Question

ADC is a right angled isosceles $△$ with right angle at D. What should be the area of $△$ AFE?

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

Answer 1

The correct option is C

$1squnits$

To find the Area of triangle AFE, we need to know AF and FE which are sides of the square ACEF.

There are 2 ways to find the sides -

Method 1 :

AC = $\sqrt{A{D}^2+C{D}^2}$ = $\sqrt{1^2+1^2}$ = $\sqrt{2}$

AF = FE = AC = $\sqrt{2}$ units

Method 2 :

AE = 2 and AFE is a right angled isosceles triangle with AF = FE

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

Let $AF=FE=x$ and $AE=\sqrt{2}x$

$\sqrt{2}x=2$

$x=\sqrt{2}$

$\Rightarrow AF=FE=\sqrt{2}$ units

$\therefore$ Area of the $△$ AFE = $\frac{1}{2}\times AF\times FE$

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

= $1squnits$