# What is the area of the shaded region? [closed]

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.

## closed as off-topic by Jaap Scherphuis, greenturtle3141, ManyPinkHats, Oray, JMPJul 14 '18 at 21:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Jaap Scherphuis, greenturtle3141, ManyPinkHats, Oray, JMP
If this question can be reworded to fit the rules in the help center, please edit the question.

• This is better suited for math SE. – greenturtle3141 Jul 14 '18 at 19:33
• It doesn't even look like a square... – Mr Pie Jul 15 '18 at 0:34
• All I can say is that by letting $A_c$ being the area of the circle and $A_t$ being the area of the triangle, I can gather that $$A_c=\pi\times 4^2\div 2 = \pi \times 16\div 2 = 8\pi$$ Since if the square has area $16$, then the side lengths must be $4$. Then we divide by $2$ because it is a semi-circle. And now, $$A_t = 4\times \frac 42\sqrt{3} = 4\times 2\sqrt{3}=8\sqrt{3}$$ Since the height of the triangle is (side length $\times \sqrt{3}\div 2$) with side length being equal to $4$ as the triangle is equilateral, so the base is equal to the rest of its sides. – Mr Pie Jul 15 '18 at 0:40
• Just saying, you should include your attempt(s)... – Mr Pie Jul 15 '18 at 0:46
• @user477343 I don't think so it's appropriate to include one's attempt on puzzling SE. Even though it's more like a mathematical problem, OP has posted it as a mathematical puzzle. – Laschet Jain Jul 15 '18 at 10:13

My solution is brute-force. Just leaving it here in case you don't get any other answers.

Let side of square be $a$.

I'm assuming $O$ is the centre of the semi-circle. Also, a line joining vertex $E$ of the equilateral triangle and point $O$ is perpendicular to $BC$. Let that perpendicular intersect the red region at point $O'$. Also let $EB$ and $EC$ intersect the semicircle at $B'$ and $C'$ respectively.

The distance EO is $a - \sqrt{3}a/2$ [height of square - height of eq. triangle]. Also, the distance $EO'$ is $a/2-length(EO)$ which is $\sqrt{3}a/2 - a/2$.

Now construct a triangle $EB'C'$. By symmetry, $EB' = EC'$. Also, the angle $B'EC' = 60^\circ$. Which means the constructed triangle is equilateral. Let height of this equilateral triangle be $h$. I'll show it's calculation later.

Following is the method to calculate area of region $R$ left by removing triangle $EB'C'$ from red region.

The area of the region $R$ can be found out by considering sector $OB'C'$. This sector belongs to the semi-circle. We can calculate the area of the triangle $OB'C'$ since $OB' = a/2$ and $OC' = a/2$ and $B'C' = (2/\sqrt{3})h$. [Since $B'C'$ is side of equilateral triangle $EB'C'$ and $h$ is it's height.]

Since we've got all sides of the triangle, we can calculate the area of the sector $OB'C'$ by applying co-sine rule to find the angle $B'OC'$. Once this area is calculated, subtract it from the area of the triangle $OB'C'$ to get area of region $R$.

To calculate $h$, we first need to find out the length of $EO' - h$: Let's call this $l$. $l$ is equal to radius of semi-circle minus height of $OB'C'$. Find the height in terms of $a$ and $h$ and later solve the the quadratic equation $(EO' - h)^2 = l^2$ for $h$. One root is positive and other is negative. Reject the negative one.

Use the fact $a = 4$ to get a numeric result.