Question

. O is a point on side PQ of a APQR such that PO = QO = RO, then
(a) RS² = PR × QR
(b) PR² + QR² = PQ²
(c) QR² = QO² + RO²
(d) PO² + RO² = PR²​

Answer 1

Step-by-step explanation:

1. O is a point on side PQ of a ∆PQR such that PO= QO=RO

then.

(I) RS²=PR×QR

(II)PR²+QR²=PQ²

(III)QR²=QO²+RO²

(IV)PO²+RO²=PR² the answer is d

Answer 2

Given:

O is a point on side PQ of a ΔPQR  

PO = QO = RO

In ΔPQR ,  

PO=OQ=RO (given)  

Now , in ΔPSR ,  

PO=SO (given)

∴∠1=∠P [Angles opposite to equal sides in a triangle are equal]

Similarly , in ∠ORQ ,

RO=OQ (given)  

∠Q=∠2

Now , in ΔPQR ,  

∠P+∠Q+∠PRQ=180°  [By Angle sum property of a triangle]

∠1+∠2+(∠1+∠2)=180°

2(∠1+∠2)=180°

∠1+∠2=90°

∠PRQ=90°

By Pythagoras theorem , we have

PR² + QR² = PQ²

The correct option is "b"

Hence the answer is PR² + QR² = PQ²