The base and altitude of a right-angled triangle are 12 cm and 5 cm respectively. The perpendicular distance of its hypotenuse from the opposite vertex is :

Answer: Option B

Area of the triangle :
$$\begin{array}{rl}& =\left(\frac{1}{2}\times 12\times 5\right)c{m}^2\\ & =30c{m}^2\end{array}$$
Hypotenuse :
$$\begin{array}{rl}& =\sqrt{{12}^2+5^2}cm\\ & =\sqrt{169}cm\\ & =13cm\end{array}$$
Let the perpendicular distance of the hypotenuse from the opposite vertex be x cm
Then,
$$\begin{array}{rl}& \Rightarrow \frac{1}{2}\times 13\times x=30\\ & \Rightarrow x=\frac{60}{13}\\ & \Rightarrow x=4\frac{8}{13}cm\end{array}$$