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 ∠QPR = 90°, then QR = ?


(a) 28 cm
(b) 25 cm
(c) 26 cm
(d) 32 cm

Answer 1

(c) 26 cm

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

PR = 10 cm

Now, applying Pythagoras theorem in right-angled triangle QPR, we get:

$$\begin{gathered} Q{R}^2=P{Q}^2+P{R}^2 \\ ={24}^2+{10}^2=576+100=676 \\ QR=\sqrt{676}=26\mathrm{cm} \end{gathered}$$