Question

Question 17

In the given figure, $\Delta MNO$ is a right-angled triangle. Its legs are 6 cm and 8 cm long. Length of perpendicular NP on the side MO is


(a) 4.8 cm
(b) 3.6 cm
(c) 2.4 cm
(d) 1.2 cm

Answer 1

Given, $\Delta MNO$ is a right angled triangle.
So, according to Pythagoras theorem,
$M{O}^2=M{N}^2+N{O}^2=6^2+8^2=36+64$
$\Rightarrow M{O}^2=100\Rightarrow MO=\sqrt{100}$
$\Rightarrow MO=10cm$
$\therefore Areaof\Delta MNO=\frac{1}{2}\times Base\times Height$
$\Rightarrow \frac{1}{2}\times MN\times NO=\frac{1}{2}\times MO\times NP$
$\Rightarrow \frac{1}{2}\times 6\times 8=\frac{1}{2}\times 10\times NP$
$\Rightarrow NP=\frac{24}{5}$
$\Rightarrow NP=4.8cm$