Question

What is the Area of the triangle ACD?

• A. $3\sqrt{2}{m}^2$
• B. $2m^2$
• C. $2\sqrt{2}{m}^2$
• D. $2\sqrt{3}{m}^2$

Answer 1

The correct option is C

$2\sqrt{2}{m}^2$

In the given figure ABC is a right-angled isosceles triangle

The sides will be in the ratio 1 : 1 : $\sqrt{2}$

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

Here, x = 2 m

$\therefore$ AC = $\sqrt{2}$x = $2\sqrt{2}$ m

Now, in $△$ ACD,

Height AC = $2\sqrt{2}$ m

Base CD = 2 m

$\therefore$ Area of $△$ ACD = $\frac{1}{2}\times Base\times Height$

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

= $2\sqrt{2}{m}^2$