(a) In the figure (1) given below, ABCD is a parallelogram. P, Q are any two
points on the sides AB and BC respectively. Prove that, area of ∆ CPD = area of ∆
AQD.
(b) In the figure (2) given below, PQRS and ABRS are parallelograms and X is any
the point on the side BR. Show that area of ∆ AXS = ½ area of ||gm PQRS.
(a) Given:
From fig (1)
||gm ABCD in which P is a point on AB and Q is a point on BC.
To prove:
area of ∆ CPD = area of ∆ AQD.
Proof:
∆ CPD and ||gm ABCD are on the same base CD and between the same parallels AB and
CD.
ar (∆ CPD) = ½ ar (||gm ABCD) …. (1)
∆ AQD and ||gm ABCD are on the same base AD and between the same parallels AD and
BC.
ar (∆AQD) = ½ ar (||gm ABCD) …. (2)
from (1) and (2)
ar (∆ CPD) = ar (∆AQD)
Hence proved.
(b) From fig (2)
Given:
PQRS and ABRS are parallelograms on the same base SR. X is any point on the side BR.
Join AX and SX.
To prove:
area of ∆ AXS = ½ area of ||gm PQRS
we know that, || gm PQRS and ABRS are on the same base SR and between the same
parallels.
So, ar ||gm PQRS = ar ||gm ABRS …. (1)
we know that ∆ AXS and || gm ABRS are on the same base AS and between the same
parallels.
So, ar ∆ AXS = ½ ar ||gm ABRS
= ½ ar ||gm PQRS [From (1)]
Hence proved.
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