Question

plzz. answer ques no 16

Answer 1

You should use properties of triangle.
Two formula , here we have to use
1. area of ∆ABC = $\bold{\frac{1}{2}absinC}$
2. CosC = $\bold{=\frac{a^2+b^2-c^2}{2ab}}$

Now come to the question ,
Question said A ∆PQR is given where ∠PQR = 45°
so, from formula no - 1
area of ∆PQR {or, ar(∆PQR)} = 1/2.PQ.QR sin∠PQR = 1/2PQ.QRsin45°
⇒ar(∆PQR) = 1/2.PQ.QR × 1/√2 = 1/2√2 .PQ.QR
⇒1/√2 = 2ar (∆PQR)/PQ.QR --------(1)

Now, apply formula no -2 for angle ∠PQR
cos∠PQR = (PQ² + QR² - PR²)/2.PQ.QR
⇒cos45° = (PQ² + QR² - PR²)/2.PQ.QR
⇒1/√2 = (PQ² + QR² - PR²)/2.PQ.QR

Now put equation (1) here,
2.ar (∆PQR)/PQ.QR = (PQ² + QR² - PR²)/2.PQ.QR
⇒4ar(∆PQR) = PQ² + QR² - PR²
⇒ PR² = PQ² + QR² - 4ar(∆PQR) , hence proved //