The Pythagorean Theorem

Discovering $a^2 + b^2 = c^2$ through squares, triangles, and geometry

A A square inside a square

Draw an outer square. Mark a point on each side at the same distance from a corner. Connect those four points to form an inner (tilted) square. You get four identical right triangles in the corners — with legs $a$ and $b$. Work through each question below to uncover the relationship between the areas.

How to read the diagram

Each side of the outer square is divided into two pieces: length $a$ (orange label) and length $b$ (green label). The inner blue square has side $c$ — the hypotenuse of each corner triangle. The right-angle markers sit at the four outer corners.

A1 Triangle legs: $a = 3$, $b = 4$ Starter ▼

The four corner triangles each have legs $a = 3$ and $b = 4$. Find:

Area of one right triangle

Total area of all four triangles

Side length of the outer square

Area of the outer square

Area of the inner square

Side length of the inner square (= $c$)

Show all answers

One triangle: $\dfrac{1}{2} \times 3 \times 4 = 6$

Four triangles: $4 \times 6 = 24$

Outer side: $a + b = 3 + 4 = 7$

Outer area: $7^2 = 49$

Inner area: $49 - 24 = 25$

Inner side: $c = \sqrt{25} = 5$

$c = 5$

A2 Triangle legs: $a = 5$, $b = 12$ Starter ▼

The four corner triangles each have legs $a = 5$ and $b = 12$. Find the six quantities as in A1.

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One triangle: $\dfrac{1}{2} \times 5 \times 12 = 30$

Four triangles: $4 \times 30 = 120$

Outer side: $5 + 12 = 17$

Outer area: $17^2 = 289$

Inner area: $289 - 120 = 169$

Inner side: $c = \sqrt{169} = 13$

$c = 13$

A3 Triangle legs: $a = 6$, $b = 8$ Easy ▼

Legs $a = 6$, $b = 8$. Find all six quantities.

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One triangle: $\dfrac{1}{2} \times 6 \times 8 = 24$

Four triangles: $4 \times 24 = 96$

Outer side: $6 + 8 = 14$

Outer area: $14^2 = 196$

Inner area: $196 - 96 = 100$

Inner side: $c = \sqrt{100} = 10$

$c = 10$

A4 Triangle legs: $a = 4$, $b = 7$ Easy ▼

Legs $a = 4$, $b = 7$. Find all six quantities. The answer for $c$ will not be a whole number — leave it as a square root.

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One triangle: $\dfrac{1}{2} \times 4 \times 7 = 14$

Four triangles: $4 \times 14 = 56$

Outer side: $4 + 7 = 11$

Outer area: $11^2 = 121$

Inner area: $121 - 56 = 65$

Inner side: $c = \sqrt{65} \approx 8.06$

$c = \sqrt{65}$

A5 Triangle legs: $a = 8$, $b = 15$ Easy ▼

Legs $a = 8$, $b = 15$. Find all six quantities.

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One triangle: $\dfrac{1}{2} \times 8 \times 15 = 60$

Four triangles: $4 \times 60 = 240$

Outer side: $8 + 15 = 23$

Outer area: $23^2 = 529$

Inner area: $529 - 240 = 289$

Inner side: $c = \sqrt{289} = 17$

$c = 17$

A6 Triangle legs: $a = 2$, $b = 5$ Medium ▼

Legs $a = 2$, $b = 5$. Find all six quantities.

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One triangle: $\dfrac{1}{2} \times 2 \times 5 = 5$

Four triangles: $4 \times 5 = 20$

Outer side: $2 + 5 = 7$

Outer area: $7^2 = 49$

Inner area: $49 - 20 = 29$

Inner side: $c = \sqrt{29} \approx 5.39$

$c = \sqrt{29}$

B Finding the hypotenuse using the construction

Now you are given a right triangle with legs $a$ and $b$. Mentally build the square-in-square construction and work out $c$ step by step. The path is always the same: outer area → subtract 4 triangles → inner area → take square root → c.

B1 Right triangle: $a = 3$, $b = 4$ — find $c$ Starter ▼

Build the square-in-square with these triangle legs. Follow each step to find the hypotenuse $c$.

Step guide:
1. Outer side $= a + b$
2. Outer area $= (a+b)^2$
3. Four triangles area $= 4 \times \frac{1}{2}ab = 2ab$
4. Inner area $= (a+b)^2 - 2ab$
5. $c = \sqrt{\text{inner area}}$

Show solution

Outer side $= 3 + 4 = 7$

Outer area $= 7^2 = 49$

Four triangles $= 2 \times 3 \times 4 = 24$

Inner area $= 49 - 24 = 25$

$c = \sqrt{25} = 5$

Hypotenuse $c = 5$

B2 Right triangle: $a = 5$, $b = 12$ — find $c$ Easy ▼

Use the square-in-square method to find $c$.

