Multiple choice

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to

  1. $\displaystyle \frac{9\sqrt{3}(1+\sqrt{3})}{\pi ^{2}}$
  2. $\displaystyle \frac{9\sqrt{3}(\sqrt{3}-1)}{\pi ^{2}}$
  3. $\displaystyle \frac{9\sqrt{3}(1+\sqrt{3})}{2\pi^{2}}$
  4. $\displaystyle \frac{9\sqrt{3}(\sqrt{3-1})}{2 \pi^{2}}$
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Explanation

The arcs are 3, 4, 5. Total circumference = 12. Radius R = 12 / (2*pi) = 6/pi. The angles subtended by the arcs are 3/12 * 360 = 90 degrees, 4/12 * 360 = 120 degrees, 5/12 * 360 = 150 degrees. Area = 1/2 * R^2 * (sin(90) + sin(120) + sin(150)) = 1/2 * (36/pi^2) * (1 + sqrt(3)/2 + 1/2) = 18/pi^2 * (3/2 + sqrt(3)/2) = 9/pi^2 * (3 + sqrt(3)) = 9*sqrt(3)/pi^2 * (sqrt(3)+1).

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