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 ?

Solution(By Examveda Team)

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)
$$\eqalign{ & = 2 \times {\text{Area (}}\vartriangle {\text{ABC)}} \cr & = 2 \times \frac{1}{2} \times BC \times AB \cr & = 3x \times 4x \cr & = 12{x^2} \cr} $$
Area of the circle :
$$\eqalign{ & = \pi {r^2} \cr & = \left( {\frac{{22}}{7} \times \frac{{5x}}{2} \times \frac{{5x}}{2}} \right) \cr & = \frac{{275{x^2}}}{{14}} \cr} $$
Area wasted :
$$\eqalign{ & = \left( {\frac{{275{x^2}}}{{14}} - 12{x^2}} \right) \cr & = \frac{{107{x^2}}}{{14}} \cr} $$
Required percentage :
$$\eqalign{ & = \left( {\frac{{107}}{{14}} \times \frac{{14}}{{275}} \times 100} \right)\% \cr & = 39\% \cr} $$