SEAB Original-style practice: A right-angled triangle has the two shorter sides 6\ cm and 8\ cm. Find the length of the hypotenuse.

The hypotenuse is the longest side, opposite the right angle. By Pythagoras: hypotenuse^2 = 6^2 + 8^2 = 36 + 64 = 100. Hypotenuse = sqrt(100) = 10\ cm. What markers reward: adding the squares of the two shorter sides, then taking the square root. This is the classic 6-8-10 triangle, so the hypotenuse is exactly 10\ cm.

SingaporeMathsSyllabus dot point

In a right-angled triangle, how are the three sides related, and how do we find a missing side?

Apply Pythagoras' theorem to find a missing side in a right-angled triangle and to solve simple problems

A focused answer to the N(A)-Level Mathematics outcome on Pythagoras' theorem. The relationship between the three sides, finding the hypotenuse or a shorter side, and real problems.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to use Pythagoras' theorem to find a missing side in a right-angled triangle and to apply it to simple real problems such as ladders, diagonals and distances. The theorem links the three sides of a right-angled triangle, and recognising which side is missing tells you whether to add or subtract.

The answer

The theorem

In a right-angled triangle, the longest side (the hypotenuse, opposite the right angle) is related to the other two sides by:

c^2 = a^2 + b^2

where $c$ is the hypotenuse and $a$ and $b$ are the two shorter sides. In words: the square of the hypotenuse equals the sum of the squares of the other two sides.

Finding the hypotenuse

When the two shorter sides are known and the hypotenuse is missing, add the squares and take the square root:

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

For shorter sides $3$ and $4$ : $c = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Finding a shorter side

When the hypotenuse and one shorter side are known, rearrange to subtract:

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

For hypotenuse $10$ and one side $6$ : $a = \sqrt{100 - 36} = \sqrt{64} = 8$ . The clue is that the hypotenuse is the largest number; if it is known, you subtract.

Identifying the hypotenuse

The hypotenuse is always opposite the right angle and is always the longest side. Spotting it correctly is the most important step, because it decides whether you add or subtract.

Surds and rounding

Often the answer is a surd such as $\sqrt{50}$ . Leave it exact if allowed, or round to the accuracy asked for (commonly $3$ significant figures) if a decimal is required.

Worked example

A ladder $5\ \text{m}$ long leans against a vertical wall, with its foot $3\ \text{m}$ from the base of the wall. How far up the wall does the ladder reach?

Step 1: Identify the sides

The ladder is the hypotenuse ( $5\ \text{m}$ , the longest side). The distance along the ground ( $3\ \text{m}$ ) is one shorter side, and the height up the wall is the missing shorter side.

Step 2: Set up Pythagoras

A shorter side is missing, so subtract:

\text{height}^2 = 5^2 - 3^2 = 25 - 9 = 16.

Step 3: Take the square root

\text{height} = \sqrt{16} = 4\ \text{m}.

Step 4: State the answer

The ladder reaches $4\ \text{m}$ up the wall. This is the familiar $3$ - $4$ - $5$ right-angled triangle.

Examples in context

Example 1. The diagonal of a rectangle. A rectangle $8\ \text{cm}$ by $6\ \text{cm}$ has a diagonal that splits it into two right-angled triangles. The diagonal is the hypotenuse, so its length is $\sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \text{cm}$ . Pythagoras turns a rectangle measurement into a single triangle calculation.

Example 2. Distance between two points. The distance formula in coordinate geometry is Pythagoras in disguise: the horizontal and vertical gaps are the two shorter sides, and the straight-line distance is the hypotenuse. This is why $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ works.

Try this

Cue. Find the hypotenuse of a right-angled triangle with shorter sides $9\ \text{cm}$ and $12\ \text{cm}$ . Compute $\sqrt{81 + 144} = \sqrt{225} = 15\ \text{cm}$ .
Cue. A right-angled triangle has hypotenuse $17\ \text{cm}$ and one side $8\ \text{cm}$ . The other side is $\sqrt{289 - 64} = \sqrt{225} = 15\ \text{cm}$ .
Cue. Is a triangle with sides $5, 12, 13$ right-angled? Check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ , so yes.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksA right-angled triangle has the two shorter sides

6\ \text{cm}

and

8\ \text{cm}

. Find the length of the hypotenuse.

Show worked answer →

The hypotenuse is the longest side, opposite the right angle. By Pythagoras:

$\text{hypotenuse}^2 = 6^2 + 8^2 = 36 + 64 = 100$ .

Hypotenuse $= \sqrt{100} = 10\ \text{cm}$ .

What markers reward: adding the squares of the two shorter sides, then taking the square root. This is the classic $6$ - $8$ - $10$ triangle, so the hypotenuse is exactly $10\ \text{cm}$ .

Original3 marksA right-angled triangle has a hypotenuse of

13\ \text{cm}

and one shorter side of

5\ \text{cm}

. Find the other shorter side.

Show worked answer →

Here a shorter side is missing, so subtract. By Pythagoras, $13^2 = 5^2 + x^2$ .

$169 = 25 + x^2$ , so $x^2 = 169 - 25 = 144$ .

$x = \sqrt{144} = 12\ \text{cm}$ .

What markers reward: arranging Pythagoras so the missing shorter side is found by subtraction, the correct value of $x^2$ , and the square root. Adding instead of subtracting is the common error when the hypotenuse is known.