The area of the square is 16 sq. units. A semicircle is inscribed on a side of the square with its diameter being that side of the square. An equilateral triangle rests with its base, on the opposite side of the square. Find the intersection area of the semicircle and the equilateral triangle.enter image description here

I was able to figure out a solution using coordinate geometry. But I want a solution without using it. (Also with minimal usage of trigonometry if possible). Please give a numerical answer.

  • $\begingroup$ Did you solve it using formulae of area like $\pi r^2 , \frac{\sqrt{3}}{4}a^2$ ...? $\endgroup$ – Entrepreneur Jul 17 '18 at 16:31
  • $\begingroup$ No I used coordinate geometry to solve it. $\endgroup$ – kaushalpranav Jul 17 '18 at 16:35
  • $\begingroup$ Use $A=\frac{\pi r^2 \theta}{360^{\circ}}$ $\endgroup$ – Entrepreneur Jul 17 '18 at 16:39

In $\triangle ABC$ shown above, $AB=4-2\sqrt3$ because the triangle's height is $2\sqrt3$. The law of sines gives $$\frac{\sin{150^\circ}}2 =\frac{\sin\angle C}{4-2\sqrt3}$$ $$\sin\angle C=\frac{4-2\sqrt3}4=0.1339$$ $\angle B$ can then be evaluated as $30^\circ-C=22.30^\circ$.

The area of the sector containing two copies of $\triangle ABC$ and the shaded area is $\pi\cdot2^2\cdot\frac{2\angle B}{360^\circ}=1.556$. $\triangle ABC$'s own area is $\frac12(BC)(BA)\sin\angle B=0.2033$. Subtracting twice this from 1.556 gives the final answer as 1.150 square units.


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