Finding the Hypotenuse

Use Pythagoras' theorem to find the longest side of a right-angled triangle from the other two: square, add, take the square root, and round only at the end.

Before you start

Pythagoras' theorem (lesson 9.1): in $\triangle ABC$ with $\angle C = 90^{\circ}$ , the hypotenuse is $AB$ and $AB^2 = AC^2 + BC^2$ .
The square root undoes squaring: if $c^2 = 25$ , then $c = \sqrt{25} = 5$ (a length is never negative). A calculator gives square roots that are not whole numbers: $\sqrt{117} = 10.816\ldots$
To 3 significant figures (3 s.f.), keep the first three figures that count: $10.816\ldots$ is $10.8$ , and $3.0805\ldots$ is $3.08$ .

The Method

When the two sides that form the right angle are known, the theorem gives the square of the hypotenuse. The square root then gives the hypotenuse itself.

Finding the hypotenuse

1. Find the right angle, and name the hypotenuse: the side opposite it.

2. Write Pythagoras' theorem with the hypotenuse on its own: $AB^2 = AC^2 + BC^2$ .

3. Put in the lengths, square each one, then add.

4. Take the square root.

5. Round only at the very end, as the question asks, and give the unit.

Example 1: A whole-number answer

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $BC = 3$ cm and $AC = 4$ cm. Find $AB$ .

Solution

$\angle C = 90^{\circ}$ , so the hypotenuse is $AB$ , the side opposite $C$ .

By Pythagoras' theorem, $AB^2 = AC^2 + BC^2$

$AB^2 = 4^2 + 3^2 = 16 + 9 = 25$

$AB = \sqrt{25} = 5$

$AB = 5$ cm

Example 2: An answer to 3 significant figures

In $\triangle XYZ$ , $\angle Y = 90^{\circ}$ , $XY = 6$ m and $YZ = 9$ m. Find $XZ$ , correct to 3 significant figures.

Solution

$\angle Y = 90^{\circ}$ , so the hypotenuse is $XZ$ .

By Pythagoras' theorem, $XZ^2 = XY^2 + YZ^2$

$XZ^2 = 6^2 + 9^2 = 36 + 81 = 117$

$XZ = \sqrt{117} = 10.816\ldots$ (the calculator's answer, kept in full)

$XZ = 10.8$ (3 s.f.)

Check: $10.8$ m is longer than both $6$ m and $9$ m, as a hypotenuse must be.

$XZ = 10.8$ m (to 3 s.f.)

Rounding the Answer

Most hypotenuses are not whole numbers. Keep every figure the calculator gives until the last step, then round as the question asks. When it does not say, Singapore textbooks give lengths correct to 3 significant figures.

Below, three square roots are each rounded three ways. Notice that 3 s.f. and 1 d.p. are not the same thing: for a number in the hundreds, 3 s.f. is a whole number.

Three square roots, each rounded three ways

$\sqrt{9.49} = 3.0805\ldots$ : $3.1$ (1 d.p.), $3.08$ (2 d.p.), $3.08$ (3 s.f.)

$\sqrt{117} = 10.816\ldots$ : $10.8$ (1 d.p.), $10.82$ (2 d.p.), $10.8$ (3 s.f.)

$\sqrt{15\,241} = 123.454\ldots\text{:}$ $123.5$ (1 d.p.), $123.45$ (2 d.p.), $123$ (3 s.f.)

Example 3: Sides with decimals

The sail of a dinghy at a sailing club is a right-angled triangle. The two edges that meet at the right angle are $1.2$ m and $3.5$ m. Find the length of its third edge.

Solution

The third edge, $x$ m, is opposite the right angle: it is the hypotenuse.

$x^2 = 1.2^2 + 3.5^2 = 1.44 + 12.25 = 13.69$

$x = \sqrt{13.69} = 3.7$

The calculator gives exactly $3.7$ , so there is nothing to round: $3.7^2 = 13.69$ .

The third edge of the sail is $3.7$ m long.

Example 4: Sides in different units

In $\triangle PQR$ , $\angle Q = 90^{\circ}$ , $PQ = 2.5$ m and $QR = 180$ cm. Find $PR$ in metres, correct to 2 decimal places.

Solution

The answer is wanted in metres, so first change $180$ cm to metres: $180 \div 100 = 1.8$ m.

