Question

Let ABCD is a square with sides of unit length. Point E and F are taken on sides AB and AD respectively so that AE$=$AF. Let P be a point inside the square ABCD.

The maximum possible area of quadrilateral CDFE is?

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{5}{8}$
• D. $\frac{3}{8}$

Answer 1

The correct option is C $\frac{5}{8}$

We have, the sides of square $ABCD$ is unit.

$\therefore$ $AB=BC=CD=DA=1unit$

Now, according to the question,

$\Rightarrow$ $ar\left(△EAF\right)=\frac{1}{2}\times x\times x$

$\therefore$ $ar\left(△EAF\right)=\frac{1}{2}{x}^2$

$\Rightarrow$ $ar\left(△EBC\right)=\frac{1}{2}\times (1-x)\times 1$

$\therefore$ $ar\left(△EBC\right)=\frac{1}{2}(1-x)$

$\Rightarrow$ $ar\left(\square ABCD\right)=1\times 1=1$

Now,

$\Rightarrow$ $ar\left(\square CDFE\right)=ar\left(\square ABCD\right)-ar\left(△EAF\right)-ar\left(△EBC\right)$

$=1-\frac{x^2}{2}-\frac{(1-x)}{2}$

$=\frac{2-x^2-(1-x)}{2}$

$=\frac{1+x-x^2}{2}$

$\Rightarrow$ Area of quadrilateral $CDFE$ $=\frac{1+x-x^2}{2}$

$=\frac{1+\frac{1}{2}-{\left(\frac{1}{2}\right)}^2}{2}$ [ Since, $x=\frac{1}{2}$ ]

$=\frac{\frac{3}{2}-\frac{1}{4}}{2}$

$=\frac{5}{8}$