The theorem and a visual proof

A right triangle has one angle exactly $90^\circ$. The side opposite that right angle is the hypotenuse; the other two sides are the legs. The theorem connects the lengths of all three.

Concept

The statement. In a right triangle with legs $a$ and $b$ and hypotenuse $c$,

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

Baudhayana's words. Around $800$ BCE the Sulba Sutras (manuals for building Vedic altars) included this idea:

"The diagonal of a rectangle produces both the areas which the longer and shorter sides produce separately."

In modern language: the square on the diagonal of a rectangle equals the sum of the squares on the two adjacent sides. The same as Pythagoras.

A visual proof , rearrangement. Here is a hands-on proof you can do today.

1. Cut four identical right triangles with legs $a, b$ and hypotenuse $c$ from paper.
2. Arrange them inside a square of side $a + b$ so that they leave a tilted square of side $c$ in the middle. Two pieces are visible: four triangles (total area $4 \cdot \tfrac{1}{2}ab = 2ab$) and the centre square of area $c^2$.

Total area of the big square: $(a + b)^2 = a^2 + 2ab + b^2$.

So $a^2 + 2ab + b^2 = 2ab + c^2$, giving $a^2 + b^2 = c^2$. ✓

A second visual proof , rearrange the same four triangles differently. Place them so they leave two rectangles of size $a \times b$ and two squares of sides $a$ and $b$. Now the total area is still $(a+b)^2$, but split as $a^2 + b^2 + 2ab$. Setting this equal to the previous expression gives the same conclusion.

Why squares? The theorem is about areas. The square on a side is the area of the square whose edge is that side. The equation $a^2 + b^2 = c^2$ is about adding areas, not lengths. So a right triangle has its three sides connected in a square-area relationship , never in a simple linear way.

Where to spot a right angle. A square corner of a room, a corner of a piece of paper, the meeting of a wall and the floor, the rungs of a step ladder, the cross of a window frame , all of these are right angles. Wherever a right angle hides, the theorem applies.

Worked examples

Example 1. Verify the theorem for the $(3, 4, 5)$ triangle.

• $3^2 + 4^2 = 9 + 16 = 25 = 5^2$. ✓

Example 2. Verify for the $(5, 12, 13)$ triangle.

• $25 + 144 = 169 = 13^2$. ✓

Example 3. Construct a right triangle with legs $6$ cm and $8$ cm. What is the hypotenuse?

• $\sqrt{36 + 64} = \sqrt{100} = 10$ cm.

Example 4. Use the rearrangement proof: if $a = 4, b = 3$, verify by computing both arrangements' total area.

• Big square side $= 7$. Area $= 49$.
• Triangles: $4 \cdot \tfrac{1}{2}(4)(3) = 24$. Centre square side $5$, area $25$.
• $24 + 25 = 49$. ✓

Try it yourself

1. Verify $a^2 + b^2 = c^2$ for $(8, 15, 17)$.
2. Verify for $(7, 24, 25)$.
3. A right triangle has legs $9$ and $12$. Find the hypotenuse.
4. Cut paper triangles and demonstrate the rearrangement proof for $(3,4,5)$.
5. Why are squares used in the theorem, not (say) circles or equilateral triangles? Discuss.
6. Find the hypotenuse of a right triangle with legs $1$ and $1$.
7. Express the diagonal of a square of side $s$ in terms of $s$.
8. Show that the diagonal of a rectangle of sides $a$ and $b$ has length $\sqrt{a^2 + b^2}$.

Activity

Build the proof. Use coloured card to cut four congruent right triangles. Place them inside a square frame in the two ways above. Take a photo of each arrangement and label the areas. This is exactly the proof that Indian, Chinese, and Greek mathematicians independently discovered millennia ago , and you have just rediscovered it.