Question

In triangle PQR, PQ = 24 cm, QR = 7 cm and <PQR = 90°. Find the radius of the inscribed circle. [4 marks]

Answer 1

In triangle PQR,

$\angle \mathrm{PQR}=90°\mathrm{Therefore},P{R}^2=P{Q}^2+Q{R}^2\Rightarrow P{R}^2={24}^2+7^2=625{\mathrm{cm}}^2\Rightarrow PR=25cm\left[1\mathrm{mark}\right]$

Let the given inscribed circle touches the sides of the given triangle at points A, B and C respectively as shown.
Clearly, OAQB is a square and so
AQ = BQ = x cm,
PA = PQ - AQ = (24 - x) cm and [0.5 marks]
RB = QR - BQ = (7 - x) cm [0.5 marks]
As tangents to a circle, from an exterior point arc equal,
PC = PA = (24 - x) cm [0.5 marks]
and RC = RB = (7 - x) cm [0.5 marks]
PR = PC + RC = 25 = (24 - x) + (7 - x)
$\Rightarrow$25 = 31 - 2x
$\Rightarrow$2x = 6
$\Rightarrow$x = 3 [1 mark]
Radius of the inscribed circle = 3 cm.