Mathematics notes, chapter 13

The Pythagoras' Theorem

Textbook Mathematics Form 1 (DLP), pages 292–305

What this chapter covers

Identify and define the hypotenuse of a right-angled triangle (13.1)
Determine the relationship between the sides of a right-angled triangle and explain the Pythagoras' theorem using the areas of squares (13.1)
Determine the length of an unknown side of a right-angled triangle and of combined geometric shapes (13.1)
Solve problems involving the Pythagoras' theorem (13.1)
Determine whether a triangle is a right-angled triangle and justify it with the converse of the Pythagoras' theorem (13.2)
Solve problems involving the converse of the Pythagoras' theorem (13.2)

Right angles are everywhere: the corners of a room, a door frame, a ladder against a wall. Builders and civil engineers use the Pythagoras' theorem to make sure the corners of a building are true right angles (textbook p. 292). This chapter has two parts: the theorem itself, and its converse.

13.1The Pythagoras' theorem

The hypotenuse

The hypotenuse is the longest side of a right-angled triangle. It is always the side opposite the right angle.

Three right-angled triangles in different positions. In each one the side opposite the right angle is drawn in orange and labelled hypotenuse — However the triangle is turned, the hypotenuse is the side that does **not** touch the right angle (textbook p. 294).

To find the hypotenuse, put your finger on the right-angle box, then look straight across the triangle. The side you see is the hypotenuse.

Calling the longest side of any triangle a hypotenuse. Only a right-angled triangle has a hypotenuse. In the triangle below, side a is the longest side but it is not a hypotenuse, because the angle opposite it is 95°, not 90°.

A triangle with an angle of 95°; the side a opposite it is the longest side

Example 1 (textbook p. 295)

Quadrilateral PSRQ split by the line PR into triangle PSR with a right angle at S and triangle PRQ with a right angle at R

Name the hypotenuse of triangle PSR and of triangle PRQ.
In triangle PSR the right angle is at S. The side opposite S is PR.
In triangle PRQ the right angle is at R. The side opposite R is PQ.

When two triangles share a side, look at one triangle at a time. PR is the hypotenuse of triangle PSR but only a shorter side of triangle PRQ.

Triangle XYZ has a right angle at Y. (a) Name the hypotenuse. (b) Which angle is opposite the hypotenuse?

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(a) XZ (the side that does not touch Y)
(b) ∠XYZ, the right angle

The relationship between the sides

Draw a square on each side of a right-angled triangle with sides 3, 4 and 5 units. Count the unit squares.

A right-angled triangle with sides 3, 4 and 5 units. A 3 by 3 square (9 unit squares) is drawn on the side of 3, a 4 by 4 square (16 unit squares) on the side of 4, and a 5 by 5 square (25 unit squares) on the hypotenuse — 9 + 16 = 25: the two small squares together fill the big square exactly (textbook p. 296).

Pythagoras' theorem: in a right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
If c is the hypotenuse and a and b are the other two sides: c² = a² + b²

Example 2 (textbook p. 296): state the relationship

(a) Triangle PQR with a right angle at Q. (b) Triangle with sides a and b meeting at the right angle and hypotenuse c

(a) The right angle is at Q, so the hypotenuse is PR: PR² = QR² + PQ²
(b) The hypotenuse is c: c² = a² + b²

Writing the hypotenuse on the wrong side, e.g. QR² = PR² + PQ². The hypotenuse is always alone on one side of the equals sign.

Writing c = a + b. The theorem is about the squares of the sides: 5 = 3 + 4 is false, but 5² = 3² + 4² is true.

Three whole numbers that fit c² = a² + b² are called a Pythagorean triple. Knowing a few saves time (textbook p. 297).

a	b	c	a² + b²	c²
3	4	5	9 + 16 = 25	25
5	12	13	25 + 144 = 169	169
6	8	10	36 + 64 = 100	100
7	24	25	49 + 576 = 625	625
8	15	17	64 + 225 = 289	289
9	12	15	81 + 144 = 225	225

Multiples of a triple are triples too: 3, 4, 5 → 6, 8, 10 → 9, 12, 15 → 15, 20, 25.

Finding an unknown side

If you know two sides of a right-angled triangle, the theorem gives the third side. First decide: is the unknown side the hypotenuse or not?

