Is the Triangle Right-Angled?

The converse of Pythagoras' theorem: from the three sides alone, decide whether a triangle has a right angle, and if it has, which angle it is.

Before you start

Pythagoras' theorem (lesson 9.1): in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. In $\triangle ABC$ with $\angle C = 90^{\circ}$ , $AB^2 = AC^2 + BC^2$ .
The hypotenuse is the longest side, and it is opposite the right angle.

The Converse of the Theorem

Pythagoras' theorem starts from a right angle and tells us something about the sides. Its converse goes the other way: it starts from the sides and tells us about the angle. Both statements are true.

Pythagoras' theorem and its converse

Pythagoras' theorem: if a triangle has a right angle, then the square of the longest side equals the sum of the squares of the other two sides.

The converse: if the square of the longest side of a triangle equals the sum of the squares of the other two sides, then the triangle is right-angled, and the right angle is opposite the longest side.

So if the square of the longest side is not equal to the sum of the other two squares, the triangle is not right-angled: a right-angled triangle would have made them equal.

Testing a triangle

1. Find the longest side and square it.

2. Square the other two sides and add.

3. Equal: the triangle is right-angled, and the right angle is opposite the longest side. Not equal: it is not right-angled.

Example 1: Right-angled or not?

Is a triangle with these sides right-angled? (a) $9$ cm, $12$ cm, $15$ cm (b) $10$ cm, $11$ cm, $12$ cm (c) $20$ m, $21$ m, $29$ m

Solution

(a) The longest side is $15$ cm. $15^2 = 225$ , and $9^2 + 12^2 = 81 + 144 = 225\text{.}$ They are equal.

(b) The longest side is $12$ cm. $12^2 = 144$ , and $10^2 + 11^2 = 100 + 121 = 221\text{.}$ $221 \neq 144$ .

(a) Right-angled (b) Not right-angled (c) Right-angled

Example 2: Which angle is the right angle?

(a) In $\triangle ABC$ , $AB = 26$ cm, $BC = 10$ cm and $AC = 24$ cm. (b) In $\triangle PQR$ , $PQ = 17$ m, $QR = 12$ m and $PR = 11$ m. Is each triangle right-angled? If it is, name the right angle.

Solution

(a) The longest side is $AB = 26$ cm. $AB^2 = 26^2 = 676$ , and $BC^2 + AC^2 = 10^2 + 24^2 = 100 + 576 = 676\text{.}$

They are equal, so $\triangle ABC$ is right-angled. The right angle is opposite $AB$ , the side that does not touch $C$ : $\angle ACB = 90^{\circ}$ .

(b) The longest side is $PQ = 17$ m. $PQ^2 = 17^2 = 289$ , and $QR^2 + PR^2 = 12^2 + 11^2 = 144 + 121 = 265\text{.}$

$265 \neq 289$ , so $\triangle PQR$ is not right-angled. (Its angle at $R$ looks close to a right angle, but it is not one: a drawing cannot decide this.)

(a) Yes: $\angle ACB = 90^{\circ}$ . (b) No, $\triangle PQR$ is not right-angled.

Decimals, Mixed Units, and Close Is Not Equal

The test works the same way with decimals. With mixed units, change every side to one unit first. And the two numbers must be exactly equal: nearly equal is not enough.

Example 3: Sides with decimals

Is a triangle with these sides right-angled? (a) $0.7$ m, $2.4$ m, $2.5$ m (b) $0.9$ m, $1.2$ m, $1.6$ m

Solution

(a) $2.5^2 = 6.25$ , and $0.7^2 + 2.4^2 = 0.49 + 5.76 = 6.25\text{.}$ Equal: right-angled.

(b) $1.6^2 = 2.56$ , and $0.9^2 + 1.2^2 = 0.81 + 1.44 = 2.25\text{.}$ $2.25 \neq 2.56$ : not right-angled.

(a) Right-angled (b) Not right-angled

Example 4: Sides in different units

A triangle has sides of $30$ cm, $0.4$ m and $500$ mm. Is it right-angled?

