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The Square in a Semicircle Puzzle: Area Explained

Type
Math puzzle
Reading
2 min

Short answer: a square inscribed in a semicircle of radius rr, with one side on the diameter, has area 4r25\dfrac{4r^2}{5}. A square inscribed in the full circle of the same radius has area 2r22r^2. So the semicircle square is exactly 25\dfrac{2}{5} of the circle square, a surprisingly clean ratio.

Contents5 sections

The solution

Let the square have side ss, sitting on the diameter and centered on the center OO of the semicircle (by symmetry it must be centered). Its top corners touch the arc.

From OO, a top corner is s2\tfrac{s}{2} to the side and ss up, and its distance to OO is the radius. By the Pythagorean theorem:

(s2)2+s2=r25s24=r2s2=4r25.\begin{aligned}\left(\frac{s}{2}\right)^2+s^2&=r^2\\ \frac{5s^2}{4}&=r^2\\ s^2&=\frac{4r^2}{5}.\end{aligned}

For a square inscribed in the full circle, the diagonal is the diameter 2r2r, so its area is (2r)22=2r2\frac{(2r)^2}{2}=2r^2. The ratio:

4r2/52r2=25.\frac{4r^2/5}{2r^2}=\frac{2}{5}.

The popular version

The puzzle often appears as: “A square of area 40 is inscribed in a semicircle. What is the area of the square inscribed in the full circle?” With the ratio above, the answer is immediate: 40÷25=10040\div\frac{2}{5}=\mathbf{100}. Check: 4r25=40\frac{4r^2}{5}=40 gives r2=50r^2=50, and 2r2=1002r^2=100.

A neat feature: you never need to know rr itself, only r2r^2, and many people get stuck trying to find the radius first.

Try these

Puzzle 1

The semicircle has radius 5. What is the area of the inscribed square?

Show Answer

4⋅255=20\frac{4\cdot 25}{5}=\mathbf{20}. The side is 20=25≈4.47\sqrt{20}=2\sqrt5\approx 4.47.

Puzzle 2

What fraction of the semicircle’s area does the square cover?

Show Answer

4r2/5πr2/2=85π≈0.509\dfrac{4r^2/5}{\pi r^2/2}=\dfrac{8}{5\pi}\approx\mathbf{0.509}, just over half.

Puzzle 3

Two equal squares stand side by side on the diameter of a semicircle of radius rr, touching each other at the center, with their outer top corners on the arc. What is the side of each square?

Show Answer

An outer top corner is at horizontal distance ss from OO and height ss, so s2+s2=r2s^2+s^2=r^2 and s=r2s=\dfrac{r}{\sqrt2}. Each square has area r22\dfrac{r^2}{2}, so together they cover r2r^2, more than the single square’s 45r2\frac45 r^2.

Frequently asked questions

What is the area of a square inscribed in a semicircle?

4r25\frac{4r^2}{5}, where rr is the radius, for a square with one side on the diameter.

What is the side length?

s=2r5≈0.894 rs=\frac{2r}{\sqrt5}\approx 0.894\,r.

Why must the square be centered?

Both top corners have to lie on the arc at the same height. Points of the arc at the same height are symmetric about the center, so the square is too.