Question

In the given right angled isosceles triangle ABC, if $AC=10cm$ then the area of triangle is ___ $c{m}^2$

Answer 1

In right angled isosceles triangle ABC, $\angle B={90}^{\circ }$

$AB=BC$

So, $\theta =\alpha$ [angles opposite to equal sides]

Also, $\theta =\alpha ={45}^{\circ }$ [Using angle sum property]

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

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

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

$\Rightarrow AB=5\sqrt{2}cm$

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

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

$\Rightarrow BC=5\sqrt{2}cm$

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

$=\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}=25c{m}^2$