Question

PA
QB
5. In Fig. 9.17, PQRS and ABRS are parallelograms
and X is any point on side BR. Show that
(i) ar (PQRS)=ar (ABRS)
X
1
(ii) ar (AXS) =
ar (PQRS)
S
R​

Answer 1

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In figure, PQRS and ABRS are parallelogram and X is any point on side BR. Show that

(i) ar(PQRS)=ar(ABRS)

(ii) ar(AXZ)=

2

1

ar(PQRS)

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Solution

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(i) Since PQRS is a parallelogram

PQ∣∣RX [Opposite sides of parallelogram are parallel]

Since ABRS is a parallelogram

AB∣∣RS [Opposite sides of parallelogram are parallel]

Since PQ∣∣RS & AB∣∣RS

We can say that PB∣∣RS

Now,

PQRS & ABRS are two parallelograms with the same base RS and between the same parallels PB and RS

∴ar(PQRS)=ar(ABRS) [Parallelogram with same base and between the same parallels are equal in area]

(ii) Since ABRS is a paralleogram

AS∣∣BR [Opposite sides of parallelogram are parallel]

△AXS and parallelogram ABRS lie on the same base AS and are between the same parallel lines AS and BR

∴Area(△AXS)=

2

1

Area(ABRS) [Area of triangle is half of parallelogram if they have the same base and parallels]

We proved in part (i)

ar(PQRS)=ar(ABRS)

∴Area(△AXS)=

2

1

Area(PQRS)

Hence proved.