The Pythagorean Theorem

In every right triangle, the three side lengths are linked by one simple equation. If you know any two sides, you can find the third, without measuring. Builders use it to check that corners are square, and you can use it to find how high a ladder reaches, how big a TV screen really is, or how much shorter it is to cut across a field.

Key ideas

Parts of a right triangle

A right triangle has one $90^\circ$ angle, marked with a small square.

• The hypotenuse is the side opposite the right angle. It is always the longest side. We usually call it $c$.
• The other two sides are the legs. They form the right angle. We usually call them $a$ and $b$.

The Pythagorean theorem

In a right triangle with legs $a$ and $b$ and hypotenuse $c$:

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

In words: the square of the hypotenuse equals the sum of the squares of the two legs.

The name comes from squares you can actually draw. Build a square on each side of the triangle. The two squares on the legs have the same total area as the square on the hypotenuse.

For a 3–4–5 triangle: $9 + 16 = 25$, so $3^2 + 4^2 = 5^2$.

Finding the hypotenuse or a leg

To find the hypotenuse, add the squares of the legs, then take the square root:

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

To find a leg, subtract the square of the other leg from the square of the hypotenuse, then take the square root:

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

A quick check: the hypotenuse must come out longer than both legs, and a leg must come out shorter than the hypotenuse.

Exact and decimal answers

Often the square root isn’t a whole number. Then you can give

• an exact answer, left as a square root, like $\sqrt{74}$ cm, or
• a decimal approximation from your calculator, like $8.60$ cm (rounded to two decimal places).

Use the exact answer if you’ll keep calculating with it, and round only at the very end.

The converse: testing for a right triangle

The theorem also works backwards. If the side lengths of a triangle fit $a^2 + b^2 = c^2$ (where $c$ is the longest side), then the triangle is a right triangle, with the right angle opposite $c$. If they don’t fit, it isn’t.

Sets of whole numbers that fit, like $3, 4, 5$ and $5, 12, 13$ and $8, 15, 17$, are called Pythagorean triples. Any multiple of a triple also works, such as $6, 8, 10$.

Builders and carpenters use the converse to check square corners: measure $3$ units along one wall and $4$ units along the other. If the diagonal between those marks is exactly $5$ units, the corner is $90^\circ$.

Looking ahead: in Grade 10 you’ll use this same theorem to find the length of a line segment on a coordinate grid, and it’s the starting point for the primary trigonometric ratios.

Where this comes from. The theorem is named after the Greek thinker Pythagoras (around 500 BCE), but people knew the relationship long before him. A Babylonian clay tablet known as Plimpton 322, written about $3800$ years ago, lists numbers connected to Pythagorean triples. The Indian Sulba Sutras, rules for building fire altars, describe it, and in China it’s called the gougu theorem.

Worked examples

Example 1: Finding the hypotenuse

Find the hypotenuse of a right triangle with legs

• (a) $6$ cm and $8$ cm
• (b) $5$ cm and $7$ cm (give an exact answer and a decimal to two places).

Solution.

(a)

$$\begin{aligned} c^2 &= 6^2 + 8^2 \\ &= 36 + 64 \\ &= 100 \\ c &= \sqrt{100} = 10 \end{aligned}$$

The hypotenuse is $10$ cm.

(b)

$$\begin{aligned} c^2 &= 5^2 + 7^2 \\ &= 25 + 49 \\ &= 74 \\ c &= \sqrt{74} \approx 8.60 \end{aligned}$$

The hypotenuse is exactly $\sqrt{74}$ cm, or about $8.60$ cm.

Check: both answers are longer than the legs. ✓

Example 2: A ladder against a wall

A $5.0$ m ladder leans against a wall. Its foot is $1.4$ m from the bottom of the wall. How high up the wall does the ladder reach?

Solution. The wall, the ground and the ladder make a right triangle. The ladder is the hypotenuse ($c = 5.0$), the ground distance is one leg ($b = 1.4$), and the height $h$ is the other leg.

$$\begin{aligned} h^2 + 1.4^2 &= 5.0^2 \\ h^2 + 1.96 &= 25 \\ h^2 &= 23.04 \\ h &= \sqrt{23.04} = 4.8 \end{aligned}$$

The ladder reaches $4.8$ m up the wall.

