# Ex.11.2 Q8 Perimeter and Area - NCERT Maths Class 7

## Question

\(\triangle ABC\) is isosceles with \(AB = AC = 7.5 \,\rm cm\) and \(BC = 9 \,\rm cm\) (Fig 11.26). The height \(AD\) from \(A\) to \(BC\), is \(6 \,\rm cm\). Find the area of \(\triangle ABC\). What will be the height from \(C\) to \(AB\) i.e., \(CE\,?\)

## Text Solution

**What is known?**

\(\triangle ABC\) is isosceles with \(AB = AC = 7.5 \,\rm cm\) and \(BC = 9 \,\rm cm\). The height \(AD\) from \(A\) to \(BC,\) is \(6 \,\rm cm.\)

**What is unknown?**

The area of \(\triangle ABC\) and the height from \(C\) to \(AB\) i.e., \(CE.\)

**Reasoning:**

First, find the area of the given triangle using base as \(9 \,\rm cm\) and height as \(6\,\rm cm.\) Now, the area of triangle is known, take \(AB = 7.5 \,\rm cm\) as the base and find height \(CE\) using the formula of area of triangle.

**Steps:**

In \(\triangle ABC\), \(AD = 6 \,\rm cm\) and \(BC = 9\,\rm cm\)

\[\begin{align}{\text{Area of triangle ABC}} &= \frac{1}{2} \times {\rm{Base}} \times {\rm{Height}}\\& = \frac{1}{2} \times {\rm{BC}} \times {\rm{AD}}\\&= \frac{1}{2} \times 9 \times 6\\&=27\;c{m^2}\end{align}\]

Now,

\[\begin{align}{\text{Area of triangle ABC}} &= \frac{1}{2} \times {\rm{AB}} \times {\rm{CE}}\\

{\rm{27}} &= \frac{1}{2} \times {\rm{7}}{\rm{.5}} \times {\rm{CE}}\\{\rm{CE}} &= \frac{{2 \times {\rm{27}}}}{{{\rm{7}}{\rm{.5}}}}\\{\rm{CE}} &= \frac{{54}}{{{\rm{7}}{\rm{.5}}}}\\{\rm{CE}} &= 7.2\;\rm{cm}\end{align}\]