Bunuel wrote:
A kite – shaped quadrilateral is cut from a circular sheet of paper such that the vertices of the kite lie on the circumference of the circle. If the lengths of the sides of the kite are in the ratio 3: 3: 4 : 4, then approximately what percentage of the area of the circular sheet of paper remains after the kite has been cut out?
(A) 39%
(B) 40%
(C) 45 %
(D) 51%
(E) 53%
The graph is labeled below, from the inscribed angle theorem the kite must consist of two right triangles. The ratios are obviously 3:4:5 so we can assume the radius is 2.5 and the area of the triangles added up is 3*4 = 12.
The area of the circle is \((\frac{5}{2})^2 * \pi \approx \frac{25}{4} * \pi \). Then the ratio that is cut out is: \(12/(\frac{25}{4} * \pi) = \frac{4*12}{25\pi} \approx \frac{4*12}{25*3*1.05} = \frac{16}{25}/1.05 = 64\%/1.05 \approx 64\% - 5\%*64\% = 64\% - 3.2\% = 60.8\%\). Then there is roughly 39.2% not cut out.
(The answers were quite close which normally isn't the case, so don't bother too much about the estimations. Most of the time 3.14 ->3 is a sufficient approximation).
Ans: A
Attachments

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