Question

In the given right angled isosceles triangle ABC, if AC = 10 cm then determine the area of triangle in square cm.

• A. 25

Answer 1

The correct option is A 25

$\text{In triangle ABC,}\angle ABC={90}^o.$
$AB=BC\text{(Since, ABC is an isosceles triangle)}$

$\text{So,}\theta =\alpha ={45}^o$

$\cos \theta =\frac{1}{\sqrt{2}}$

$\frac{AB}{AC}=\frac{1}{\sqrt{2}}$

$AB=\frac{AC}{\sqrt{2}}$

$AB=5\sqrt{2}cm$

$\text{And,}\sin \theta =\frac{1}{\sqrt{2}}$

$\frac{BC}{AC}$=$\frac{1}{\sqrt{2}}$

$BC=5\sqrt{2}cm$

$\text{Area of triangle}=\frac{1}{2}(\text{Base}\times \text{Height})=\frac{1}{2}(5\sqrt{2}\times 5\sqrt{2})=25c{m}^2$