Question

The triangle BDC is right angled. If BD measures $2\sqrt{2}$ cm, what is the area of the square?

• A. 2 $c{m}^2$
• B. $2\sqrt{2}c{m}^2$
• C. 4 $c{m}^2$
• D. $\sqrt{2}c{m}^2$

Answer 1

The correct option is C

4 $c{m}^2$

BC = CD ( Sides of the square)

Since BDC is a right-angled isosceles triangle,

BC : CD : BD = 1 : 1 : $\sqrt{2}$

Let BC be x

BD = $\sqrt{2}$x

$\sqrt{2}$x = $2\sqrt{2}$

x = 2

Area of the square = $Sid{e}^2$ = $2^2$ = 4 $c{m}^2$