Question

PQRS and ABRS are parallelogram and x is any point on br show that 1) area(PQRS)=area(ABRS)
2) area(AXS)=1/2 area(PQRS)

Answer 1

Answer:

Step-by-step explanation:

Given PQRS and ABRS are parallelogram and X is any point on BR. we have to prove that  

1) $area(PQRS)=area(ABRS)$

2) $area(AXS)=\frac{1}{2}area(PQRS)$

In ∆ASP and ΔBRQ  

∠SPA = ∠RQB [Corresponding angles]

∠PAS = ∠QBR [Corresponding angles]

PS = QR [Opposite sides of the parallelogram PQRS]

By AAS rule, ∆ASP≅ΔBRQ  

∴ ar(ASP) = ar(BRQ)      (Congruent triangles have equal area)

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

                           = ar(QRB) + ar(ASRQ]  = ar(ABRS)  

So, ar(PQRS)=ar(ABRS)

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

$ar(AXS)=\frac{1}{2}ar(ABRS)$  

⇒ $ar(AXS)=\frac{1}{2}ar(PQRS)$   (∵ar(PQRS)=ar(ABRS))  

Hence Proved.