$\angle Q = 90^{\circ}$ , so the hypotenuse is $PR$ .

$PR^2 = PQ^2 + QR^2 = 2.5^2 + 1.8^2 = 6.25 + 3.24 = 9.49$

$PR = \sqrt{9.49} = 3.0805\ldots = 3.08$ (2 d.p.)

Without changing the units, $2.5^2 + 180^2 = 32\,406.25$ would give a "hypotenuse" of about $180$ , which mixes metres with centimetres and is meaningless.

$PR = 3.08$ m (to 2 d.p.)

Is the Answer Reasonable?

A quick check

The hypotenuse is longer than each of the other two sides.

It is shorter than the two of them added together, since the straight path between two corners is shorter than the path round the other two sides.

Example 5: Checking an answer

Wei Jie works out the hypotenuse of a right-angled triangle whose other two sides are $9$ cm and $12$ cm, and gets $2.08$ cm. (a) Without working it out, explain why his answer cannot be right. (b) Find the hypotenuse.

Solution

(a) The hypotenuse must be longer than $12$ cm, the longer of the other two sides. $2.08$ cm is shorter than both, so it cannot be right.

(b) $c^2 = 9^2 + 12^2 = 81 + 144 = 225\text{,}$ so $c = \sqrt{225} = 15$ .

Check: $15$ cm is more than $12$ cm and less than $9 + 12 = 21$ cm.

(a) $2.08$ cm is shorter than the other two sides. (b) The hypotenuse is $15$ cm.

Common Mistakes

Adding before squaring. For sides $3$ cm and $4$ cm, $(3 + 4)^2 = 49$ gives $7$ cm, which is the two sides added together. Square each side first: $9 + 16 = 25$ , so $5$ cm.
Forgetting the square root. $c^2 = 25$ is not the answer; $c = \sqrt{25} = 5$ . An answer bigger than the two sides added together must be wrong; often the square root has been missed.
Rounding too early. In a triangle with $c^2 = 52.25$ , rounding $c^2$ to $52$ gives $\sqrt{52} = 7.21$ , but $\sqrt{52.25} = 7.228\ldots$ , which is $7.23$ to 3 s.f. Keep the full value and round once, at the end.
Mixing units. Change every length to the same unit before squaring.

Practice Questions

Try each question before you open its solution. Give lengths correct to 3 significant figures where they are not exact, unless the question says otherwise.

Question 1

In $\triangle ABC$ , $\angle B = 90^{\circ}$ , $AB = 28$ cm and $BC = 45$ cm. Find $AC$ .

ShowHide solution

$\angle B = 90^{\circ}$ , so the hypotenuse is $AC$ .

$AC^2 = AB^2 + BC^2 = 28^2 + 45^2 = 784 + 2025 = 2809$

$AC = \sqrt{2809} = 53$

$AC = 53$ cm

Question 2

Find $DF$ in the triangle drawn below.

ShowHide solution

The right angle is at $E$ , so the hypotenuse is $DF$ , the side along the bottom.

$DF^2 = DE^2 + EF^2 = 8^2 + 15^2 = 64 + 225 = 289$

$DF = \sqrt{289} = 17$

$DF = 17$ cm

Question 3

The two sides that form the right angle of a triangle are $4$ cm and $7$ cm. Find the hypotenuse.

ShowHide solution

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

$c = \sqrt{65} = 8.062\ldots = 8.06$ (3 s.f.)

$8.06$ cm

Question 4

In $\triangle XYZ$ , $\angle Y = 90^{\circ}$ , $XY = 35$ m and $YZ = 28$ m. Find $XZ$ .

ShowHide solution

$\angle Y = 90^{\circ}$ , so the hypotenuse is $XZ$ .

$XZ^2 = 35^2 + 28^2 = 1225 + 784 = 2009$

$XZ = \sqrt{2009} = 44.82\ldots = 44.8$ (3 s.f.)

$XZ = 44.8$ m

Question 5

In $\triangle PQR$ , $\angle R = 90^{\circ}$ , $PR = 2.4$ cm and $QR = 0.7$ cm. Find $PQ$ .

ShowHide solution

$\angle R = 90^{\circ}$ , so the hypotenuse is $PQ$ .

