Question

In PQRS and ABRS are parallelogram and x is any point on side BR show that

Answer 1

ar(PQRS ) = ar(ABRS) (because they lie on the same base SR and between the same parallel lines ) .....1

ar(triangle AXS) =1/2 ar( ABRS ) ....2

From 1 and 2
Ar(triangle AXS) = 1/2 ar(PQRS)

Hence Proved.

Answer 2

Step-by-step explanation:

Given :

PQRS and ABRS are parallelograms and X is any point on side BR.

To prove :

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

(ii) ar (AXS) = 1/2 ar (PQRS)

Proof :

(i) In ∆ASP and BRQ,

we have ∠SPA = ∠RQB [Corresponding angles] ...(1)

∠PAS = ∠QBR [Corresponding angles] ...(2)

∴ ∠PSA = ∠QRB [Angle sum property of a triangle] ...(3)

Also, PS = QR [Opposite sides of the parallelogram PQRS] ...(4)

So, ∆ASP ≅ ∆BRQ [ASA axiom, using (1), (3) and (4)]

Therefore, area of ∆PSA = area of ∆QRB [Congruent figures have equal areas] ...(5)

Now, ar (PQRS) = ar (PSA) + ar (ASRQ] = ar (QRB) + ar (ASRQ] = ar (ABRS)

So, ar (PQRS) = ar (ABRS) Proved ✔

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS and between same parallels AS and BR

∴ area of ∆AXS = 1/ 2 area of ABRS

⇒ area of ∆AXS = 1/ 2 area of PQRS [ ar (PQRS) = ar (ABRS]

⇒ ar of (AXS) = 1/ 2 ar of (PQRS) Proved ✔

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