Question

$△ABC$ is isosceles with $AB=AC=7.5cm$ and $BC=9cm$. The height $AD$ from $A$ to $BC$, is $6cm$. Find the area of $△ABC$. What will be the height from $C$ to $AB$ i.e.,$CE$?

Answer 1

Given:

$△ABC$ is an isosceles triangle with $AB=AC=7.5cm$.

$BC=9cm$

$AD\perp BC$ and $CE\perp AB$

Height from $A$ to $CD=AD=6cm$

Height from $C$ to $AB=CE=?$

Since, Area of the triangle $=\frac{1}{2}\times \text{Base}\times \text{Corresponding height}$

$Rightarrow$ Area of the triangle $ABC=\frac{1}{2}\times BC\times AD$ (or) $\frac{1}{2}\times AB\times CE$

$\Rightarrow \frac{1}{2}\times BC\times AD=\frac{1}{2}\times AB\times CE$

$\Rightarrow BC\times AD=AB\times CE$

$\Rightarrow \left(9cm\right)\times \left(6cm\right)=\left(7.5cm\right)\times CE$

$\Rightarrow 54c{m}^2=\left(7.5cm\right)\times CE$

$\Rightarrow \frac{54c{m}^2}{7.5cm}=CE$

$\Rightarrow 7.2cm=CE$

i.e., Height from $C$ to $AB$ is $7.2cm$