Finding	Do this	Formula
the hypotenuse (longest side)	square, add, square root	c = √(a² + b²)
a shorter side	square, subtract, square root	a = √(c² − b²)

Example 3 (textbook p. 297): find x

(a) Right-angled triangle with sides 12 cm and 16 cm and hypotenuse x cm. (b) Right-angled triangle with hypotenuse 26 cm, a side of 24 cm and a side of x cm

(a) x is the hypotenuse, so add.
x² = 12² + 16²
     = 144 + 256
     = 400
x = √400 = 20
(b) 26 cm is the hypotenuse, so subtract.
26² = x² + 24²
x² = 26² − 24²
     = 676 − 576
     = 100
x = √100 = 10

Forgetting the last step. x² = 400 means x = √400 = 20, not 400.

Adding when the unknown is a shorter side. In (b), √(26² + 24²) = 35.38 is longer than the hypotenuse, which is impossible. Check: the hypotenuse must be the longest side.

Combined shapes

Some diagrams have two right-angled triangles joined together. Find the shared side first, then use it in the second triangle.

Example 4 (textbook p. 297): calculate the length of PQ

(a) Triangle RSP with RS = 3 cm, SP = 4 cm and a right angle at S, joined to triangle PRQ with RQ = 12 cm and a right angle at R — (a)

(b) Triangle SRQ with SR = 8 cm, RQ = 6 cm and a right angle at R, joined to triangle SQP with SP = 15 cm and a right angle at Q — (b)

(a) Triangle PSR: PR² = 3² + 4² = 9 + 16 = 25, so PR = 5 cm (you can leave it as 25 for now).
Triangle PRQ: PQ² = PR² + RQ² = 25 + 12² = 25 + 144 = 169
PQ = √169 = 13 cm
(b) Triangle SRQ: QS² = 6² + 8² = 36 + 64 = 100
Triangle SQP: SP is the hypotenuse, so PQ² = SP² − QS² = 15² − 100 = 225 − 100 = 125
PQ = √125 = 11.18 cm (2 decimal places)

Keep the squared value (like QS² = 100) and use it directly in the next triangle. It saves a step and avoids rounding errors.

Calculate x. Give your answer correct to two decimal places if necessary.
(a) The two shorter sides are 10 cm and 24 cm; the hypotenuse is x cm.
(b) The hypotenuse is 15 cm and one side is 9 cm; the other side is x cm.
(c) The two shorter sides are 4 cm and 7 cm; the hypotenuse is x cm.

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(a) x² = 10² + 24² = 100 + 576 = 676, x = √676 = 26
(b) x² = 15² − 9² = 225 − 81 = 144, x = 12
(c) x² = 4² + 7² = 16 + 49 = 65, x = √65 = 8.06

Solving problems

Draw (or find) the right-angled triangle in the story, mark the right angle, label the sides you know, and decide whether the unknown is the hypotenuse.

Example 5 (textbook p. 298): the fireman's ladder

A child is trapped on the third floor, 6 m above the ground. The base of the ladder is 4.5 m from the wall. How long is the ladder?

A ladder leaning against a wall. The top of the ladder is 6 m above the ground and its base is 4.5 m from the wall; the wall meets the ground at a right angle

Understand: height 6 m, distance from the wall 4.5 m, find the ladder.
Plan: the wall and the ground meet at a right angle, so the ladder is the hypotenuse.
Solve: ladder² = 4.5² + 6² = 20.25 + 36 = 56.25, ladder = √56.25 = 7.5 m
Check: 7.5² = 56.25 ✓, and 7.5 m is longer than 6 m and 4.5 m ✓

Example 6 (textbook p. 299): perimeter of a combined shape

PVRQ and RUTS are squares with sides 12 cm and 16 cm. Calculate the perimeter of the whole diagram.

Square PVRQ of side 12 cm and square RUTS of side 16 cm meeting at R. VU is joined, forming a right-angled triangle VRU

VR = 12 cm and RU = 16 cm, and ∠VRU = 90°.
VU² = 12² + 16² = 144 + 256 = 400, VU = 20 cm
Perimeter = VU + UT + TS + SR + RQ + QP + PV = 20 + 16 + 16 + 16 + 12 + 12 + 12 = 104 cm

Adding the inside lines VR and RU to the perimeter. The perimeter is only the outside boundary.

Ship A is 34 km north of ship B. Ship C is 10 km west of ship A. Calculate the distance between ship B and ship C, correct to two decimal places.

B at the bottom, A 34 km north of B, C 10 km west of A; a dashed line joins B and C

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North and west are at right angles, so ∠BAC = 90° and BC is the hypotenuse.
BC² = 34² + 10² = 1 156 + 100 = 1 256
BC = √1 256 = 35.44 km

A 7.2 m rope is tied to the tip of a flag pole and to a spot on the ground 4.5 m from the base of the pole. Calculate the height of the pole, correct to two decimal places.

