A kite-shaped quadrilateral of the largest possible area is cut from a circular sheet of paper. If the lengths of the sides of the kite are in the ratio 3 : 3 : 4 : 4, what percentage of the circular sheet is wasted ?

Answer: Option B

Clearly, the longer diagonal of the kite is the diameter of the circle
Also,
∠ABC = 90° (angle in a semi-circle)
Let AB = AD = 3x and BC = CD = 4x
Then,
$$AC=\sqrt{A{B}^2+B{C}^2}=5x$$
Area of the kite = 2 × area (ΔABC)
$$\begin{array}{rl}& =2\times \text{Area (}△\text{ABC)}\\ & =2\times \frac{1}{2}\times BC\times AB\\ & =3x\times 4x\\ & =12x^2\end{array}$$
Area of the circle :
$$\begin{array}{rl}& =\pi r^2\\ & =\left(\frac{22}{7}\times \frac{5x}{2}\times \frac{5x}{2}\right)\\ & =\frac{275x^2}{14}\end{array}$$
Area wasted :
$$\begin{array}{rl}& =\left(\frac{275x^2}{14}-12x^2\right)\\ & =\frac{107x^2}{14}\end{array}$$
Required percentage :
$$\begin{array}{rl}& =\left(\frac{107}{14}\times \frac{14}{275}\times 100\right)\%\\ & =39\%\end{array}$$