In a triangle PQR, ∠PQR = 90°, PQ = 10 cm and PR = 26 cm, then what is the value (in cm) of inradius of incircle?

Answer: Option B

Solution

Answer: Option B

Solution:

$$\begin{array}{rl}& QR=\sqrt{{\left(26\right)}^2-{\left(10\right)}^2}\\ & =\sqrt{676-100}\\ & =\sqrt{576}\\ & =24\\ & \text{Incirde radius in a right angled triangle }\left(r\right)\\ & =\frac{\text{P}+\text{B}-\text{H}}{2}\\ & =\frac{10+24-26}{2}\\ & =\frac{8}{2}\\ & =4\text{ cm}\end{array}$$