A C iB

( ) ( )

( )

1

2

1

( ) ( )

2

...( )

Now, ∆DPQ and ∆DPC are on the same base DP and

between the same parallels lines PD and QC.

( ) ( ) i. i ar DPQ ar DPC

From (i) and (ii),

(

1

( ) ( )

2

1

)

)

( (

2

)

ar BDP ar DPQ ar ABC

BPQ aar r ABC

Hence, it is proved.

Question: 5

ABCD is a square. E and F are respectively the

midpoints of BC and CD. If R is the mid-point of EF

(Fig. 9.15), prove that

Area (

∆

AER) =Area (

∆

AFR).

Fig. 9.15

Solution:

Since R is the midpoint of EF, AR is the median of

∆AEF.

Then, AR will divide ∆AEF into two triangles of equal

areas.

So, Area (AER) = Area (AFR)

Hence, it is proved.

Question: 6

O is any point on the diagonal PR of a parallelogram

PQRS (Fig. 9.16). Prove that

Area (

∆

PSO)= Area (

∆

PQO).

Fig. 9.16

Solution:

Let us join SQ which intersects PR at M.

M is the midpoint of SQ (∵Diagonals of a

parallelogram bisect each other)

So, in ∆PSQ, PM is a median.

∴ Area (∆PSM) = Area (∆PQM) ……… (i) (Median of a

triangle divides it into two triangles of equal areas).

Also, in ∆OSQ, OM is a median.

∴ Area (∆OSM) = Area (∆OQM) …... (ii) (Median of a

triangle divides it into two triangles of equal areas).

On adding (i) and (ii),

Area (∆PSM) + Area (∆OSM) = Area (∆PQM) + Area

(∆OQM)

⇒Area (∆PSO) = Area (∆PQO)

Hence it is proved.

Question: 7

ABCD is a parallelogram in which BC is produced to E

such that CE = BC (Fig. 9.17). AE intersects CD at F.

If Area (DFB) =3 cm

2

, find the area of the

parallelogram ABCD.

Fig. 9.17

Solution:

In ∆EAB, C is the midpoint of BE (∵BC=CE)

And CF

∥

AB (∵ABCD is a ∥gm)

So, by the midpoint theorem,

AF=FE…….. (i) and

1 1

2 2

FC AB CD

DF FC ii

Now, in

ADF

and

CFE

AD=CE (

∵

AD=BC=CE)

DF=FC (Using (ii))

AF=FE (Using (i))

So, by SSS congruence,

ADF ECF

Area Area ADF ECF

But,

2

3Area ADF Area DFB cm

(Since ∆ADF and ∆DFB lie on same base DF and

between same parallel lines DF and AB)

So,

2

3Area ECF cm

……… (iii)

Now, in

,FEB BC CE

(given)

∴FC is a median

Thus,

Area Area FCB ECF cm

2

3

... (iv) (Using (iii))

Now, ∆DCB and parallelogram ABCD lie on same base

DC and between same parallel lines DC and AB.

So,

+

1

2

2

2

2 3 3

2 6 1

(

Area DCB Area parallelogram ABCD

Area parallelogram ABCD Area DCB

Area parallelogram ABCD

Area DFB Area FCB Using iv

Area parallelogram ABCD

2

2cm

Question: 8

In trapezium ABCD, AB || DC and L is the mid-point of

BC. Through L, a line PQ || AD has been drawn which

meets AB in P and DC produced in Q

(Fig.9.18). Prove that

Area (ABCD) = Area (APQD).

Fig. 9.18

Solution:

In

and ,CLQ BLP

LQC LPB

(Alternate interior angles

as

AB DC AB DQ

)

LCQ LBP

(Alternate interior angles)

CL=LB (L is the midpoint of BC)

Thus,

BLPCLQ

(By AAS congruence rule)

∴ Area (∆CLQ) = Area (∆BLP) ….. (i) (Congruent

triangles have equal areas)

Now,

Area (ABCD) = Area (APLCD) + Area (BLP)

= Area (APLCD) + Area (CLQ)

(Using (i))

= Area (APQD)

Hence, it is proved.

