Question

Help me in solving question number 2

Answer 1

since ΔPQR is right <s Δ
and PQ=QR=x(say)
by Pythagoras Theorem
PR^2=PQ^2+QR^2
(√28)^2=x^2+x^2
28=2x^2
14=x^2
x=√14
PQ=QR=√14