Pythagoras’ Theorem

Finding missing lengths in right-angled triangles

GCSE Geometry And Measures
Pythagoras’ Theorem – cover
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Pythagoras’ Theorem

Finding lengths in right-angled triangles

Introduction

Pythagoras’ Theorem is one of the most useful results in GCSE Geometry & Measures. It lets you find the missing side of a right-angled triangle when you know the other two. You’ll meet it in questions about ladders against walls, ramps, diagonals of rectangles, roofs, navigation, and even coordinate geometry.

The big idea is a relationship between the areas of squares drawn on the three sides of a right-angled triangle. For the longest side (the hypotenuse) the area of its square equals the sum of the areas of the squares on the other two sides. In number form, that becomes a very famous formula you’ll use over and over again.

Why it matters: It turns “can’t measure directly” problems into simple calculations. If a triangle has a right angle, you can connect the three side lengths with one neat equation.

Key Vocabulary

  • Right-angled triangle — a triangle with one 90° angle.
  • Hypotenuse — the side opposite the right angle. It is the longest side.
  • Legs (or perpendicular sides) — the two sides that meet at the right angle.
  • Pythagoras’ Theorem — in any right-angled triangle with legs \(a\) and \(b\) and hypotenuse \(c\): \(\;a^2 + b^2 = c^2\).
  • Square root — the value which, when squared, gives the original number; e.g. \(\sqrt{25}=5\).
  • Exact form — leaving answers with a surd (e.g. \(\sqrt{50}=5\sqrt{2}\)) instead of a rounded decimal.
  • Rounding — giving an approximate value to a required accuracy (e.g. 1 dp, 2 dp, or to the nearest mm).

Core Ideas

  • If a triangle is right-angled with legs \(a\), \(b\) and hypotenuse \(c\), then \(\;a^2 + b^2 = c^2\).
  • To find the hypotenuse \(c\): \(c = \sqrt{a^2+b^2}\).
  • To find a leg when the hypotenuse and the other leg are known, rearrange: \(a = \sqrt{c^2-b^2}\) or \(b = \sqrt{c^2-a^2}\).
  • The theorem only works for right-angled triangles. Check for a 90° angle before using it.
  • In coordinate geometry, the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) — this is Pythagoras in disguise.
  • Answers can be left in exact surd form (especially in Higher) or rounded to the accuracy requested.

Step-by-Step Method

  1. Check it’s right-angled. Look for a small square marking at the angle or the word “right-angled”.
  2. Label the sides. Mark the hypotenuse \(c\) (opposite the right angle). The other two are legs \(a\) and \(b\).
  3. Write the formula. \(a^2 + b^2 = c^2\).
  4. Substitute the known values.
  5. Rearrange (if you’re finding a leg, move terms across).
  6. Square root at the end to find the unknown length.
  7. Units and rounding. Attach the correct units (cm, m, km) and round as required (e.g. 1 dp).
Calculator note: If allowed, use your calculator for squaring and square-roots. If a non-calculator paper asks for exact value, simplify the surd (e.g. \(\sqrt{72}=6\sqrt{2}\)).

Worked Examples — Foundation

Example F1 — Find the hypotenuse

[diagram: right-angled triangle with legs 6 cm and 8 cm]

A right-angled triangle has legs 6 cm and 8 cm. Find the hypotenuse.

\(c=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ cm}\).

Example F2 — Find a missing leg

[diagram: right-angled triangle with hypotenuse 13 cm and one leg 5 cm]

The hypotenuse is 13 cm and one leg is 5 cm. Find the other leg.

Let the missing leg be \(b\). Then \(5^2+b^2=13^2 \Rightarrow 25+b^2=169 \Rightarrow b^2=144 \Rightarrow b=12\text{ cm}\).

Example F3 — Diagonal of a rectangle

[diagram: rectangle 9 cm by 12 cm; diagonal drawn]

Find the diagonal of a 9 cm by 12 cm rectangle.

The diagonal is the hypotenuse of a right-angled triangle with legs 9 and 12: \(d=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\text{ cm}\).

Example F4 — Ladder against a wall

[diagram: wall vertical, ground horizontal, ladder length unknown as hypotenuse; foot 1.5 m from wall, top 3.6 m high]

A ladder reaches 3.6 m up a wall. Its foot is 1.5 m from the wall. How long is the ladder?

\(L=\sqrt{3.6^2+1.5^2}=\sqrt{12.96+2.25}=\sqrt{15.21}\approx 3.90\text{ m}\) (to 2 dp).

Worked Examples — Higher

Example H1 — Exact surd answer

[diagram: right-angled triangle with legs 7 cm and 3 cm]

Find the exact length of the hypotenuse when the legs are 7 cm and 3 cm.

