Question

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

i.e. $AB=BC=CD=DA=1unit$

Now, According to given question

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

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

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

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

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

Now,

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

$ar\left(\square CDFE\right)=1-\frac{x^2}{2}-\frac{(1-x)}{2}$

$ar\left(\square CDFE\right)=\frac{2-x^2-(1-x)}{2}$

$ar\left(\square CDFE\right)=\frac{(1+x-x^2)}{2}$

We know that,

$x=\frac{1}{2}$ Then,

$ar\left(\square CDFE\right)=\frac{(1+\frac{1}{2}-{\left(\frac{1}{2}\right)}^2)}{2}$

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

$=\frac{5}{8}$

Hence, this is the answer.