POPStudying

Cambridge 0580 IGCSE  ·  Topic C6.1

C6.1 Pythagoras' Theorem

Image  ·  914 × 270

01 — Explanation

What Does Pythagoras' Theorem Say?

In any right-angled triangle, the square of the hypotenuse cc (the longest side, always opposite the right angle) equals the sum of the squares of the other two sides:

a2+b2=c2a^2+b^2=c^2

This only ever applies to right-angled triangles, and it relates lengths — never angles.

Exam Tip

Always check the diagram for a right-angle marker before reaching for Pythagoras — applying it to a triangle that isn't right-angled gives a confidently wrong answer.

02 — Explanation

How Do I Find the Hypotenuse?

Add the squares of the two shorter sides, then square root:

c=a2+b2c = \sqrt{a^2+b^2}

A right-angled triangle has shorter sides 6 cm6\text{ cm} and 8 cm8\text{ cm}:

c=62+82=36+64=100=10 cmc = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10\text{ cm}

(6,8,10)(6,8,10) is one of several "Pythagorean triples" — sets of three whole numbers that satisfy a2+b2=c2a^2+b^2=c^2 exactly. Recognising common triples like (3,4,5)(3,4,5) and its multiples can be a useful sanity check on a calculated answer.

03 — Explanation

How Do I Find One of the Shorter Sides?

Subtract the known shorter side's square from the hypotenuse's square, then square root — the opposite operation to finding the hypotenuse:

a=c2−b2a = \sqrt{c^2-b^2}

A triangle has hypotenuse 13 cm13\text{ cm} and one shorter side 5 cm5\text{ cm}:

x=132−52=169−25=144=12 cmx = \sqrt{13^2-5^2} = \sqrt{169-25} = \sqrt{144} = 12\text{ cm}

Exam Tip

Automatically adding, regardless of which side is missing, is the most common error here. Add only when the hypotenuse is the unknown; subtract (larger square minus smaller) whenever a shorter side is the unknown.

04 — Explanation

How Do I Apply This to a Real-Life Problem?

Identify which given length plays the role of the hypotenuse before doing anything else — in context questions, it's often described rather than obviously labelled.

A ladder 4.5 m4.5\text{ m} long leans against a wall, its foot 1.2 m1.2\text{ m} from the base. How high up the wall does it reach, to 3 significant figures?

  1. The ladder is the hypotenuse; the distance from the wall is one shorter side; the height is the unknown shorter side.
  2. Apply the subtraction rule:
h=4.52−1.22=20.25−1.44=18.81h = \sqrt{4.5^2 - 1.2^2} = \sqrt{20.25-1.44} = \sqrt{18.81}
  1. Evaluate:
h=4.3372…≈4.34 m(3 s.f.)h = 4.3372\ldots \approx 4.34\text{ m} \quad (3\text{ s.f.})

05 — Explanation

How Do I Use Pythagoras With an Isosceles Triangle?

An isosceles triangle can be split into two identical right-angled triangles by drawing a line from its apex straight down to the midpoint of its base — this perpendicular line becomes a shared shorter side, and half of the original base becomes the other.

An isosceles triangle has two equal sides of 10 cm10\text{ cm} and a base of 12 cm12\text{ cm}. Find its height.

  1. Splitting the base in half gives two right-angled triangles, each with hypotenuse 10 cm10\text{ cm} and one shorter side 6 cm6\text{ cm} (half of 1212).
  2. h=102−62=100−36=64=8 cmh = \sqrt{10^2-6^2} = \sqrt{100-36} = \sqrt{64} = 8\text{ cm}

Exam Tip

It's the base that gets halved to form the right-angled triangle, not the equal sloping sides — halving the wrong length is a common slip in this style of question.

06 — Explanation

Exam-Style Question

A rectangular field measures 24 m24\text{ m} by 10 m10\text{ m}. Find the length of its diagonal.

Solution

The diagonal of a rectangle is the hypotenuse of the right-angled triangle formed by two adjacent sides:

diagonal=242+102=576+100=676=26 m\text{diagonal} = \sqrt{24^2+10^2} = \sqrt{576+100} = \sqrt{676} = 26\text{ m}

Don't stop at 676676 — the theorem gives the squared length first, so the final square root step is essential.