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

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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 \(AD\).

 Video Solution
Perimeter And Area
Ex 11.2 | Question 7

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  

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

  
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