Question

In ∆PQR, PT QR such that ∆PQT ~ ∆RQP then find QRP​

Answer 1

Answer:

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.

(I) Prove ΔPQR∼Δ SPR

(II) Find the length of QR and PS

(III)

areaofΔSPR

aareaofΔPQ.

Answer 2

In ∆PQR ,∠PQT=∠PQR   where  PT bisects ∠QPR

Step-by-step explanation:

In triangle PQR

given that - PT  is perpendicular to QR

such that triangle PQT ~ triangle RQP

then angle QPR is to be determined.

In Δ PQR,

PT⊥QR and ΔPQR≈ΔRQP

∠PTR=∠PTQ (each angle is 90°)

PQ=PR (Given)

side PQ= side QR (Common side)

Thus, △PQT≅△RQP

Hence, ∠PQT=∠PQR   where  PT bisects ∠QPR

If PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR

(I  In △PQR and △SPR

∠PRQ=∠SRP  (Common angle)

∠QPR=∠PSR   (Given that)

∠PQR=∠PSR   (Properties of triangle )

∴△PQR∼△SPR  (By AAA).