Question

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

• A. 35
• B. 45
• C. 25
• D. 32

Answer 1

The correct option is C

25

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

AB=BC

So,$\theta$ = $\alpha$

$\cos$ $\theta$ = $\frac{1}{\sqrt{2}}$ at $\theta$ =${45}^{\circ }$

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

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

AB=$5\sqrt{2}$ cm

And $\sin \theta$ =$\frac{1}{\sqrt{2}}$ at $\alpha$ =${45}^{\circ }$

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

BC=$5\sqrt{2}$cm

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

=$\frac{1}{2}$($5\sqrt{2}$ $\times$ $5\sqrt{2}$)

=25 $c{m}^2$