Solution

Change every side to centimetres: $0.4$ m $= 40$ cm and $500$ mm $= 50$ cm. The sides are $30$ cm, $40$ cm and $50$ cm.

The longest is $50$ cm. $50^2 = 2500$ , and $30^2 + 40^2 = 900 + 1600 = 2500\text{.}$ Equal.

Without changing the units, $500$ would wrongly be taken as the longest side and $500^2$ compared with $30^2 + 0.4^2$ , which tells us nothing.

Yes. The right angle is opposite the $500$ mm side.

Example 5: Close is not equal

A triangle has sides of $8$ cm, $15$ cm and $16.9$ cm. Is it right-angled?

Solution

The longest side is $16.9$ cm. $16.9^2 = 285.61$ , and $8^2 + 15^2 = 64 + 225 = 289\text{.}$

$285.61$ and $289$ are close, but they are not equal. The triangle is not right-angled. (The triangle with sides $8$ , $15$ and $17$ is right-angled; this one is slightly different, and its largest angle is a little less than $90^{\circ}$ .)

No, it is not right-angled, because $8^2 + 15^2 \neq 16.9^2$ .

Using the Converse

Builders and carpenters use the converse to check a corner: they measure along the two edges and across, and test the three lengths. It also finds right angles hidden inside other figures.

Example 6: Checking a corner

A carpenter making a rectangular frame measures $90$ cm along one edge from a corner, $120$ cm along the other edge from the same corner, and $150$ cm straight across between the two marks. Is the corner a right angle?

Solution

The three lengths form a triangle. The longest is $150$ cm: $150^2 = 22\,500$ .

$90^2 + 120^2 = 8100 + 14\,400 = 22\,500\text{.}$ Equal.

So the triangle is right-angled, with the right angle opposite the $150$ cm side: at the corner of the frame.

Yes, the corner is a right angle.

Example 7: A right angle inside a quadrilateral

In the quadrilateral $ABCD$ , $AB = 3$ cm, $BC = 4$ cm, $CD = 12$ cm, $DA = 13$ cm and $\angle ABC = 90^{\circ}$ . (a) Find $AC$ . (b) Show that $\angle ACD = 90^{\circ}$ . (c) Find the area of $ABCD$ .

Solution

(a) In $\triangle ABC$ , $\angle B = 90^{\circ}$ , so by Pythagoras' theorem $AC^2 = 3^2 + 4^2 = 9 + 16 = 25$ and $AC = 5$ cm.

(b) In $\triangle ACD$ the sides are $AC = 5$ cm, $CD = 12$ cm and $DA = 13$ cm. The longest is $DA$ : $13^2 = 169$ , and $5^2 + 12^2 = 25 + 144 = 169\text{.}$

By the converse, $\triangle ACD$ is right-angled, with the right angle opposite $DA$ : $\angle ACD = 90^{\circ}$ .

(c) Area of $\triangle ABC = \frac{1}{2} \times 3 \times 4 = 6$ cm². In $\triangle ACD$ the sides $AC$ and $CD$ meet at the right angle, so its area is $\frac{1}{2} \times 5 \times 12 = 30$ cm².

Area of $ABCD = 6 + 30 = 36$ cm².

(a) $AC = 5$ cm (b) $5^2 + 12^2 = 13^2$ , so $\angle ACD = 90^{\circ}$ . (c) $36$ cm²

Common Mistakes

Not using the longest side. For $5$ , $12$ , $13$ , testing $13^2 + 5^2 = 194$ against $12^2 = 144$ wrongly says "not right-angled". Always compare the square of the longest side with the other two squares added.
Naming the wrong angle. The right angle is opposite the longest side. If $AB$ is longest, the right angle is $\angle C$ , the angle that does not touch $AB$ .
Deciding from the look of a drawing. $\triangle PQR$ in Example 2 looks almost right-angled and is not. Only the numbers decide.
Accepting close as equal. $289$ and $285.61$ are not equal.
Mixing units. Change all three sides to one unit before squaring.

