Question

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​

Answer 1

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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