\(c=\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58}\). 58 has no square factors \(>1\), so the exact answer is \(\sqrt{58}\) cm.

Example H2 — Embedded triangle (roof)

[diagram: isosceles roof cross-section forming two congruent right-angled triangles; half-span 4 m, rise 1.8 m]

A symmetrical roof has a span of 8 m and a rise (vertical height) of 1.8 m at the centre. Find the rafter length.

Half-span is 4 m. Rafter is hypotenuse: \(r=\sqrt{4^2+1.8^2}=\sqrt{16+3.24}=\sqrt{19.24}\approx 4.39\text{ m}\) (2 dp).

Example H3 — Distance on a coordinate grid

[diagram: coordinate plane; points A(−2, 3), B(5, −1)]

Find the distance \(AB\) between \(A(-2,3)\) and \(B(5,-1)\).

Horizontal change \(=5-(-2)=7\). Vertical change \(=-1-3=-4\). \(AB=\sqrt{7^2+(-4)^2}=\sqrt{49+16}=\sqrt{65}\) (exact), \(\approx 8.06\) (2 dp).

Common Mistakes & Fixes

  • Using the wrong side as hypotenuse. Fix: The hypotenuse is always opposite the right angle and is the longest side.
  • Forgetting to square before adding. Fix: Do \(a^2+b^2\), not \(a+b\).
  • Square root too early. Fix: Add the squares first, then square-root once at the end.
  • Applying Pythagoras to non-right triangles. Fix: Only use when there’s a 90° angle (or you can split into right triangles).
  • Rounding errors. Fix: Keep extra decimals in working; round only in the final answer to the stated accuracy.
  • Units missing or mixed. Fix: Convert to the same units before calculating and attach the unit at the end.

Practice Questions — Foundation

  1. [diagram: right-angled triangle] Legs are 5 cm and 12 cm. Find the hypotenuse (to 1 dp).
  2. [diagram: right-angled triangle] Hypotenuse is 10 cm, one leg is 6 cm. Find the other leg.
  3. Find the diagonal of a 7 cm by 24 cm rectangle.
  4. A ramp rises 0.9 m over a horizontal distance of 2.4 m. Find the ramp length (to 2 dp).
  5. A square has side 11 cm. Find the length of its diagonal (exact form and 2 dp).
  6. A TV measures 40 inches diagonally. If its height is 19.6 inches, find its width (to 1 dp).
  7. A path goes 300 m east then 400 m north. How far is the finishing point from the start (straight-line) in metres?
  8. A right-angled triangle has hypotenuse 26 cm and one leg 24 cm. Find the other leg.

Practice Questions — Higher

  1. Give an exact surd: legs 2 cm and 8 cm. Find the hypotenuse.
  2. The points \(P(3,5)\) and \(Q(-1,-4)\). Find \(PQ\) in exact form and as a decimal (2 dp).
  3. A rectangular field is 52 m by 120 m. A fence runs along the diagonal. Find its length (nearest metre).
  4. A triangle has sides 9 cm, 12 cm, and \(x\) cm. If the triangle is right-angled with hypotenuse \(x\), find \(x\).
  5. A staircase has a total rise of 2.8 m and horizontal run of 3.6 m. What length of stringer (the side support) is needed? Give 2 dp.
  6. A phone screen is a 16:9 rectangle with 6.0 inch diagonal. Find its width and height to 1 dp.
  7. The vector from \(A(1,2)\) to \(B(10,8)\) is \((9,6)\). Use Pythagoras to find \(|\overrightarrow{AB}|\).

Challenge Questions

  1. A 13 m ladder must clear a 2.5 m high garden wall that is 4.0 m from the house. Will the ladder reach the house roof if the foot is at the wall? If so, how high up the house will it reach (2 dp)?
  2. A square tile has side \(a\). A smaller square is cut from one corner so that its diagonal equals \(a\). Find the side length of the small square in exact form.
  3. On a coordinate grid, triangle \(ABC\) has \(A(0,0)\), \(B(12,0)\), and \(C(12,5)\). (a) Show that \(\angle ABC\) is a right angle. (b) Find the length \(AC\). (c) A point \(D\) lies directly above \(A\) so that \(AD=13\). Find the length \(CD\) (exact form).

Quick Revision Sheet

Key Formulas

  • \(a^2+b^2=c^2\) (right-angled triangles only).
  • \(c=\sqrt{a^2+b^2}\) (to find hypotenuse).
  • \(a=\sqrt{c^2-b^2}\) (to find a leg).
  • Distance formula: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).

Steps

  1. Confirm the right angle.
  2. Identify the hypotenuse (opposite the right angle).
  3. Substitute into \(a^2+b^2=c^2\).
  4. Add, then square-root once at the end.
  5. Round sensibly and include units.

