# Ex.6.5 Q3 The Triangle and Its Properties - NCERT Maths Class 7

## Question

A \(\rm{}15\, m\) long ladder reached a window \(\rm{}12\, m\) high from the ground on placing it against a wall at a distance \(a\). Find the distance of the foot of the ladder from the wall.

## Text Solution

**What is known ?**

Length of ladder is \(\rm{}15\, m\) and the height of the window from the ground is \(\rm{}12\,m.\)

**What is unknown?**

The distance of the foot of the ladder from the wall.

**Reasoning:**

This question is based on the concept of right – angled triangle. As it is clear from the figure

that the ladder is kept slanted on the wall so, consider length of the ladder as hypotenuse

i.e.: \(AB =\rm{}15\,m\) and as it is kept slanted on the wall so, we consider wall as perpendicular

i.e.: \(AC = \rm{}12\, m.\) Now, you must find out the distance of the foot of the ladder from the wall

i.e.: \(BC = a.\) Now by applying Pythagoras theorem in triangle \(ABC,\) we can find out \(BC. \)

For better visual understanding draw a right-angled triangle consider ladder as hypotenuse, wall as perpendicular and distance between the foot of the ladder and wall as base.

**Steps:**

Given, Length of ladder \(AB = \rm{}15\, m \)

Length of wall \(AC = \rm{}12\, m\)

To find \((BC) =\) distance of the foot of the ladder from the wall.

According to Pythagoras theorem,

\(\begin{align}&{{\left(\text{Hypotenuse} \right)}^2} \\ &= {\rm{ }}{{\left(\text {Perpendicular} \right)}^2} + {\rm{ }}{{\left( \text{Base} \right)}^2} \end{align} \)

\(\begin{align} {{\left( {AB} \right)}^2} &= {{\left( {AC} \right)}^2} + {{\left( {BC} \right)}^2}\\{{\left( {15} \right)}^2} &={{\left( {12} \right)}^2} + {{\left( a \right)}^2}\\225 &= 144 + {{\left( a \right)}^2}\\225-144 &= {{\left( a \right)}^2}\\81 &= {a^2}\\a &= 9 m\end{align}\)

Therefore, the distance of the foot of the ladder from the wall is \(\rm{}9 \,m. \)

**Useful Tip:**

Whenever you encounter problem of this kind, it is better to understand its visually.