Finding a Shorter Side

When the hypotenuse and one other side of a right-angled triangle are known, Pythagoras' theorem gives the third side: subtract the squares, then take the square root.

Before you start

Pythagoras' theorem (lesson 9.1): in $\triangle ABC$ with $\angle C = 90^{\circ}$ , $AB^2 = AC^2 + BC^2$ , where $AB$ is the hypotenuse.
If $x^2 = 144$ and $x$ is a length, then $x = \sqrt{144} = 12$ ; the other solution, $-12$ , is not a length.
A quadratic equation such as $x^2 - 2x - 15 = 0$ is solved by factorising (chapter 5): $(x - 5)(x + 3) = 0$ , so $x = 5$ or $x = -3$ .

Rearranging the Theorem

The theorem says the square of the hypotenuse is the sum of the other two squares. So when the hypotenuse and one other side are known, the square of the missing side is what is left when the known square is taken away from the square of the hypotenuse.

Finding a shorter side

$(\text{shorter side})^2 = (\text{hypotenuse})^2 - (\text{other side})^2$

In $\triangle ABC$ with $\angle C = 90^{\circ}$ : $AC^2 = AB^2 - BC^2$ and $BC^2 = AB^2 - AC^2$ .

The square of the hypotenuse always comes first: take the other square away from it.

Example 1: A whole-number answer

In $\triangle ABC$ , $\angle B = 90^{\circ}$ , $AC = 13$ cm and $BC = 5$ cm. Find $AB$ .

Solution

$\angle B = 90^{\circ}$ , so the hypotenuse is $AC = 13$ cm. The side to find, $AB$ , is a shorter side.

$AB^2 = AC^2 - BC^2 = 13^2 - 5^2 = 169 - 25 = 144$

$AB = \sqrt{144} = 12$

Check: $12$ cm is shorter than the hypotenuse, $13$ cm, and $5^2 + 12^2 = 169 = 13^2\text{.}$

$AB = 12$ cm

Another way is to write the theorem as it stands, put in the numbers, and solve the equation for the unknown. It gives the same working, and it is a safe way when you are unsure which square to subtract.

Example 2: Solving an equation

In $\triangle PQR$ , $\angle Q = 90^{\circ}$ , $PR = 25$ cm and $PQ = 15$ cm. Find $QR$ .

Solution

$\angle Q = 90^{\circ}$ , so the hypotenuse is $PR$ , and $PR^2 = PQ^2 + QR^2$ .

$25^2 = 15^2 + QR^2$

$625 = 225 + QR^2$

$QR^2 = 625 - 225 = 400$

$QR = \sqrt{400} = 20$

$QR = 20$ cm

Example 3: An answer to 3 significant figures

In $\triangle XYZ$ , $\angle X = 90^{\circ}$ , $YZ = 25$ cm and $XY = 17$ cm. Find $XZ$ , correct to 3 significant figures.

Solution

$\angle X = 90^{\circ}$ , so the hypotenuse is $YZ = 25$ cm.

$XZ^2 = YZ^2 - XY^2 = 25^2 - 17^2 = 625 - 289 = 336$

$XZ = \sqrt{336} = 18.330\ldots = 18.3$ (3 s.f.)

Check: $18.3$ cm is shorter than the hypotenuse, $25$ cm.

$XZ = 18.3$ cm (to 3 s.f.)

Is the Answer Reasonable?

A quick check

A shorter side must come out shorter than the hypotenuse.

An answer longer than the hypotenuse is wrong: most often the squares were added instead of subtracted, or the square root was forgotten. A negative number under the square root means they were subtracted the wrong way round.

Example 4: Spot the error

A right-angled triangle has hypotenuse $13$ cm and another side $5$ cm. (a) Aisyah writes $x^2 = 13^2 + 5^2 = 194$ , so $x = 13.9$ cm. (b) Ben writes $x^2 = 5^2 - 13^2$ . Explain each mistake.

Solution

(a) $13.9$ cm is longer than the hypotenuse, $13$ cm, which is impossible. Aisyah added the squares; she should have subtracted.

(b) $5^2 - 13^2 = 25 - 169 = -144\text{.}$ The square of a length cannot be negative. Ben took the square of the hypotenuse away from the smaller square, the wrong way round.

Correctly, $x^2 = 13^2 - 5^2 = 169 - 25 = 144\text{,}$ so $x = 12$ .

(a) She added instead of subtracting. (b) He subtracted the wrong way round. The side is $12$ cm.

Sides Given in Algebra

Sometimes the sides are given in terms of an unknown $x$ . Write the theorem with the hypotenuse on its own, simplify, and solve. Find the sides from $x$ only at the end, and round only the final answers.

