Question

In the given figure, O is a point inside a ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm and QR = 26 cm, prove that ∆PQR is right-angled.

Answer 1

Applying Pythagoras theorem in right-angled triangle POR, we have:
$$\begin{gathered} {\mathrm{PR}}^2={\mathrm{PO}}^2+{\mathrm{OR}}^2 \\ \Rightarrow {\mathrm{PR}}^2=6^2+8^2=36+64=100 \\ \Rightarrow \mathrm{PR}=\sqrt{100}=10\mathrm{cm} \end{gathered}$$

In ∆ PQR,

$$\begin{gathered} {\mathrm{PQ}}^2+{\mathrm{PR}}^2={24}^2+{10}^2=576+100=676 \\ \mathrm{and}{\mathrm{QR}}^2={26}^2=676 \\ \therefore {\mathrm{PQ}}^2+{\mathrm{PR}}^2={\mathrm{QR}}^2 \end{gathered}$$

Therefore, by applying Pythagoras theorem, we can say that ∆PQR is right-angled at P.