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The rope is the hypotenuse.
height² = 7.2² − 4.5² = 51.84 − 20.25 = 31.59
height = √31.59 = 5.62 m

13.2The converse of Pythagoras' theorem

The theorem says: right angle → c² = a² + b². The converse works the other way round: if the sides fit c² = a² + b², the triangle must have a right angle. So we can test a triangle using only its side lengths (textbook p. 300).

Three triangles with sides 6 and 8 and a third side c. c = 9: 81 is less than 100, acute angle opposite c. c = 10: 100 equals 100, right angle opposite c. c = 12: 144 is greater than 100, obtuse angle opposite c — Keep the sides 6 and 8 and change the longest side c. Only c = 10 gives a right angle.

Compare (c = longest side)	Angle opposite c
c² < a² + b²	acute (less than 90°)
c² = a² + b²	right angle (90°)
c² > a² + b²	obtuse (between 90° and 180°)

Converse of Pythagoras' theorem: if c² = a² + b², then the angle opposite side c is a right angle.

Example 7 (textbook p. 301): is it a right-angled triangle?

(a) Triangle DEF with DE = 24 cm, EF = 7 cm and DF = 25 cm. (b) Triangle STU with ST = 12 cm, TU = 18 cm and SU = 20 cm

(a) The longest side is 25 cm. 25² = 625 and 24² + 7² = 576 + 49 = 625
625 = 625, so DEF is a right-angled triangle (the right angle is at E, opposite DF).
(b) The longest side is 20 cm. 20² = 400 and 18² + 12² = 324 + 144 = 468
400 ≠ 468, so STU is not a right-angled triangle.

Testing with the wrong side as c. Always square the longest side on its own. For 27 m, 45 m, 35 m, the longest side is 45 m, not the last number written.

Deciding by looking. A triangle that "looks" right-angled may not be. You must show the calculation and write a conclusion.

Example 8 (textbook p. 301): find x

Triangle PQS with PQ = 21 cm, QS = 20 cm and PS = 29 cm, joined to triangle SQR. Angle QSR = 28°, angle PQR = 155° below Q, and x at R

20² + 21² = 400 + 441 = 841 = 29², so by the converse, ∠PQS = 90°.
Angles at a point add up to 360°: ∠SQR = 360° − 90° − 155° = 115°
Angles in triangle SQR: x = 180° − 115° − 28° = 37°

Example 9 (textbook p. 302): Sheila's straws

Sheila has straws 15 cm, 20 cm and 25 cm long. Can she make a right-angled triangle frame?
15² + 20² = 225 + 400 = 625 and 25² = 625. They are equal, so yes: the angle opposite the 25 cm straw is a right angle.

Builders use this to check a corner: mark 3 m along one wall and 4 m along the other. If the diagonal between the marks is exactly 5 m, the corner is a right angle.

Can these lengths form a right-angled triangle? Show your working.
(a) 9 cm, 40 cm, 41 cm (b) 27 m, 45 m, 35 m (c) 2.5 cm, 6 cm, 6.5 cm
(d) A 2.5 m ladder leans against a wall. Its base is 1.5 m from the wall and its top is 2 m up the wall. Is the wall vertical?

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(a) 41² = 1 681; 9² + 40² = 81 + 1 600 = 1 681. Equal, so yes.
(b) 45² = 2 025; 27² + 35² = 729 + 1 225 = 1 954. Not equal, so no (the angle opposite 45 m is obtuse).
(c) 6.5² = 42.25; 2.5² + 6² = 6.25 + 36 = 42.25. Equal, so yes.
(d) 2.5² = 6.25; 1.5² + 2² = 2.25 + 4 = 6.25. Equal, so the wall makes a right angle with the ground: it is vertical.

Summary

	Pythagoras' theorem	Converse of Pythagoras' theorem
You know	The triangle has a right angle	The three side lengths
You find	The length of a missing side	Whether there is a right angle
Rule	c² = a² + b² (c = hypotenuse)	If c² = a² + b², the angle opposite c is 90°
Example	Sides 6 and 8: c = √100 = 10	15, 20, 25: 225 + 400 = 625 = 25², right angle

Remember	Key fact
Hypotenuse	Longest side, opposite the right angle
Finding the hypotenuse	Square, add, square root
Finding a shorter side	Square, subtract, square root
Combined shapes	Find the shared side first
Converse test	Square the longest side alone; compare with the sum of the other two squares
Check	The hypotenuse must be longer than each of the other sides