Show solution

Outer area $= (5+12)^2 = 17^2 = 289$

Four triangles $= 2 \times 5 \times 12 = 120$

Inner area $= 289 - 120 = 169$

$c = \sqrt{169} = 13$

$c = 13$

B3 Right triangle: $a = 1$, $b = 1$ — find $c$ Easy ▼

Both legs are equal ($a = b = 1$). Use the method to find $c$. You should get a famous irrational number.

Show solution

Outer area $= (1+1)^2 = 4$

Four triangles $= 2 \times 1 \times 1 = 2$

Inner area $= 4 - 2 = 2$

$c = \sqrt{2} \approx 1.414$

$c = \sqrt{2}$ (the diagonal of a unit square)

B4 Right triangle: $a = 7$, $b = 24$ — find $c$ Easy ▼

Use the method. The answer will be a whole number.

Show solution

Outer area $= (7+24)^2 = 31^2 = 961$

Four triangles $= 2 \times 7 \times 24 = 336$

Inner area $= 961 - 336 = 625$

$c = \sqrt{625} = 25$

$c = 25$

B5 Right triangle: $a = 9$, $b = 12$ — find $c$ Medium ▼

Use the square-in-square method. The answer is a whole number.

Hint: $961 - 216 = ?$ Is the inner area a perfect square?

Show solution

Outer area $= (9+12)^2 = 21^2 = 441$

Four triangles $= 2 \times 9 \times 12 = 216$

Inner area $= 441 - 216 = 225$

$c = \sqrt{225} = 15$

$c = 15$

B6 Right triangle: $a = 3$, $b = 5$ — find $c$ Medium ▼

The answer will not be a whole number. Express $c$ as a square root and as a decimal (to 2 decimal places).

Show solution

Outer area $= (3+5)^2 = 64$

Four triangles $= 2 \times 3 \times 5 = 30$

Inner area $= 64 - 30 = 34$

$c = \sqrt{34} \approx 5.83$

C Deriving the Pythagorean theorem

Instead of computing with specific numbers, let's keep $a$ and $b$ as symbols and work through the same five steps. We will arrive at a famous formula valid for every right triangle.

C1 Express the outer square area in terms of $a$ and $b$ Easy ▼

The outer square has side $a + b$. Write its area in two ways: as $(a+b)^2$, and also by expanding the bracket fully.

Hint: $(a+b)^2 = (a+b)(a+b)$. Use FOIL.

Show solution

Outer area $= (a+b)^2$

$= a^2 + 2ab + b^2$

Outer area $= a^2 + 2ab + b^2$

C2 Express the area of all four triangles in terms of $a$ and $b$ Easy ▼

Each right triangle has legs $a$ and $b$. Write the total area of all four triangles.

Show solution

One triangle area $= \dfrac{1}{2}ab$

Four triangles $= 4 \times \dfrac{1}{2}ab = 2ab$

Four triangles area $= 2ab$

C3 Express the inner square area — then derive $c^2 = a^2 + b^2$ Medium ▼

The inner square has side $c$. Using your answers from C1 and C2, write the inner area two ways:
(i) as $c^2$
(ii) as outer area minus four triangles.
Set them equal and simplify to get the Pythagorean theorem.

Show derivation

Inner area $= c^2$ (side of inner square is $c$)

Inner area $=$ outer area $-$ four triangles $= (a^2 + 2ab + b^2) - 2ab$

$= a^2 + b^2$

Therefore: $\quad c^2 = a^2 + b^2$

$$\boxed{c^2 = a^2 + b^2}$$

The two middle terms $+2ab$ and $-2ab$ cancel perfectly — that is the magic of this construction.

C4 Verify with three numbers from Section A Easy ▼

Check that $c^2 = a^2 + b^2$ holds for A1, A2, and A5.

Show verification

A1: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ ✓

A2: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ✓

A5: $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ ✓

The formula $c^2 = a^2 + b^2$ is confirmed in all three cases.

D Practice: using $a^2 + b^2 = c^2$ directly

Now apply the formula directly — no need to draw the full construction each time. Some exercises ask for $c$, others ask you to find a missing leg $a$ or $b$. Where indicated, verify your answer using the geometric method.

Three forms of the formula

$$c = \sqrt{a^2 + b^2} \qquad a = \sqrt{c^2 - b^2} \qquad b = \sqrt{c^2 - a^2}$$

D1 Find $c$: $a = 6,\; b = 8$ Starter ▼

A right triangle has legs 6 and 8. Find the hypotenuse $c$.