Question: 9

If the mid-points of the sides of a quadrilateral are

joined in order, prove that the area of the

parallelogram so formed will be half of the area of the

given quadrilateral (Fig. 9.19).

[Hint: Join BD and draw perpendicular from A on BD.]

Fig. 9.19

Solution:

Let ABCD be a quadrilateral and P, F, R and S are the

midpoints of BC, CD, DA and AB, respectively.

We need to prove that

Area Area

1

2

PFRS ABCD

Let us join AP and AC.

AP divides ∆ABC into two triangles of equal area.

1

( ) ( ) ..... ( )

2

ar ABar ABP C i

Similarly, the median PS divides ∆ABP into two

triangles of equal area.

1

( ) ( ) ..... ( )

2

arar BSP ABP ii

From eq. (i) and (ii), we get

1

( ) ( ) ..... ( )

4

ar ar ABS B iP C i i

Similarly,

1

( ) ( ) ... . (

4

. )ar ar ADCF iD vR

Adding (iii) and (iv), we have

( [ar ar(

( [Area ..... (v)

1

( ) ( ) )]

4

1

( ) ( )]

)

)

4

ar ar ABC ADC

ar ar ABCD

BSP DRF

BSP DRF

Similarly,

( [ara ..... (vi)r

1

(

)

4

)) ( ]ar AASR P DF BC C

Adding eq. (v) and (vi),

( ar(

ar +a

[Area ..... ii)

r

(v

( ) ( )

1

( )]

2

) )

BSP DRF ASR PCFar

ABCD

Now, adding the Area (SPFR) on both the sides, we

get

( (

[ar +ar(PFRS)

ar(ABCD)

+a

[ar

r

+ar(PFRS)

[ar = ar(PFRS)

( ) ( )

1

( )

) ) ( )

]

2

1

( )]

2

1

( )]

2

BSP DRF ASR PCFar ar ar

ABCD

ABCD

AB

ar PFRS

CD

Hence, it is proved.

EXERCISE 9.4

Question: 1

A point E is taken on the side BC of a parallelogram

ABCD. AE and DC are produced to meet at F. Prove

that Area (∆ADF) = Area (ABFC).

Solution:

Diagonal of a parallelogram divides it into two

triangles of equal area.

∴Area (∆ABC) = Area (∆ACD)....... (i)

Also,

Area (∆ABC) = Area (∆ABF)………..... (ii)

(Triangles on the same base AB and between the

same parallels AB and CF are equal in area)

From (i) and (ii), we have

Area (∆ACD) = Area (∆ABF) ... (iii)

Now,

Area (∆ADF) = Area (∆ACF) + Area (∆ACD)

⇒Area (∆ADF) = Area (∆ACF) + Area (∆ABF) (From

(iii))

⇒Area (∆ADF) =Area (quadrilateral ABFC)

Hence, it is proved.

Question: 2

The diagonals of a parallelogram ABCD intersect at a

point O. Through O, a line is drawn to intersect AD at

P and BC at Q. Show that PQ divides the

parallelogram into two parts of equal area.

Solution:

In

APO

and

CQO

,

POA COQ

(Vertically opposite angles)

OA = OC (Diagonals of parallelogram bisect each

other)

OCQPAO

(Alternate interior angles)

So,

POA QOC

(By ASA Congruence Rule)

∴Area (∆POA) = Area (∆QOC)......... (i)

The diagonal AC of parallelogram ABCD divides it into

two triangles of equal area.

∴Area (∆ABC) = Area (∆ADC) …….. (ii)

On adding eq. (i) and (ii), we get

2×Area (∆POA) + Area (∆ABC) = 2×Area (∆COQ) +

Area (∆ADC)

⇒Area (∆ABC) +Area (∆POA) +Area (∆POA) = Area

(∆ADC) + Area (∆COQ) + Area (∆COQ)

⇒Area (∆ABC) + Area (∆POA) − Area (∆COQ)

= Area (∆ADC) + Area (∆COQ) − Area (∆POA)

⇒Area (quadrilateral ABQP) = Area (quadrilateral

DCQP)

Hence, it is proved.

