Question

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​

Answer 1

Answer:

4) is correct

please mark me as brainliest

Answer 2

Given:

A ∆PQR

A point O on PQ such that PO= QO=RO (refer figure)

To find:

Relation between the square of sides of the triangle

Solution:

We know that angles opposite to equal sides are equal,

So in ∆ POR, ∠1 = ∠2       -(1)

Similarly, in ∆QOR, ∠3 = ∠4       -(2)

Also, since the sum of all angles of a triangle = 180°

In ∆PQR, ∠P + ∠Q + ∠R =180°

OR ∠P + ∠Q + (∠1 + ∠3) = 180°

Substituting from (1) and (2),

∠1 + ∠3 + (∠1 + ∠3) = 180°

or  2 (∠1 + ∠3) = 180°

or ∠1 + ∠3 = 90°

⇒ ∠R = 90° i.e ΔPQR is a right angled triangle

Applying Pythagoras Theorem in ΔPQR,

PQ² = PR²+QR²

Hence, option (ii) )PR²+QR²=PQ² is correct