A
7. In APQR, PQ > PR; Let's cut off the line segment PS equal to the length of PR
from the side PQ. Let's join two points R and S. Let's prove that
(1) ZPSR = 3(ZPQR+ ZPRQ (11) ZQRS = 3 (ZPRQ - ZPQR) I
of BAS meets BC at D. Let's cut off a line
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Answer:
Given that:
QP=8cm,PR=6cm and SR=3cm
(I) In △PQR and △SPR
∠PRQ=∠SRP (Common angle)
∠QPR=∠PSR (Given that)
∠PQR=∠PSR (Properties of triangle )
∴△PQR∼△SPR (By AAA)
(II)
SP
PQ
=
PR
QR
=
SR
PR
(Properties of similar triangles)
⇒
SP
8cm
=
3cm
6cm
⇒SP=4cm and
⇒
6cm
QR
=
3cm
6cm
⇒QR=12cm
(III)
ar(△SPR)
ar(△PQR)
=
SP
2
PQ
2
=
4
2
8
2
=4
Step-by-step explanation:
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