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Year 8 Maths Pythagoras

Finding the Hypotenuse

20 practice questions 2 video lessons Theory + worked examples
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Theory

The hypotenuse is the side opposite the right angle; it is always the longest side of a right-angled triangle. Pythagoras’ theorem says \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse. This Year 8 maths page shows how to find the hypotenuse as an exact answer, a surd or a rounded decimal.

Find the right angle first. The side that does not touch it is the hypotenuse, however the triangle is turned. It is the longest side.

Pythagoras’ theorem: the area of the square drawn on the hypotenuse equals the sum of the areas of the squares on the other two sides. With \(c\) as the hypotenuse, \(c^2 = a^2 + b^2\).

To find the hypotenuse: write \(c^2 = a^2 + b^2\), substitute the two shorter sides, add, then take the square root. A length is positive, so only the positive root is used.

Answer forms. If the root works out exactly (a whole number, or a decimal that stops) that is the exact answer. Otherwise the exact answer keeps the root sign, such as \(\sqrt{61}\), a surd; a rounded decimal is an approximation.

Estimate first. The hypotenuse is longer than each of the other two sides and shorter than the two added. Same units: change every length to one unit before you square. Scaling: sides \(15\) and \(20\) are \(5\) times \(3\) and \(4\), so the hypotenuse is \(5 \times 5 = 25\).

Not a triangle? Draw one inside the shape: a diagonal splits a rectangle in two; a line from a top corner of a right trapezium leaves a rectangle and a right-angled triangle. Where two right-angled triangles are joined, find the shared side in the first, then use it in the second.

Naming the sides of a right-angled triangle A right-angled triangle with the right angle at the bottom left. The shorter sides are a and b; the side opposite the right angle is c, the hypotenuse. a b c
The right angle is at the bottom left, so the hypotenuse \(c\) is the side opposite it, and it is the longest side.
Squares on the sides of a 3, 4, 5 triangle A right-angled triangle with sides 3, 4 and 5. Squares are drawn on each side: areas 9 and 16 on the shorter sides and 25 (red) on the hypotenuse. 16 9 25
A triangle with sides \(3\), \(4\) and \(5\). The square on the hypotenuse has area \(25\), and \(9 + 16 = 25\).

Pythagoras' theorem (c is the hypotenuse):

\[ c^2 = a^2 + b^2 \]
c 2= a 2+ b 2
\[ c = \sqrt{a^2 + b^2} \]
c =( a 2+ b 2)
Pythagorean triadCheck
\(3, 4, 5\)\(9 + 16 = 25\)
\(5, 12, 13\)\(25 + 144 = 169\)
\(8, 15, 17\)\(64 + 225 = 289\)
\(7, 24, 25\)\(49 + 576 = 625\)
Add, never subtract, for the hypotenuse. Check your answer: the hypotenuse must be longer than both of the other sides.
Square first, then add. \(c^2 = a^2 + b^2\) does not mean \(c = a + b\). Sides \(15\) and \(20\) give \(c = 25\), not \(35\).
Simplest form. Take out a square factor: \(\sqrt{500} = \sqrt{100 \times 5} = 10\sqrt{5}\).

How to find the hypotenuse

  1. Find the right angle; the side opposite it is \(c\).
  2. Change all lengths to the same unit.
  3. Write \(c^2 = a^2 + b^2\) and substitute.
  4. Add, then take the square root (exact, surd or rounded).
  5. Check: \(c\) is longer than each shorter side.
Example 1 — A whole-number answer
Find the hypotenuse.
Right-angled triangle with sides 21 cm and 28 cm A right-angled triangle with shorter sides 21 cm (upright) and 28 cm (base); the hypotenuse is c. 21 cm 28 cm c
Solution
\(c^2\)\(=\)\(21^2 + 28^2\)
\(\)\(=\)\(441 + 784\)
\(\)\(=\)\(1225\)
\(c\)\(=\)\(\sqrt{1225} = 35\)
c =441+784=1225, c =35

\(35\) cm. Check by scaling: \(21 = 3 \times 7\) and \(28 = 4 \times 7\), so \(c = 5 \times 7 = 35\).