$PQ^2 = 2.4^2 + 0.7^2 = 5.76 + 0.49 = 6.25$

$PQ = \sqrt{6.25} = 2.5$ (exact)

$PQ = 2.5$ cm

Question 6

The two sides that form the right angle of a triangle are $6$ cm and $75$ mm. Find the hypotenuse in centimetres, correct to 1 decimal place.

ShowHide solution

$75$ mm $= 75 \div 10 = 7.5$ cm.

$c^2 = 6^2 + 7.5^2 = 36 + 56.25 = 92.25$

$c = \sqrt{92.25} = 9.604\ldots = 9.6$ (1 d.p.)

$9.6$ cm

Question 7

The two sides that form the right angle of a triangle are $10$ cm and $11$ cm. Find the hypotenuse correct to (a) 1 decimal place, (b) 2 decimal places, (c) 3 significant figures.

ShowHide solution

$c^2 = 10^2 + 11^2 = 100 + 121 = 221\text{,}$ so $c = \sqrt{221} = 14.866\ldots$

(a) $14.9$ (b) $14.87$ (c) $14.9$

(a) $14.9$ cm (b) $14.87$ cm (c) $14.9$ cm

Question 8

For a right-angled triangle whose other two sides are $1.5$ m and $2$ m, Farid gives the hypotenuse as $3.5$ m. (a) Explain why his answer is not reasonable. (b) Find the hypotenuse.

ShowHide solution

(a) $3.5$ m is $1.5 + 2$ : the two sides added together. The hypotenuse must be shorter than that.

(b) $c^2 = 1.5^2 + 2^2 = 2.25 + 4 = 6.25\text{,}$ so $c = \sqrt{6.25} = 2.5$ .

Check: $2.5$ m is more than $2$ m and less than $3.5$ m.

(a) $3.5$ m is the two sides added together, and the hypotenuse must be shorter than that. (b) $2.5$ m

Question 9

Hui Min works out the hypotenuse of a right-angled triangle whose other two sides are $6$ cm and $8$ cm. She writes $c^2 = 36 + 64 = 100$ , so the hypotenuse is $100$ cm. Explain her mistake and give the right answer.

ShowHide solution

$100$ is $c^2$ , the square of the hypotenuse, not the hypotenuse. She has not taken the square root.

Her answer is also unreasonable: $100$ cm is far longer than $6 + 8 = 14$ cm.

$c = \sqrt{100} = 10$

She forgot the square root. The hypotenuse is $10$ cm.

Question 10

A wheelchair ramp at the void deck of an HDB block rises $0.3$ m over a distance of $3.6$ m measured along the ground. Find the length of the ramp's sloping surface.

ShowHide solution

The sloping surface $AB$ is opposite the right angle at $C$ , so it is the hypotenuse.

$x^2 = 3.6^2 + 0.3^2 = 12.96 + 0.09 = 13.05$

$x = \sqrt{13.05} = 3.612\ldots = 3.61$ (3 s.f.)

Check: $3.61$ m is a little more than $3.6$ m, as expected for a gentle slope.

The sloping surface is $3.61$ m long.

Question 11

A wire is fixed to a flagpole $4.2$ m above the ground and pegged to the level ground $1.8$ m from the foot of the pole. The pole is upright. Find the length of the wire.

ShowHide solution

The pole is upright on level ground, so the angle at its foot is $90^{\circ}$ . The wire is opposite it: the hypotenuse.

$x^2 = 4.2^2 + 1.8^2 = 17.64 + 3.24 = 20.88$

$x = \sqrt{20.88} = 4.569\ldots = 4.57$ (3 s.f.)

The wire is $4.57$ m long.

Question 12

In $\triangle ABC$ , $\angle C = 90^{\circ}$ and $AC = BC = 5$ cm. Find $AB$ .

ShowHide solution

$AB^2 = AC^2 + BC^2 = 5^2 + 5^2 = 25 + 25 = 50$

$AB = \sqrt{50} = 7.071\ldots = 7.07$ (3 s.f.)

$AB = 7.07$ cm

Key Points

Name the hypotenuse first: the side opposite the right angle.
The square of the hypotenuse is the sum of the squares of the other two sides: square each side, add, then take the square root.
Keep the calculator's full value and round only at the last step; lengths are usually given to 3 significant figures.
Change all lengths to the same unit before squaring.
Check: the hypotenuse is longer than each of the other two sides and shorter than the two added together.