S
9. PQRST is a pentagon
in which TF, QG and
SH are perpendiculars
drawn on the diagonal
PR from the points T, Q
and S respectively. The
following measurements
are given: PR = 12 cm, PH = 8 cm, PG = 5 cm,
PF = 3 cm, TF = 4 cm, SH = 6 cm, QG = 3.5 cm.
Find the area of the pentagon.
Answers
Answer:
Triangles on the same base and between same parallels are equal in area.
Step-by-step explanation:
From the above question,
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)
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