Show solution

$c^2 = 6^2 + 8^2 = 36 + 64 = 100$

$c = \sqrt{100} = 10$

$c = 10$

Geometric check: outer area $=(6+8)^2=196$, four triangles $=2\times6\times8=96$, inner $=100$, $c=\sqrt{100}=10$ ✓

D2 Find $c$: $a = 15,\; b = 20$ Easy ▼

Legs 15 and 20. Find $c$.

Hint: Notice that $15 = 3 \times 5$ and $20 = 4 \times 5$. Can you spot the pattern from A1?

Show solution

$c^2 = 15^2 + 20^2 = 225 + 400 = 625$

$c = \sqrt{625} = 25$

$c = 25$

This is the 3-4-5 triple scaled by 5: $(3\times5, 4\times5, 5\times5)$.

D3 Find $c$: $a = 7,\; b = 7$ Easy ▼

An isosceles right triangle has both legs equal to 7. Find $c$.

Show solution

$c^2 = 7^2 + 7^2 = 49 + 49 = 98$

$c = \sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$

$c = 7\sqrt{2} \approx 9.90$

D4 Find the missing leg: $c = 13,\; b = 5$, find $a$ Medium ▼

A right triangle has hypotenuse 13 and one leg 5. Find the other leg $a$.

Hint: Rearrange: $a^2 = c^2 - b^2$.

Show solution

$a^2 = 13^2 - 5^2 = 169 - 25 = 144$

$a = \sqrt{144} = 12$

$a = 12$

D5 Find the missing leg: $c = 17,\; a = 8$, find $b$ Medium ▼

Hypotenuse 17, one leg 8. Find the other leg $b$.

Show solution

$b^2 = 17^2 - 8^2 = 289 - 64 = 225$

$b = \sqrt{225} = 15$

$b = 15$

D6 Verify A4 with the formula: $a = 4,\; b = 7$ Easy ▼

In A4 you found $c = \sqrt{65}$ using the geometric method. Confirm this directly with $c^2 = a^2 + b^2$.

Show solution

$c^2 = 4^2 + 7^2 = 16 + 49 = 65$

$c = \sqrt{65}$

Matches A4 exactly. ✓

D7 Verify A6 with the formula: $a = 2,\; b = 5$ Easy ▼

In A6 you found $c = \sqrt{29}$. Confirm with the formula.

Show solution

$c^2 = 2^2 + 5^2 = 4 + 25 = 29$

$c = \sqrt{29}$

Matches A6. ✓

D8 Is it a right triangle? $a = 9,\; b = 40,\; c = 41$ Medium ▼

Check whether a triangle with sides 9, 40, and 41 is a right triangle by testing $a^2 + b^2 = c^2$.

Show solution

$9^2 + 40^2 = 81 + 1600 = 1681$

$41^2 = 1681$

$81 + 1600 = 1681$ ✓ — yes, it is a right triangle (the 9-40-41 triple).

D9 Ladder problem: ladder 10 m, wall 8 m — how far from wall? Hard ▼

A 10 m ladder leans against a vertical wall. The top of the ladder touches the wall at 8 m above the ground. How far is the base of the ladder from the wall?

Hint: The ladder is the hypotenuse ($c = 10$), the wall height is one leg ($b = 8$). Find the base $a$.

Show solution

$a^2 = c^2 - b^2 = 10^2 - 8^2 = 100 - 64 = 36$

$a = \sqrt{36} = 6$ m

The base is 6 m from the wall.

D10 Diagonal of a rectangle: $12 \times 5$ Hard ▼

A rectangle is 12 cm long and 5 cm wide. Find the length of its diagonal. Then verify using the square-in-square geometric method.

Show solution

Formula: diagonal $= \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ cm

Geometric verification:

Outer area $= (12+5)^2 = 289$

Four triangles $= 2 \times 12 \times 5 = 120$

Inner area $= 289 - 120 = 169 = 13^2$ ✓

Diagonal $= 13$ cm

D11 Challenge: three squares on the sides Challenge ▼

Draw squares on each of the three sides of a right triangle with legs 3 and 4. Show that the area of the square on the hypotenuse equals the sum of the areas of the squares on the two legs. This is the original geometric meaning of the Pythagorean theorem.

Show solution

Square on leg $a = 3$: area $= 3^2 = 9$

Square on leg $b = 4$: area $= 4^2 = 16$

Sum $= 9 + 16 = 25$

Hypotenuse $c = 5$, so square on $c$: area $= 5^2 = 25$

$9 + 16 = 25$ ✓ — the square on $c$ equals the sum of squares on $a$ and $b$.

In symbols: the three squares have areas $a^2$, $b^2$, and $c^2$. The theorem says $a^2 + b^2 = c^2$ — the areas of the two smaller squares add up to the area of the largest square.