Question

In the given figure, O is a point inside a $\Delta PQR$ such that $\angle POR={90}^o,OP=6cm$ and $OR=8cm$. If $PQ=24cmandQR=26cm$, prove that $\Delta PQR$ is right - angled.

Answer 1

In ΔPQR, $\angle QPR={90}^o$, PQ = 24 cm, and QR = 26 cm

In ΔPOR, PO = 6 cm, QR = 8 cm and $\angle POR={90}^o$

In ΔPOR,

$P{R}^2=P{O}^2+O{R}^2$

$P{R}^2=(6^2+8^2)c{m}^2=(36+64)c{m}^2=100c{m}^2$

$PR=\sqrt{100}cm=10cm$

In ΔPQR,

By Pythagoras theorem, we have

$Q{R}^2=Q{P}^2+P{R}^2$

$(26{)}^{2}c{m}^2=({24}^2+{10}^2)c{m}^2$

$676c{m}^2=(576+100)c{m}^2$

$676c{m}^2=676c{m}^2$

Hence, $Q{R}^2=Q{P}^2+P{R}^2$

(Sum of square of two sides equal to square of the greatest side)

Hence, ΔPQR is a right triangle which is right angled at P.