Question

PQRST is a pentagon. TX is drawn parallel to SP which meets PQ produced at X. RY is drawn parallel to SQmeets PQproduced at Y.Show that ar(PQRST)=ar of (SXY

Answer 1

this is the answer okay...........

Answer 2

Concept we will be using:
(i)Triangles on the same base and between same parallels are equal in area.

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Given: PQRST is pentagon. TX || SP And RY || SQ (Please refer to the attachment)

RTP: ar(PQRST) = ar(SXY)

Proof:
ΔSPX and ΔSPY  are triangles on the same base SP and between same parallels TX and SP because TX || SP

Therefore, ar(ΔSPX) =ar(ΔSPT)
 

ΔSPX and ΔSPT  are triangles on the same base SP and between same parallels TX and SP (Given: TX || SP)

Therefore, ar(ΔSPX) =ar(ΔSPT)


ΔSQY and ΔSQR  are triangles on the same base SQ and between same parallels RY and SQ (Given: RY || SQ)

Therefore, ar(ΔSQY) =ar(ΔSQR)

And,
ar(ΔSXY)
=ar(ΔSPX) + ar(ΔSQY) + ar(ΔSPQ)


ar(PQRST)
= ar(ΔSPT) + ar(ΔSQR)+ ar(ΔSPQ)
Plug in ar(ΔSPT)=ar(ΔSPX) and  ar(ΔSQR)=ar(ΔSQY):
=ar(ΔSPX) + ar(ΔSQY) + ar(ΔSPQ)
=ar(ΔSXY)


Therefore, ar(PQRST)  = ar(SXY)  (Proved)