Example 2 — Find the exact length
Find the exact length of the hypotenuse.
Right-angled triangle with sides 12 cm and 5 cm A right-angled triangle with an upright side of 12 cm and a base of 5 cm, right angle at the bottom right; the hypotenuse is x. 12 cm 5 cm x
Solution
\(x^2\)\(=\)\(12^2 + 5^2\)
\(\)\(=\)\(144 + 25 = 169\)
\(x\)\(=\)\(\sqrt{169} = 13\)
x =144+25=169, x =13

\(x = 13\) cm

Example 3 — Correct to 1 decimal place
Find the length of the hypotenuse correct to \(1\) decimal place.
Right-angled triangle with sides 5 cm and 6 cm A right-angled triangle with an upright side of 5 cm and a base of 6 cm; the hypotenuse is x. 5 cm 6 cm x
Solution
\(x^2\)\(=\)\(5^2 + 6^2\)
\(\)\(=\)\(25 + 36 = 61\)
\(x\)\(=\)\(\sqrt{61}\)
x =25+36=61, x =61≈7.8

\(x = \sqrt{61} = 7.81\ldots \approx 7.8\) cm

Example 4 — An answer as a surd
In \(\triangle PQR\) the right angle is at \(Q\), \(PQ = 5\) cm and \(QR = 8\) cm. Find \(PR\) as a surd.
Solution
\(PR^2\)\(=\)\(5^2 + 8^2\)
\(\)\(=\)\(25 + 64 = 89\)
PR =25+64=89, PR =89

\(81 < 89 < 100\), so \(89\) is not a square number: \(PR = \sqrt{89}\) cm (about \(9.4\) cm).

Example 5 — Estimate, then round
Find \(h\), correct to \(1\) decimal place.
Right-angled triangle with sides 9 m and 14 m A right-angled triangle with shorter sides 9 m and 14 m; the hypotenuse is h. 9 m 14 m h
Solution
\(h^2\)\(=\)\(9^2 + 14^2\)
\(\)\(=\)\(81 + 196 = 277\)
\(h\)\(=\)\(\sqrt{277} = 16.6433\ldots\)
h =81+196=277, h ≈16.6

Estimate: between \(14\) m and \(23\) m. \(h \approx 16.6\) m, which fits.

Example 6 — Two triangles in one figure
\(ABCD\) is made of two right-angled triangles, with the right angles at \(B\) and \(C\). All lengths are in cm.
(i) Find \(AC\).
(ii) Find \(AD\) and the perimeter of \(ABCD\).
Two right-angled triangles Quadrilateral ABCD split by AC. Triangle ABC has a right angle at B with AB 24 cm and BC 32 cm. Triangle ACD has a right angle at C with CD 9 cm. A B C D 24 32 9
Solution
\(AC^2\)\(=\)\(24^2 + 32^2\)
\(\)\(=\)\(1600\)
\(AD^2\)\(=\)\(40^2 + 9^2\)
\(\)\(=\)\(1681\)
\(24 + 32 + 9 + 41\)\(=\)\(106\)
AC =1600=40 AD =1681=41 perimeter 106

(i) \(AC = 40\) cm
(ii) \(AD = 41\) cm; perimeter \(106\) cm

Example 7 — Units and a trapezium
(i) A gate is \(1.5\) m wide and \(80\) cm tall. A brace runs from corner to corner. How long is the brace, in cm?
(ii) A right trapezium has parallel sides \(21\) cm and \(14\) cm and height \(24\) cm. Find its slanted side.
Solution

(i) Same units: \(1.5\) m \(= 150\) cm.
(ii) A line down from the top corner leaves a right-angled triangle with sides \(21 - 14 = 7\) cm and \(24\) cm.

\(\sqrt{150^2 + 80^2}\)\(=\)\(\sqrt{28\,900} = 170\)
\(\sqrt{7^2 + 24^2}\)\(=\)\(\sqrt{625} = 25\)
(22500+6400)=28900=170 cm (49+576)=25 cm

(i) \(170\) cm
(ii) \(25\) cm

Common pitfalls

Subtracting for the hypotenuse. The hypotenuse uses addition: c² = a² + b².
Forgetting the square root. If \(c^2 = 625\), then \(c = 25\), not \(625\).
Mixed units. Change \(1.5\) m to \(150\) cm before squaring.
Picking the wrong side. The hypotenuse is opposite the right angle, not always the slanted-looking side.

Frequently asked questions

What is the hypotenuse?

The side opposite the right angle in a right-angled triangle. It is the longest side.

What is Pythagoras' theorem?

In a right-angled triangle, \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse.

How do you find the hypotenuse?

Square the two shorter sides, add, then take the square root: \(\sqrt{5^2 + 12^2} = 13\).

When do you leave the answer as a surd?

When the question asks for an exact answer and the number under the root is not a square number, such as \(\sqrt{61}\).

How can you check your answer?

The hypotenuse must be longer than each of the other sides and shorter than the two added together.