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

(i) ar (PQRS) = ar (ABRS) 

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

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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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