Example 5: One side twice the other

The hypotenuse of a right-angled triangle is $60$ cm. One of the other sides is twice as long as the third side. Find the lengths of the two shorter sides, correct to 1 decimal place.

Solution

Let the shorter side be $x$ cm. The other is $2x$ cm.

$x^2 + (2x)^2 = 60^2$

$x^2 + 4x^2 = 3600$ , so $5x^2 = 3600$

$x^2 = 720$ , so $x = \sqrt{720} = 26.832\ldots$

The longer side is $2x = 2 \times 26.832\ldots = 53.665\ldots$

To 1 d.p.: $x = 26.8$ and $2x = 53.7$ . Doubling the rounded $26.8$ would give $53.6$ , which is wrong: double the full value, then round.

The two shorter sides are $26.8$ cm and $53.7$ cm.

Example 6: Sides $x$ , $x + 7$ and $x + 8$

The sides of a right-angled triangle are $x$ cm, $(x + 7)$ cm and $(x + 8)$ cm. Form an equation in $x$ , solve it, and find the three sides.

Solution

The longest side, $(x + 8)$ cm, is the hypotenuse.

$x^2 + (x + 7)^2 = (x + 8)^2$

$x^2 + x^2 + 14x + 49 = x^2 + 16x + 64$

$x^2 - 2x - 15 = 0$

$(x - 5)(x + 3) = 0$ , so $x = 5$ or $x = -3$ .

A length cannot be negative, so $x = 5$ . The sides are $5$ cm, $5 + 7 = 12$ cm and $5 + 8 = 13$ cm.

Check: $5^2 + 12^2 = 25 + 144 = 169 = 13^2\text{.}$

$x = 5$ ; the sides are $5$ cm, $12$ cm and $13$ cm.

Common Mistakes

Adding instead of subtracting. For hypotenuse $13$ cm and side $5$ cm, $13^2 + 5^2 = 194$ gives $13.9$ cm, longer than the hypotenuse. Take the smaller square away: $169 - 25 = 144$ .
Subtracting the wrong way round. $5^2 - 13^2$ is negative. The square of the hypotenuse always comes first.
Using a rounded value again. In Example 5, $2 \times 26.8 = 53.6$ is wrong; $2 \times 26.832\ldots = 53.7\text{.}$ Work with the full value and round only the final answers.
Keeping a negative solution. $x = -3$ solves the equation, but a side cannot be $-3$ cm.

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 $\triangle PQR$ , $\angle R = 90^{\circ}$ , $PQ = 17$ m and $QR = 8$ m. Find $PR$ .

ShowHide solution

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

$PR^2 = PQ^2 - QR^2 = 17^2 - 8^2 = 289 - 64 = 225$

$PR = \sqrt{225} = 15$

$PR = 15$ m

Question 2

Find $x$ in the triangle drawn below.

ShowHide solution

The right angle is at $E$ , so the hypotenuse is $DF = 10$ cm.

$x^2 = 10^2 - 6^2 = 100 - 36 = 64$

$x = \sqrt{64} = 8$

$x = 8$

Question 3

The hypotenuse of a right-angled triangle is $12$ cm and one of the other sides is $7$ cm. Find the third side.

ShowHide solution

$x^2 = 12^2 - 7^2 = 144 - 49 = 95$

$x = \sqrt{95} = 9.746\ldots = 9.75$ (3 s.f.)

Check: $9.75$ cm is shorter than $12$ cm.

$9.75$ cm

Question 4

In $\triangle KLM$ , $\angle L = 90^{\circ}$ , $KM = 6.5$ m and $KL = 2.5$ m. Find $LM$ .

ShowHide solution

$\angle L = 90^{\circ}$ , so the hypotenuse is $KM$ .

$LM^2 = KM^2 - KL^2 = 6.5^2 - 2.5^2 = 42.25 - 6.25 = 36$

$LM = \sqrt{36} = 6$

$LM = 6$ m

Question 5

The hypotenuse of a right-angled triangle is $1.3$ m and one of the other sides is $50$ cm. Find the third side in centimetres.

ShowHide solution

$1.3$ m $= 1.3 \times 100 = 130$ cm.

$x^2 = 130^2 - 50^2 = 16\,900 - 2500 = 14\,400$

$x = \sqrt{14\,400} = 120$

$120$ cm

Question 6

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $AB = 41$ cm and $AC = 9$ cm. Write Pythagoras' theorem for the triangle, put in the lengths, and solve for $BC$ .