Practice Questions

Try each question before you open its solution. Square the longest side, add the squares of the other two, and compare.

Question 1

Is a triangle with these sides right-angled? (a) $7$ cm, $24$ cm, $25$ cm (b) $6$ cm, $8$ cm, $11$ cm (c) $12$ m, $35$ m, $37$ m

ShowHide solution

(a) $25^2 = 625$ ; $7^2 + 24^2 = 49 + 576 = 625\text{.}$ Equal.

(b) $11^2 = 121$ ; $6^2 + 8^2 = 36 + 64 = 100\text{.}$ Not equal.

(a) Yes (b) No (c) Yes

Question 2

In $\triangle DEF$ , $DE = 33$ cm, $EF = 56$ cm and $DF = 65$ cm. Show that the triangle is right-angled and name the right angle.

ShowHide solution

The longest side is $DF$ : $DF^2 = 65^2 = 4225$ , and $DE^2 + EF^2 = 33^2 + 56^2 = 1089 + 3136 = 4225\text{.}$

Equal, so the triangle is right-angled, with the right angle opposite $DF$ : at $E$ .

$\angle DEF = 90^{\circ}$

Question 3

In $\triangle LMN$ , $LM = 39$ cm, $MN = 15$ cm and $LN = 36$ cm. Is the triangle right-angled? If so, which angle is $90^{\circ}$ ?

ShowHide solution

The longest side is $LM$ : $LM^2 = 39^2 = 1521$ .

$MN^2 + LN^2 = 15^2 + 36^2 = 225 + 1296 = 1521\text{.}$ Equal.

The right angle is opposite $LM$ : the angle at $N$ .

Yes: $\angle LNM = 90^{\circ}$ .

Question 4

In $\triangle XYZ$ , $XY = 20$ m, $YZ = 28$ m and $XZ = 19$ m. Is the triangle right-angled?

ShowHide solution

The longest side is $YZ$ : $YZ^2 = 28^2 = 784$ .

$XY^2 + XZ^2 = 20^2 + 19^2 = 400 + 361 = 761\text{.}$

$761 \neq 784$ .

No, $\triangle XYZ$ is not right-angled.

Question 5

Is a triangle with these sides right-angled? (a) $0.6$ m, $0.8$ m, $1$ m (b) $1.1$ cm, $6$ cm, $6.1$ cm

ShowHide solution

(a) $1^2 = 1$ ; $0.6^2 + 0.8^2 = 0.36 + 0.64 = 1\text{.}$ Equal.

(b) $6.1^2 = 37.21$ ; $1.1^2 + 6^2 = 1.21 + 36 = 37.21\text{.}$ Equal.

(a) Yes (b) Yes

Question 6

A triangle has sides of $0.5$ m, $120$ cm and $1.3$ m. Is it right-angled? If so, which side is opposite the right angle?

ShowHide solution

In centimetres the sides are $50$ cm, $120$ cm and $130$ cm.

$130^2 = 16\,900$ , and $50^2 + 120^2 = 2500 + 14\,400 = 16\,900\text{.}$ Equal.

Yes. The right angle is opposite the $1.3$ m side.

Question 7

A triangle has sides of $10$ cm, $24$ cm and $26.1$ cm. Siew Ling says it is right-angled because $10$ , $24$ , $26$ is a Pythagorean triple. Is she right?

ShowHide solution

The longest side is $26.1$ cm: $26.1^2 = 681.21$ .

$10^2 + 24^2 = 100 + 576 = 676\text{.}$

$676 \neq 681.21$ : close, but not equal. The sides are not $10$ , $24$ and $26$ .

No. The triangle is not right-angled.

Question 8

To test a triangle with sides $20$ cm, $48$ cm and $52$ cm, Ravi works out $52^2 + 20^2 = 3104$ and $48^2 = 2304$ , and says the triangle is not right-angled. Explain his mistake and give the correct conclusion.

ShowHide solution

He must compare the square of the longest side, $52$ cm, with the sum of the squares of the other two. He used $48$ cm as the side on its own.

