Question

In the given figure, what is the area of the triangle ACD?

• A. $2m^2$
• B. $\sqrt{2}{m}^2$
• C. $1m^2$
• D. $2\sqrt{2}{m}^2$

Answer 1

The correct option is C

$1m^2$

To find the area of the $△$ ACD, we need to find the length of the side AC.

ABC is a right-angled isosceles triangle.

The sides are in the ratio 1 : 1 : $\sqrt{2}$.

Let the sides be x, x , $\sqrt{2}$x.

Here x = $\sqrt{2}$

$\therefore$ The Length of AC = $\sqrt{2}$x = $\sqrt{2}\times \sqrt{2}$ = 2 m

Area of $△$ ACD = $\frac{1}{2}\times Base\times Height$ (Base = CD, Height = AC)

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

= $1m^2$