Question: 3

The medians BE and CF of a triangle ABC intersect at

G. Prove that the area of ∆GBC= area of the

quadrilateral AFGE.

Solution:

We know that a median of a triangle divides it into

two triangles of equal area.

As CF is the median of ∆ABC,

Area Area Area ( ) (

1

) .

2

BCF ACF ABC i

Also,

BE is a median of ∆ABC,

Area Area Area ( )

1

2

ABE BCE ABC ii

From eq. (i) and (ii), we have

Area Area

Area Area

Area Area

Area Area

( )

ABE BCF

BFG AFGE

BFG BGC

AFGE BGC

Hence, it is proved.

Question: 4

In Fig. 9.24,

||CD AE

and

||CY BA

. Prove that

Area (∆CBX) = Area (∆AXY).

Solution:

It is given that

| .|CY BA

Now, triangles on the same base and between the

same parallels are equal in area.

∴ Area (∆AYC) = Area (∆BCY)

(As both are on the same base CY and between the

same parallel lines CY and BA)

Area Area

Area Area

Area Area

( ) ( )

( ) ( )

( ) ( )

AXY CXY

BXC CXY

AXY BXC

Hence, it is proved.

Question: 5

ABCD is a trapezium in which

| ,|AB DC

30DC cm

and

50AB cm

. If X and Y are, respectively the

mid-points of AD and BC, prove that

7

( ) ( )

9

ar DCYX ar XYBA

Solution:

It is given that ABCD is a trapezium with

| ,|AB DC

30DC cm

and

50AB cm

.

Let us join DY and produce it to meet AB produced at

M.

In ∆DCY and ∆YBM,

CYD BYM

(Vertically opposite angles)

CY = BY (Y is the midpoint of BC)

DCY MBY

(Alternate opposite angles)

Thus

CYD BYM

(By ASA congruence rule)

So, DY = YM and DC = BM=30cm (By CPCT)

Now, in ∆DAM,

X and Y is the midpoints of AD and DM.

∴Using the midpoint theorem,

||XY AM

and

cm.

1 1 1 1

( ) (50 30) 80 40

2 2 2 2

XY AM AB BM

It is given that X and Y are, respectively the

mid-points of AD and BC.

∴Trapezium DCYX and ABYX are of same height, say h

cm.

Now,

Area of a trapezium

=

1

2

(Sum of parallel sides)×(Distance between them)

So, Area of trapezium DCYX =

1

2

(30+40)

×

h

=

1

2

×

70 h

Similarly, Area of trapezium ABYX =

1

2

(50+40)

×

h

=

1

2

×

90 h

Thus,

70 7

( ) 90 9

7

( )

9

h

ar

ar DCYX

ar D

ABYX h

ar ABYXCYX

Hence, it is proved.

Question: 6

In ∆ABC, L and Mare the points on AB and AC,

respectively such that

||LM BC

. Prove that

Area (LOB) = Area (MOC).

Solution:

It is given that

||LM BC

.

So, ∆BCL and ∆CBM are on the same base BC and

between the same parallel lines LM and BC).

Therefore,

ar ar

ar ar ar ar

ar ar

( ) ( )

(( ) ( ) ( ) )

( ) ( )

BCL CBM

LOB BOC MOC BOC

LOB MOC

Hence, it is proved.

Question: 7

In Fig. 9.25, ABCDE is any pentagon. BP drawn

parallel to AC meets DC produced at P and EQ drawn

parallel to AD meets CD produced at Q. Prove that

Area (ABCDE) = Area (APQ)

Fig. 9.25

Solution:

It is given that

BP AC

and

EQ AD

.

Now,

∆ABC and ∆APC lie on the same base AC and

between the same parallel lines AC and BP.

So, Area (∆ABC) = Area (∆APC) …. (i)

Similarly,

∆ADE and ∆ADQ lie on the same base AD and

between the same parallel lines AD and EQ.