ShowHide solution

$AB^2 = AC^2 + BC^2$

$41^2 = 9^2 + BC^2$

$1681 = 81 + BC^2$

$BC^2 = 1681 - 81 = 1600\text{,}$ so $BC = \sqrt{1600} = 40$

$BC = 40$ cm

Question 7

A right-angled triangle has hypotenuse $15$ cm and another side $9$ cm. Jun Hao finds the third side to be $17.5$ cm. Without repeating his working, explain why he must be wrong. Then find the third side.

ShowHide solution

$17.5$ cm is longer than the hypotenuse, $15$ cm. No side of a right-angled triangle is longer than the hypotenuse. (He added the squares: $\sqrt{225 + 81} = \sqrt{306} = 17.49\ldots\text{)}$

$x^2 = 15^2 - 9^2 = 225 - 81 = 144\text{,}$ so $x = 12$ .

His side is longer than the hypotenuse. The third side is $12$ cm.

Question 8

Mei flies a kite at East Coast Park. The string is $30$ m long and pulled straight. The kite is directly above a point on the ground $18$ m from the end of the string, which Mei holds at ground level. How high is the kite above the ground?

ShowHide solution

The kite is directly above $G$ , so $\angle G = 90^{\circ}$ and the string $MK$ is the hypotenuse.

$h^2 = 30^2 - 18^2 = 900 - 324 = 576$

$h = \sqrt{576} = 24$

The kite is $24$ m above the ground.

Question 9

A straight playground slide is $3.4$ m long. Its foot is $3$ m, measured along the level ground, from the point directly below its top. How high is the top of the slide above the ground?

ShowHide solution

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

$h^2 = 3.4^2 - 3^2 = 11.56 - 9 = 2.56$

$h = \sqrt{2.56} = 1.6$

The top of the slide is $1.6$ m above the ground.

Question 10

The hypotenuse of a right-angled triangle is $20$ cm. One of the other sides is three times as long as the third side. Find the two shorter sides.

ShowHide solution

Let the shorter side be $x$ cm; the other is $3x$ cm.

$x^2 + (3x)^2 = 20^2$ , so $10x^2 = 400$ and $x^2 = 40$ .

$x = \sqrt{40} = 6.324\ldots = 6.32$ (3 s.f.)

$3x = 3 \times 6.324\ldots = 18.973\ldots = 19.0$ (3 s.f.)

$6.32$ cm and $19.0$ cm

Question 11

The sides of a right-angled triangle are $x$ cm, $(x + 2)$ cm and $(x + 4)$ cm. Find $x$ and the three sides.

ShowHide solution

The hypotenuse is the longest side, $(x + 4)$ cm.

$x^2 + (x + 2)^2 = (x + 4)^2$

$x^2 + x^2 + 4x + 4 = x^2 + 8x + 16$

$x^2 - 4x - 12 = 0$ , so $(x - 6)(x + 2) = 0$ and $x = 6$ or $x = -2$ .

A length is positive, so $x = 6$ . Check: $6^2 + 8^2 = 36 + 64 = 100 = 10^2\text{.}$

$x = 6$ ; the sides are $6$ cm, $8$ cm and $10$ cm.

Question 12

In a right-angled triangle the hypotenuse is $(x + 1)$ cm and the other two sides are $x$ cm and $7$ cm. Find $x$ .

ShowHide solution

$x^2 + 7^2 = (x + 1)^2$

$x^2 + 49 = x^2 + 2x + 1$

The $x^2$ terms cancel: $49 = 2x + 1$ , so $2x = 48$ and $x = 24$ .

Check: the sides are $7$ , $24$ and $25$ , and $49 + 576 = 625 = 25^2\text{.}$

$x = 24$

Key Points

To find a shorter side, take the square of the other side away from the square of the hypotenuse, then take the square root.
Or write the theorem with the hypotenuse on its own, put in the lengths, and solve.
A shorter side is always less than the hypotenuse; an answer that is longer is wrong, most often because the squares were added or the square root was forgotten.
When the sides are in terms of $x$ , the longest expression is the hypotenuse; reject a negative solution.
Work with full values and round only the final answers: $2 \times 26.832\ldots = 53.7\text{,}$ not $2 \times 26.8 = 53.6$ .

Before you start

Rearranging the Theorem

Finding a shorter side

Example 1: A whole-number answer

Example 2: Solving an equation

Example 3: An answer to 3 significant figures

Is the Answer Reasonable?

A quick check

Example 4: Spot the error

Sides Given in Algebra

Example 5: One side twice the other

Example 6: Sides xxx, x+7x + 7x+7 and x+8x + 8x+8

Common Mistakes

Practice Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Key Points

Example 6: Sides $x$ , $x + 7$ and $x + 8$