GeometryFoundationYear 8-10

The Pythagorean Theorem and Its Applications

In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the two shorter sides: a squared plus b squared equals c squared. Use it to find an unknown side when two sides are known, and to check whether a triangle is right-angled.

Dr. Pankaj Jha7 min readPublished 2026-08-28

Before you begin

You should be comfortable squaring numbers, finding square roots, and identifying the hypotenuse in a right-angled triangle.

The intuition

The theorem says the area of the square on the hypotenuse equals the combined area of the squares on the other two sides. This only works for right-angled triangles.

Definitions

  • Hypotenuse: the longest side, opposite the right angle.
  • Legs: the two shorter sides that form the right angle.
  • Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2, where cc is the hypotenuse.

Step-by-step method

  1. Check the triangle is right-angled.
  2. Identify the hypotenuse.
  3. Substitute the two known sides into a2+b2=c2a^2 + b^2 = c^2.
  4. Solve for the unknown side. Take the square root, keeping the exact form where required.

Worked examples

Foundation example

A right-angled triangle has legs 3 cm3\text{ cm} and 4 cm4\text{ cm}. Find the hypotenuse. c2=32+42=9+16=25c^2 = 3^2 + 4^2 = 9 + 16 = 25 c=25=5 cmc = \sqrt{25} = 5\text{ cm}

Developing example

A right-angled triangle has hypotenuse 13 cm13\text{ cm} and one leg 5 cm5\text{ cm}. Find the other leg. a2+52=132a^2 + 5^2 = 13^2 a2=169−25=144a^2 = 169 - 25 = 144 a=144=12 cma = \sqrt{144} = 12\text{ cm}

Advanced example

An isosceles right-angled triangle has two equal legs of length 6 cm6\text{ cm}. Find the hypotenuse in exact form. c2=62+62=36+36=72c^2 = 6^2 + 6^2 = 36 + 36 = 72 c=72=36×2=62 cmc = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\text{ cm}

Common mistakes

  • Using the hypotenuse as a leg: the hypotenuse is always the longest side, opposite the right angle.
  • Forgetting the square root: c2=25c^2 = 25 means c=5c = 5, not c=25c = 25.
  • Applying to non-right-angled triangles: the theorem only works when one angle is 90∘90^\circ.

Concept checklist

  • I can identify the hypotenuse.
  • I can find the hypotenuse or a leg from the other two sides.
  • I can leave answers in exact (surd) form.
  • I can check whether a triangle is right-angled.

Practice questions

  1. Find the hypotenuse of a right-angled triangle with legs 8 cm8\text{ cm} and 6 cm6\text{ cm}.
  2. Find the missing leg if the hypotenuse is 17 cm17\text{ cm} and one leg is 8 cm8\text{ cm}.
  3. Find the hypotenuse of an isosceles right-angled triangle with legs 5 cm5\text{ cm} each, in exact form.
  4. Check whether a triangle with sides 5,12,135, 12, 13 is right-angled.
  5. A ladder reaches 12 m12\text{ m} up a wall with its base 5 m5\text{ m} from the wall. Find the ladder length.
  6. Find the diagonal of a square with side 10 cm10\text{ cm} in exact form.

Challenge questions

  1. A rectangle has sides 4 cm4\text{ cm} and 9 cm9\text{ cm}. Find the exact length of its diagonal.
  2. Show that a triangle with sides 7,24,257, 24, 25 is right-angled and state which side is the hypotenuse.

Frequently asked questions

Does the theorem work for any triangle? No — only for right-angled triangles. For other triangles, use the cosine rule.

What is an exact form? An exact answer uses surds (like 626\sqrt{2}) or fractions instead of a rounded decimal.

  • Similarity, congruence and scale factors
  • Sine, cosine and tangent
  • Sine rule, cosine rule and triangle area
  • Understanding surds and simplifying radicals
Show worked solutions

Worked solutions

1. c2=64+36=100c^2 = 64 + 36 = 100, c=10 cmc = 10\text{ cm}. 2. a2=289−64=225a^2 = 289 - 64 = 225, a=15 cma = 15\text{ cm}. 3. c2=25+25=50c^2 = 25 + 25 = 50, c=50=52 cmc = \sqrt{50} = 5\sqrt{2}\text{ cm}. 4. 52+122=25+144=169=1325^2 + 12^2 = 25 + 144 = 169 = 13^2. Yes, right-angled (hypotenuse 1313). 5. c2=144+25=169c^2 = 144 + 25 = 169, c=13 mc = 13\text{ m}. 6. d2=100+100=200d^2 = 100 + 100 = 200, d=200=102 cmd = \sqrt{200} = 10\sqrt{2}\text{ cm}.

Challenge 1. d2=16+81=97d^2 = 16 + 81 = 97, d=97 cmd = \sqrt{97}\text{ cm}. Challenge 2. 72+242=49+576=625=2527^2 + 24^2 = 49 + 576 = 625 = 25^2. Right-angled; hypotenuse is 2525.

Mathematically reviewed by Dr. Pankaj Jha

Last reviewed: 2026-08-28 · Date modified: 2026-08-28

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