Question

In figure $ ∆ MNO$ is a right-angled triangle.

Its legs are $ 6 cm$ and $ 8 cm$ long.

Length of perpendicular $ NP$ on the side $ MO$ is

6 cm 8 cm

• A. $4.8cm$
• B. $3.6cm$
• C. $2.4cm$
• D. $1.2cm$

Answer 1

The correct option is A

$4.8cm$

Step 1: Given Data

$∆MNO$ is a right-angled triangle

$MN=6cm$ and $NO=8cm$

Step 2: Solve for the length of perpendicular $NP$

It is given that $△MNO$ is a right angle .

So by Pythagoras theorem

$$\begin{gathered} \Rightarrow M{O}^2=M{N}^2+N{O}^2 \\ \Rightarrow M{O}^2=6^2+8^2 \\ \Rightarrow MO=\sqrt{100} \\ \Rightarrow MO=10cm \end{gathered}$$

Area of $△MNO$,

$$\begin{gathered} \Rightarrow \frac{1}{2}\times NO\times NM=\frac{1}{2}\times MO\times NP \\ \Rightarrow 8\times 6=10\times NP \\ \Rightarrow NP=\frac{48}{10}=4.8 \end{gathered}$$

Hence the value of $NP$ is $4.8cm$ and therefore option (A) is correct.