# Ex.6.5 Q9 Triangles Solution - NCERT Maths Class 10

## Question

A ladder \(10\)\(\rm{}m\) long reaches a window \(8\)\(\rm{}m\) above the ground. Find the distance of the foot of the ladder from base of the wall.

**Diagram**

## Text Solution

**Reasoning:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

**Steps:**

\(AB\) is height of the windows from the ground = \(8\)\(\rm{}m\)

\(AC\) is the length of the ladder = \(10\)\(\rm{}m\)

\(BC\) is the foot of the ladder from the base of ground \(= ?\)

Since \(\Delta ABC\) is right angled triangle \(\,(\angle ABC={{90}^{0}})\)

\[\begin{align} \Rightarrow B C^{2} &=A C^{2}-A B^{2} \text { (Pythagoras theorem) } \\ B C^{2}&=10^{2}-8^{2} \\ B C^{2}&=100-64 \\ B C^{2} &=36 \\ B C &=6 \rm{}m \end{align}\]

The distance of the foot of the ladder from the base of the wall =\(6\)\(\rm{}m\)