Question

∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 9.17). The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?

Answer 1

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

$=\frac{1}{2}\times BC\times AD$

$\frac{1}{2}\times 9\times 6=27\;cm^2$

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

$27=\frac{1}{2}\times7.5\times CE$

$CE =7.2\;cm$