$52^2 = 2704$ , and $20^2 + 48^2 = 400 + 2304 = 2704\text{.}$ Equal.

He did not use the longest side. The triangle is right-angled, with the right angle opposite the $52$ cm side.

Question 9

A builder checks that a wall meets the floor at a right angle. He marks a point $60$ cm up the wall and a point on the floor $45$ cm from the foot of the wall, and measures $75$ cm between the two marks. Does the wall meet the floor at a right angle? Give a reason.

ShowHide solution

The three lengths form a triangle with sides $45$ cm, $60$ cm and $75$ cm. The longest is $75$ cm.

$75^2 = 5625$ , and $45^2 + 60^2 = 2025 + 3600 = 5625\text{.}$ Equal.

By the converse of Pythagoras' theorem, the triangle is right-angled, with the right angle opposite the $75$ cm side: where the wall meets the floor.

Yes, the wall meets the floor at a right angle.

Question 10

At a community garden, Mr Lim ties a loop of rope with knots that divide it into $12$ equal parts, each $52$ cm long. He pegs it out as a triangle with sides of $3$ , $4$ and $5$ parts. (a) Find the length of each side of the triangle. (b) Show that the triangle has a right angle.

ShowHide solution

(a) $3 \times 52 = 156$ cm, $4 \times 52 = 208$ cm and $5 \times 52 = 260$ cm.

(b) $260^2 = 67\,600$ , and $156^2 + 208^2 = 24\,336 + 43\,264 = 67\,600\text{.}$ Equal, so the triangle is right-angled, with the right angle opposite the $260$ cm side.

(Each side is $52$ times $3$ , $4$ or $5$ , and $3^2 + 4^2 = 5^2$ , which is why it works for any length of part.)

(a) $156$ cm, $208$ cm, $260$ cm (b) $156^2 + 208^2 = 260^2$

Question 11

In the quadrilateral $PQRS$ , $PQ = 3$ cm, $QR = 6$ cm, $RS = 6$ cm, $SP = 9$ cm and $\angle PQR = 90^{\circ}$ . (a) Find $PR$ . (b) Show that $\angle PRS = 90^{\circ}$ . (c) Find the area of $PQRS$ .

ShowHide solution

(a) $PR^2 = 3^2 + 6^2 = 9 + 36 = 45\text{,}$ so $PR = \sqrt{45} = 6.708\ldots = 6.71$ cm (3 s.f.).

(b) In $\triangle PRS$ : $SP^2 = 9^2 = 81$ , and $PR^2 + RS^2 = 45 + 36 = 81\text{.}$ Equal, so the right angle is opposite $SP$ : $\angle PRS = 90^{\circ}$ . Use $PR^2 = 45$ itself: the rounded $6.71^2 = 45.02\ldots$ would not match exactly.

(c) Area $= \frac{1}{2} \times 3 \times 6 + \frac{1}{2} \times 6.708\ldots \times 6 = 9 + 20.12\ldots = 29.12\ldots = 29.1$ cm² (3 s.f.).

(a) $6.71$ cm (b) $PR^2 + RS^2 = 45 + 36 = 81 = SP^2$ (c) $29.1$ cm²

Question 12

A triangle has sides of $6$ cm, $8$ cm and $x$ cm, where $x$ cm is the longest side. Find the value of $x$ for which the triangle is right-angled.

ShowHide solution

The triangle is right-angled exactly when $x^2 = 6^2 + 8^2 = 36 + 64 = 100\text{.}$

$x = \sqrt{100} = 10$

$x = 10$

Key Points

Pythagoras' theorem: in a right-angled triangle, the square of the longest side equals the sum of the squares of the other two. Its converse: if they are equal, the triangle is right-angled. Both are true.
To test a triangle, square the longest side and compare it with the sum of the squares of the other two.
Equal: right-angled, with the right angle opposite the longest side. Not equal: not right-angled.
Change all sides to one unit first; close is not equal; a drawing never decides.
The converse checks corners in building, and finds right angles inside other figures.