Common Traps

  • Using a non-right triangle.
  • Forgetting to square first.
  • Square-rooting each term separately.
  • Rounding too early.

Conclusion & Next Steps

Pythagoras’ Theorem unlocks missing lengths in right-angled situations: ladders, ramps, diagonals and distances on grids. Next, you’ll use it with trigonometry (SOH-CAH-TOA) to find missing angles and to solve more complex mixed problems.

  • Next topics: Right-angle trigonometry, 3D Pythagoras (space diagonals), similar triangles.
  • Exam habit: Sketch a quick diagram and mark the right angle and hypotenuse before calculating.

Exam Tips

  • Highlight the right angle and circle the hypotenuse first.
  • Write the formula before numbers: \(a^2+b^2=c^2\).
  • Only round at the end; keep at least 3–4 dp in the calculator during working.
  • Give exact surds when asked for “exact value”.
  • Check your answer is sensible: the hypotenuse should be the longest side.

Answer Key

Foundation Answers

  1. \(\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13.0\text{ cm}\) (1 dp: 13.0 cm).
  2. \(b=\sqrt{10^2-6^2}=\sqrt{100-36}=\sqrt{64}=8\text{ cm}\).
  3. \(\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25\text{ cm}\).
  4. \(\sqrt{0.9^2+2.4^2}=\sqrt{0.81+5.76}=\sqrt{6.57}\approx 2.56\text{ m}\) (2 dp).
  5. Diagonal \(=11\sqrt{2}\text{ cm}\) (exact); \(\approx 15.56\text{ cm}\) (2 dp).
  6. \(w=\sqrt{40^2-19.6^2}=\sqrt{1600-384.16}=\sqrt{1215.84}\approx 34.86\) inches (1 dp: 34.9 in).
  7. \(\sqrt{300^2+400^2}=\sqrt{90000+160000}=\sqrt{250000}=500\text{ m}\).
  8. \(\sqrt{26^2-24^2}=\sqrt{676-576}=\sqrt{100}=10\text{ cm}\).

Higher Answers

  1. \(\sqrt{2^2+8^2}=\sqrt{4+64}=\sqrt{68}=2\sqrt{17}\) (exact).
  2. \(PQ=\sqrt{(−1-3)^2+(−4-5)^2}=\sqrt{(−4)^2+(−9)^2}=\sqrt{16+81}=\sqrt{97}\approx 9.85\) (2 dp).
  3. \(\sqrt{52^2+120^2}=\sqrt{2704+14400}=\sqrt{17104}\approx 130.8\text{ m}\) (nearest metre: 131 m).
  4. Right-angled with hypotenuse \(x\): \(x^2=9^2+12^2=81+144=225\Rightarrow x=15\text{ cm}\).
  5. \(\sqrt{2.8^2+3.6^2}=\sqrt{7.84+12.96}=\sqrt{20.8}\approx 4.56\text{ m}\) (2 dp).
  6. Let scale factor \(k\): width \(=16k\), height \(=9k\), diagonal \(=\sqrt{(16k)^2+(9k)^2}=k\sqrt{256+81}=k\sqrt{337}=6\). So \(k=6/\sqrt{337}\). Width \(=16k=\dfrac{96}{\sqrt{337}}\approx 5.23\) in; height \(=9k=\dfrac{54}{\sqrt{337}}\approx 2.94\) in (1 dp: 5.2 in by 2.9 in).
  7. \(|\overrightarrow{AB}|=\sqrt{9^2+6^2}=\sqrt{81+36}=\sqrt{117}=3\sqrt{13}\approx 10.82\).

Challenge Answers (outlines)

  1. Foot at wall means right triangle with one leg 4.0 m (to the wall) and ladder 13 m as hypotenuse. Height above ground at the wall: \(\sqrt{13^2-4^2}=\sqrt{169-16}=\sqrt{153}\approx 12.37\text{ m}\). Since the wall is only 2.5 m high, the ladder easily clears it and reaches the house at about 12.37 m up (measured vertically from ground). (Interpretation: the “house side” height reached is \(\approx 12.37\) m.)
  2. If the small square’s diagonal equals \(a\), then its side is \(a/\sqrt{2}=\dfrac{a\sqrt{2}}{2}\) (exact).
  3. (a) \(\overrightarrow{BA}=(0-12,0-0)=(-12,0)\), \(\overrightarrow{BC}=(12-12,5-0)=(0,5)\). Dot product \(=0\Rightarrow\) right angle. (b) \(AC=\sqrt{(12-0)^2+(5-0)^2}=\sqrt{144+25}=\sqrt{169}=13\). (c) \(D(0,13)\). Then \(CD=\sqrt{(12-0)^2+(5-13)^2}=\sqrt{144+64}=\sqrt{208}=4\sqrt{13}\) (exact).
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