What is the area of a square inside a semicircle?

Question

This problem is all over Twitter and Instagram right now:

A square is inscribed in a semicircle of radius $R$ . Find the area of the square in terms of $R$ .

The semicircle sits on top of the square. Two corners of the square touch the diameter (base) and the other two corners touch the circular arc.

I tried setting up coordinates: Put the semicircle centered at $(0,0)$ with radius $R$ . The equation is $y = \sqrt{R^2 - x^2}$ .

If the square has side length $s$ , the top-right corner is at $(\frac{s}{2}, s)$ .

Plugging into the circle equation:

(\frac{s}{2})^2 + s^2 = R^2

\frac{s^2}{4} + s^2 = R^2

\frac{5s^2}{4} = R^2

s^2 = \frac{4}{5}R^2

So the area is $\frac{4}{5}R^2$ .

But a friend says the answer is $\frac{4}{5}R^2$ ONLY if the square is oriented with its base on the diameter. What if the square is rotated? Is there a general formula?

Alex Kim · Accepted Answer

s 2 = 4 5 R 2 is correct for the standard orientation. For a rotated square inscribed in a semicircle, the problem changes significantly. A square of side length s rotated by angle θ inscribed in a semicircle of radius R would require solving: Two corners on the diameter: ( x 1 , 0 ) and ( x 2 , 0 ) with ∣ x 1 − x 2 ∣ = s cos ⁡ θ Top corner on the arc: ( x 1 + x 2 2 , s sin ⁡ θ ) satisfying x 2 + y 2 = R 2 This is much more complex and the area would depend on θ, with maximum at θ = 0. A more interesting variant: A square inscribed in a QUARTER-circle (both sides touching the axes). Put the square in the corner with sides on the axes and the opposite corner on the quarter-circle arc. Then the diagonal of the square equals the radius, so s 2 = R, giving area s 2 = R 2 / 2. This is a common Olympiad problem.

What is the area of a square inside a semicircle?

1 Answer

What is the area of a square inside a semicircle?

1 Answer