Check: $4.8^2 + 1.4^2 = 23.04 + 1.96 = 25 = 5^2$. ✓ And $4.8$ m is shorter than the $5.0$ m ladder, as it must be.

Example 3: Is it a right triangle?

Decide whether each set of side lengths makes a right triangle.

• (a) $9$ cm, $12$ cm, $15$ cm
• (b) $7$ cm, $9$ cm, $11$ cm

Solution. In each case, compare the sum of the squares of the two shorter sides with the square of the longest side.

(a) $9^2 + 12^2 = 81 + 144 = 225$ and $15^2 = 225$. They’re equal, so this is a right triangle. (It’s the $3, 4, 5$ triple multiplied by $3$.)

(b) $7^2 + 9^2 = 49 + 81 = 130$ and $11^2 = 121$. Since $130 \ne 121$, this is not a right triangle.

Example 4: A composite shape

The figure shows an isosceles trapezoid. Its parallel sides are $10$ cm and $16$ cm, and its height is $4$ cm. Find the length $x$ of each slanted side, and the perimeter of the trapezoid.

The dashed height cuts off a right triangle with legs $3$ cm and $4$ cm.

Solution. Look for a right triangle hiding in the shape. Draw the height from a top corner straight down. It cuts off a right triangle at each end.

The bottom is $6$ cm longer than the top ($16 - 10 = 6$). Because the trapezoid is isosceles, that extra length is split equally between the two ends, so each triangle has a bottom leg of $6 \div 2 = 3$ cm. Its other leg is the height, $4$ cm, and its hypotenuse is the slanted side $x$:

$$\begin{aligned} x^2 &= 3^2 + 4^2 \\ &= 9 + 16 = 25 \\ x &= 5 \end{aligned}$$

Each slanted side is $5$ cm. The perimeter is

$$P = 10 + 16 + 5 + 5 = 36 \text{ cm}$$

Common mistakes

Adding when you should subtract. When you’re finding a leg, subtract: $a^2 = c^2 - b^2$. In Example 2, adding would give $\sqrt{25 + 1.96} \approx 5.19$ m, which is longer than the ladder itself. That’s impossible, so the check catches it.

Using the wrong side as the hypotenuse. The hypotenuse is opposite the right angle and is always the longest side. In the converse test, always put the longest side on its own: compare $a^2 + b^2$ with the longest side squared.

Forgetting the square root. $c^2 = 100$ means $c = 10$, not $100$. The last step is always to take the square root.

Squaring the sum instead of summing the squares. $3^2 + 4^2 = 9 + 16 = 25$, but $(3 + 4)^2 = 49$. Square each side first, then add.

Rounding too early. If a problem has more than one step, keep the exact value (or all the calculator digits) until the end. Rounding in the middle can make the final answer wrong.

Using the theorem on a triangle that isn’t right-angled. $a^2 + b^2 = c^2$ is only true for right triangles. In a composite shape, draw a height or a diagonal to make a right triangle first, as in Example 4.

Practice

1. (Warm-up) The legs of a right triangle are $9$ cm and $12$ cm. Find the hypotenuse.

Solution

$$c^2 = 9^2 + 12^2 = 81 + 144 = 225 \quad\Rightarrow\quad c = \sqrt{225} = 15$$

The hypotenuse is $15$ cm.

2. (Warm-up) A right triangle has hypotenuse $13$ m and one leg $5$ m. Find the other leg.

Solution

$$a^2 = 13^2 - 5^2 = 169 - 25 = 144 \quad\Rightarrow\quad a = \sqrt{144} = 12$$

The other leg is $12$ m.

3. (Warm-up) Is a triangle with sides $8$ cm, $15$ cm and $17$ cm a right triangle?

Solution

$8^2 + 15^2 = 64 + 225 = 289$ and $17^2 = 289$. They are equal, so yes, it’s a right triangle, with the right angle opposite the $17$ cm side.

