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 ?
Options:
A . 34%
B . 39%
C . 42%
D . 47%
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) $$\eqalign{ & = 2 \times {\text{Area (}}\vartriangle {\text{ABC)}} \cr & = 2 \times \frac{1}{2} \times BC \times AB \cr & = 3x \times 4x \cr & = 12{x^2} \cr} $$ 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 : $$\eqalign{ & = \left( {\frac{{107}}{{14}} \times \frac{{14}}{{275}} \times 100} \right)\% \cr & = 39\% \cr} $$
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