Question

● Proper Image✔

● No Spam❌

○ Other Wise Reported ​

Answer 1

$\Large\bf\blue{In\:\triangle{PQR}\:\&\:\triangle{QST}\:}$

$\longmapsto\:\:\bf\red{\angle{PRQ}\:=\:\angle{STQ}\:=\:90°}$

$\longmapsto\:\:\bf\pink{\angle{PQR}\:=\:\angle{SQT}\:}(equal)$

$:\implies\:\:\bf\purple{\triangle{PQR}\:\sim\:\triangle{QST}\:}$

$\longrightarrow\:\:\bf{\dfrac{QR}{QT}\:=\:\dfrac{QP}{QS}\:}$

$\longrightarrow\:\:\bf\orange{QR\times{QS}\:=\:QP\times{QT}\:}(Proved)$

Answer 2

Step-by-step explanation:

Given:

Figure given below in attachment.

ΔPQR,

∠R = 90°

ΔSQT,

∠T = 90°

To Prove That:

QR * QS = QP * QT

Proof:

In ΔPQR and ΔSQT,

∠PQR = ∠SQT   (Common)

∠PRQ = ∠STQ  (Right Angles)

ΔPQR ≈ ΔSQT (By A.A Similarity)

$\frac{QR}{QT} =\frac{QP}{QS} (C.P, S.T)$

Cross Multiplying we get,

QR * QS = OP * QT

#Hence Proved