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kingdede8811
23.06.2019 •
Mathematics
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)
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Ответ:
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)](/tpl/images/0007/2668/62c41.png)
2)![area(AXS)=\frac{1}{2}area(PQRS)](/tpl/images/0007/2668/e7777.png)
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(PQRS)=ar(ABRS))
Hence Proved.
Ответ:
Thank you!!
Step-by-step explanation:
:))