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

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\(\triangle ABC\) 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., \(CE\,?\)

 Video Solution
Perimeter And Area
Ex 11.2 | Question 8

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., \(CE.\)


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.


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

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

\[\begin{align}& = \frac{1}{2} \times {\rm{BC}} \times {\rm{AD}}\\&= \frac{1}{2} \times 9 \times 6\\&=27\;c{m^2}\end{align}\]


Area of triangle \(ABC\)  \(=\) \(\frac{1}{2}\) \(AB\)  \(\times\) \(CE \)

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

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