So, Area (∆ADE) = Area (∆ADQ) …. (ii)

Now, Area (ABCDE) = Area (∆ABC) + Area (∆ACD) +

Area (∆ADE)

⇒Area (ABCDE) = Area (∆APC) + Area (∆ACD)

+Area (∆ADQ) (From (i) and (ii))

⇒Area (ABCDE) = Area (∆APQ)

Hence, it is proved.

Question: 8

If the medians of a ∆ABC intersect at G, show that

Area (AGB) = Area (AGC) = Area (BGC) =

1

3

Area (ABC)

Solution:

Let say AD, BE and CF are the medians of the given

triangle ABC.

It is given that the medians of ∆ABC intersect at G.

We know that, the median of a triangle divides it into

two triangles of equal area.

In ∆ABC, BE is the median.

Area (∆ABE) = Area (∆BEC)……….... (i)

In ∆GAC, GE is the median.

Area (∆GAE) = Area (∆GCE)……...... (ii)

Subtracting (ii) from (i), we get

Area (∆ABE) - Area (∆GAE) = Area (∆BEC) - Area

(∆GCE)

Area (∆AGB) = Area (∆BGC)……..… (iii)

Similarly, Area (∆AGB) = Area (∆AGC)……..... (iv)

From (iii) and (iv), we get

Area (∆AGB) = Area (∆BGC) = Area (∆AGC)…. (v)

Now, Area (∆AGB) + Area (∆BGC) + Area (∆AGC) =

Area (∆ABC)

⇒Area (∆AGB) + Area (∆AGB) + Area (∆AGB) = Area

(

∆

ABC) (Using (v))

⇒3Area (∆AGB) = Area (∆ABC)

⇒Area (∆AGB) =

1

3

Area (∆ABC)……….... (vi)

From (v) and (vi), we get

Area (∆AGB) = Area (∆BGC) =Area (∆AGC) =

1

3

Area (∆ABC)

Hence, it is proved.

Question: 9

In Fig. 9.26, X and Y are the mid-points of AC and AB

respectively, QP || BC and CYQ and BXP are straight

lines. Prove that Area (ABP)=Area(ACQ).

Fig. 9.26

Solution:

Let us join QB and PC.

It is given that X and Y are the mid-points of AC and

AB respectively.

Therefore, in ∆ABC,

XY BC

and

1

2

XY BC

….. (i)(Midpoint theorem)

Now, in ∆ABP, Y is midpoint of AB and

( ),

1

.

2

XY AP XY BC QP

XY AP ii

(By using the midpoint theorem)

From (i) and (ii),

AP= BC

Also,

AP BC

(∵

BC QP

is given)

Hence, quadrilateral ABCP is a parallelogram.

Now, BP is a diagonal of parallelogram ABCP and

diagonal of a parallelogram divides it into two

triangles of equal area.

So, Area (∆ABP) = Area (∆BPC) …. (iii)

Similarly, we can show that quadrilateral QBCA is

parallelogram and hence,

Area (∆AQC) = Area (∆QBC) …. (iv) [As QC is the

diagonal]

Now, Area (∆QBC) = Area (∆BPC)… (v)

(∵Both the triangles lie on the same base BC and

between the same parallel lines QP and BC)

From (iii), (iv) and (v), we have

Area (∆ABP) = Area (∆ACQ)

Hence, it is proved.

Question: 10

In Fig. 9.27, ABCD and AEFD are two parallelograms.

Prove that

Area (PEA) = Area (QFD) [Hint: Join PD].

Solution:

It is given that ABCD is a parallelogram.

So,

.AB CD AP DQ

Also, AEFD is a parallelogram.

AD EF AD EQ AD PQ

Now, in quadrilateral APQD,

and ADAP DQ PQ

So, APQD is a parallelogram.

Now, APQD and AEFD are two parallelograms that lie

on the same base AD and between the same parallel

lines AD and EQ.

So, Area (∥gm APQD) = Area (∥gm AEFD)

⇒Area (∆FQD) + Area (Trapezium APFD) = Area

(∆PEA) + Area (Trapezium APFD)

⇒Area (∆FQD) =Area (∆PEA)

Hence, it is proved.