Question

In the given figure, $\angle PQR={90}^o$. T is the midpoint of side QR.

Prove that $P{R}^2=4P{T}^2-3P{Q}^2$.

Answer 1

As per Pythagorean theorem
$P{T}^2=P{Q}^2+Q{T}^2$....(in triangle QPT)

$\Rightarrow Q{T}^2=P{T}^2-P{Q}^2$

And $P{R}^2=P{Q}^2+Q{R}^2$ ( in the triangle APR and given T is the mide point QR then QR=2QT) put value

$\Rightarrow P{R}^2=P{Q}^2+(2QT{)}^2$

$\Rightarrow P{R}^2=P{Q}^2+4Q{T}^2$

$\Rightarrow P{R}^2=P{Q}^2+4(P{T}^2-P{Q}^2)$

$=P{Q}^2+4P{T}^2-4P{Q}^2$

$\Rightarrow P{R}^2=4P{T}^2-3P{Q}^2$.