Question

Question 8

$\Delta$ 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 = $\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times BC\times AD$
$=\frac{1}{2}\times 9\times 6=27c{m}^2$
Again, area of triangle = $\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times AB\times CE$
$\Rightarrow 27=\frac{1}{2}\times 7.5\times CE$
$\Rightarrow CE=\frac{27\times 2}{7.5}=7.2cm$
Thus, height from C to AB i.e. CE is 7.2 cm.