4. (Core) Find the missing side. Give an exact answer and a decimal rounded to two places.

• (a) Legs $4$ cm and $6$ cm; find the hypotenuse.
• (b) Hypotenuse $10$ cm and one leg $7$ cm; find the other leg.

Solution

(a) $c^2 = 4^2 + 6^2 = 16 + 36 = 52$, so $c = \sqrt{52} \approx 7.21$ cm.

(b) $a^2 = 10^2 - 7^2 = 100 - 49 = 51$, so $a = \sqrt{51} \approx 7.14$ cm.

Check: $7.21$ is longer than both legs, and $7.14$ is shorter than the hypotenuse. ✓

5. (Core) A TV screen is $120$ cm wide and $68$ cm tall. TV sizes are given by the diagonal of the screen.

• (a) Find the diagonal in centimetres, to one decimal place.
• (b) TVs are sold in inches. Use $1 \text{ in} = 2.54 \text{ cm}$ to find the diagonal in inches, to the nearest inch.

Solution

(a) The width, height and diagonal form a right triangle, with the diagonal as the hypotenuse:

$$d^2 = 120^2 + 68^2 = 14\,400 + 4624 = 19\,024 \quad\Rightarrow\quad d = \sqrt{19\,024} \approx 137.9 \text{ cm}$$

(b) $137.93 \div 2.54 \approx 54.3$, so it’s about a $54$-inch TV.

6. (Core) A rectangular park is $80$ m long and $60$ m wide. Instead of walking along two sides to reach the opposite corner, you cut straight across the diagonal. How much shorter is your walk?

Solution

Along the two sides: $80 + 60 = 140$ m.

Across the diagonal:

$$d^2 = 80^2 + 60^2 = 6400 + 3600 = 10\,000 \quad\Rightarrow\quad d = 100 \text{ m}$$

The shortcut saves $140 - 100 = 40$ m.

7. (Core) A wheelchair ramp rises $0.5$ m over a horizontal distance of $6$ m. How long is the ramp surface, to the nearest centimetre?

Solution

The rise and the horizontal distance are the legs, and the ramp surface is the hypotenuse:

$$L^2 = 6^2 + 0.5^2 = 36 + 0.25 = 36.25 \quad\Rightarrow\quad L = \sqrt{36.25} \approx 6.0208$$

The ramp is about $6.02$ m long.

8. (Challenge) A house is $8$ m wide. Its roof has two equal slanted sides that meet at a peak $3$ m above the tops of the walls, right above the middle of the house. The house is $12$ m long (front to back).

• (a) Find the length of each slanted side of the roof (from the top of the wall to the peak).
• (b) Shingles cover both rectangular roof surfaces. Find the total roof area.

Solution

(a) The peak is above the middle, so each half of the roof spans $8 \div 2 = 4$ m horizontally and rises $3$ m. These are the legs of a right triangle:

$$s^2 = 4^2 + 3^2 = 16 + 9 = 25 \quad\Rightarrow\quad s = 5 \text{ m}$$

(b) Each roof surface is a rectangle $5$ m by $12$ m, and there are two of them:

$$A = 2 \times 5 \times 12 = 120 \text{ m}^2$$

9. (Challenge) A box is $3$ cm wide, $4$ cm deep and $12$ cm tall. How long is the longest straight stick that fits inside, going from a bottom corner to the opposite top corner?

Solution

Use the theorem twice.

Step 1: the diagonal of the bottom. The bottom is a $3$ cm by $4$ cm rectangle:

$$d^2 = 3^2 + 4^2 = 25 \quad\Rightarrow\quad d = 5 \text{ cm}$$

Step 2: up to the top corner. The bottom diagonal ($5$ cm) and the height ($12$ cm) are the legs of a second right triangle that stands up inside the box:

$$L^2 = 5^2 + 12^2 = 25 + 144 = 169 \quad\Rightarrow\quad L = 13 \text{ cm}$$

The longest stick is $13$ cm.