# Ex.11.2 Q7 Perimeter and Area - NCERT Maths Class 7

## Question

\(\triangle ABC\) is right angled at\(\, A \,\) (Fig 11.25). \(AD\) is perpendicular to \(BC\). If \(AB = 5 \,\rm cm\), \(BC = 13 \,\rm cm\) and \(AC = 12 \,\rm cm\), Find the area of \(\triangle ABC\). Also find the length of \(AD\).

## Text Solution

**What is known?**

\(\triangle ABC\) is right angled at \(A\), \(AD\) is perpendicular to side \(BC\). Length of three sides of the triangle is also given.

**What is unknown?**

The area of \(\triangle ABC\) and the length of perpendicular \(AD\).

**Reasoning:**

First, find the area of triangle using base \((5 \,\rm cm)\) and height \((12 \,\rm cm)\). Now, the area of the triangle is known, and the base is given as \(BC = 13 \,\rm cm\) and you must find the height \(AD\) by using the formula of area of triangle.

**Steps:**

In right angled \(\triangle BAC\),

\(AB = 5 \,\rm cm\), \(BC = 13 \,\rm cm\) and \(AC = 12 \,\rm cm\).

Area of triangle \(BAC\) \(=\) \(\frac{1}{2}\) Base \(\times\) Height

\[\begin{align}&= \frac{1}{2} \times 5 \times 12\\&= 30\;\rm{c{m^2}}\end{align}\]

Area of triangle \(ABC\) \(=\) \(\frac{1}{2}\) Base \(\times\) Height

\[\begin{align}30 &= \frac{1}{2} \times 13 \times {\rm{AD}}\\{\rm{AD}} &= \frac{{30 \times 2}}{{13}}\\{\rm{AD}} &= \frac{{60}}{{13}}\\{\rm{AD}} &= 4.61\;cm\end{align}\]