Question

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

Answer 1

In $\Delta$ABC, AD$=6$cm and BC$=9$cm

Area of triangle$=\frac{1}{2}\times$base$\times$height$=\frac{1}{2}\times$BC$\times$AD

$=\frac{1}{2}\times 9\times 6=27$ $c{m}^2$

Again, Area of triangle$=\frac{1}{2}\times$base$\times$height$=\frac{1}{2}\times$AB$\times$CE
$\Rightarrow 27=\frac{1}{2}\times 7.5\times$CE

$\Rightarrow$ CE$=\frac{27\times 2}{7.5}$

$\Rightarrow$ CE$=7.2$cm

Thus, height from C to AB i.e., CE is $7.2$cm.