Semicircles on a Right Triangle: Lunes of Hippocrates

Quick answer: If semicircles are drawn outward on the two shorter sides of a right-angled triangle, and a semicircle is drawn on the hypotenuse over the triangle, the two crescent-shaped regions (called lunes) left outside the hypotenuse semicircle have a total area equal to the area of the triangle. The $\pi$ terms cancel because of Pythagoras' theorem. For a triangle with legs $6\text{ cm}$ and $8\text{ cm}$, the lunes have a total area of $\frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2$.

Around the 5th century BCE, the Greek mathematician Hippocrates of Chios worked on the famous problem of "squaring the circle": constructing a square with the same area as a circle. That problem was later proved impossible, but Hippocrates found something remarkable: certain curved regions bounded by circular arcs, called lunes, have exactly the same area as a straight-sided figure.

In CBSE Class 10 Mathematics, Chapter 11 (Areas Related to Circles), the semicircles on the sides of a right triangle question is a well-known higher-order thinking (HOTS) problem that combines circle areas with Pythagoras' theorem.

What You Will Learn

• What a lune is
• How the semicircles are drawn on the sides of a right triangle
• A step-by-step proof that the total area of the two lunes equals the area of the triangle
• Why $\pi$ cancels out
• A solved example with sides $6\text{ cm}$ and $8\text{ cm}$

1. The Figure

Take a right-angled triangle $ABC$ with $\angle A = 90^\circ$. Let:

• $AB = c$ and $AC = b$ be the two shorter sides (legs)
• $BC = a$ be the hypotenuse

Draw three semicircles:

$S_3$ passes through $A$ because the angle in a semicircle is a right angle.

     S1 on AB      S2 on AC
      .-~-.         .-~-.
    (lune 1)      (lune 2)
       B ---------- A
        \         /
         \  ABC  /
          \     /
           \   /
             C
  S3 is drawn on BC around the triangle.
  The lunes are the parts of S1 and S2
  that lie outside S3.

A lune is a crescent-shaped region bounded by two circular arcs. Here, each lune is the part of a small semicircle ($S_1$ or $S_2$) that lies outside the large semicircle $S_3$.

2. Proof: Area of the Two Lunes Equals the Area of the Triangle

Statement

If semicircles are drawn on the three sides of a right triangle as described above, the total area of the two lunes equals the area of the triangle.

$\text{Area(lune 1)} + \text{Area(lune 2)} = \text{Area}(\triangle ABC)$

Proof

Step 1: Area of a semicircle in terms of its diameter. A semicircle of diameter $d$ has radius $\frac{d}{2}$, so:

$\text{Area} = \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8}$

So the three semicircles have areas:

$\text{Area}(S_1) = \frac{\pi c^2}{8} \qquad \text{Area}(S_2) = \frac{\pi b^2}{8} \qquad \text{Area}(S_3) = \frac{\pi a^2}{8}$

Step 2: Area of the triangle.

$\text{Area}(\triangle ABC) = \frac{1}{2}bc$

Step 3: Area of the whole figure. The whole figure is made of the triangle and the two small semicircles:

$\text{Whole figure} = \frac{1}{2}bc + \frac{\pi c^2}{8} + \frac{\pi b^2}{8} = \frac{1}{2}bc + \frac{\pi}{8}\left(b^2 + c^2\right)$

Step 4: Use Pythagoras' theorem. Since $\angle A = 90^\circ$:

$b^2 + c^2 = a^2$

So:

$\text{Whole figure} = \frac{1}{2}bc + \frac{\pi a^2}{8}$

Step 5: Remove the large semicircle. The two lunes are what is left when the large semicircle $S_3$ is removed from the whole figure:

$\text{Area of the two lunes} = \left(\frac{1}{2}bc + \frac{\pi a^2}{8}\right) - \frac{\pi a^2}{8} = \frac{1}{2}bc$

Hence proved: the total area of the two lunes equals $\frac{1}{2}bc$, the area of the triangle. All the $\pi$ terms cancel.

3. Solved Example

Question

In a right triangle $ABC$, $\angle A = 90^\circ$, $AB = 6\text{ cm}$ and $AC = 8\text{ cm}$. Semicircles are drawn on $AB$, $AC$ and $BC$ as described above. Find the total area of the two lunes.

Solution

Method 1: Use the result. The total area of the lunes equals the area of the triangle:

$\text{Area} = \frac{1}{2} \times 6 \times 8 = 24\text{ cm}^2$

Method 2: Check step by step.

Step 1: Hypotenuse.

$BC = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}$

Step 2: Areas of the semicircles.

$\text{Area}(S_1) = \frac{1}{2}\pi(3)^2 = \frac{9\pi}{2} \qquad \text{Area}(S_2) = \frac{1}{2}\pi(4)^2 = 8\pi \qquad \text{Area}(S_3) = \frac{1}{2}\pi(5)^2 = \frac{25\pi}{2}$

Step 3: Whole figure.

$24 + \frac{9\pi}{2} + 8\pi = 24 + \frac{25\pi}{2}$

Step 4: Subtract the large semicircle.

$\left(24 + \frac{25\pi}{2}\right) - \frac{25\pi}{2} = 24\text{ cm}^2$

Answer: The total area of the two lunes is $24\text{ cm}^2$.

4. Summary Table

5. Exam Tips and Common Mistakes

Exam Tip: Keep $\pi$ as a symbol throughout. There is no need to substitute $\frac{22}{7}$ or $3.14$, because the $\pi$ terms cancel.

Exam Tip: Clearly state where you use Pythagoras' theorem, $b^2 + c^2 = a^2$. That is the key step in the proof.

Common Mistake: Trying to find the area of each lune separately. You only need their total, which comes straight from subtracting the large semicircle from the whole figure.

Common Mistake: Using the diameter instead of the radius in $\frac{1}{2}\pi r^2$. For $AB = 6\text{ cm}$, the radius is $3\text{ cm}$.

Frequently Asked Questions

What are the lunes of Hippocrates?

They are crescent-shaped regions bounded by two circular arcs. In this problem, they are the parts of the semicircles on the two legs of a right triangle that lie outside the semicircle on the hypotenuse.

Why is the area of the two lunes equal to the area of the triangle?

By Pythagoras' theorem, the two small semicircles together have the same area as the large semicircle, so when the large one is removed, only the triangle's area is left.

Why does π cancel out in this problem?

The π terms from the small semicircles add up to the same amount as the π term of the large semicircle, so they cancel when you subtract.

What is the area of the lunes for a right triangle with legs 6 cm and 8 cm?

24 cm², which is the area of the triangle.