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

Solution (By Examveda Team)

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