Finding a Shorter Side
When the missing side is not the hypotenuse, Pythagoras' theorem is rearranged to \(b^2 = c^2 - a^2\). In NSW Year 8 maths you take the square of the known shorter side away from the square of the hypotenuse, then find the square root (NESA outcome MA4‑PYT‑C‑01).
You will pick out the hypotenuse first, find shorter sides with whole-number, decimal and surd answers, and round to one decimal place. You will check that each answer is shorter than the hypotenuse and fix a student's working that added instead of subtracting. The height of a playground slide, the width of a pool and the wires holding up a mast are all shorter sides. It is part of Year 8 Pythagoras.
Every question with a fully worked solution.
- Finding a Shorter Side - Video - Finding a Shorter Side - Example 1 Watch
Theory
When the unknown side is not the hypotenuse, subtract: take the square of the known shorter side away from the square of the hypotenuse, \(b^2 = c^2 - a^2\). This Year 8 maths page shows how to find a shorter side as a whole number, an exact surd or a rounded decimal.
The theorem does not change. Only the side you are chasing changes. Start from \(c^2 = a^2 + b^2\), where \(c\) is the hypotenuse, and take \(a^2\) from both sides: \(b^2 = c^2 - a^2\).
Two ways to write it. You can write the theorem first and then rearrange, \(x^2 + 24^2 = 25^2\), so \(x^2 = 25^2 - 24^2\). Or go straight to the subtraction. Both give the same answer.
Step 1. Find the hypotenuse: the side opposite the right angle, and the longest side. Step 2. Write \(b^2 = c^2 - a^2\) and substitute, with the hypotenuse squared first. Step 3. Subtract, then take the square root. A length is positive, so only the positive root is used.
Check. A shorter side must come out shorter than the hypotenuse. If it does not, you added instead of subtracting.
Answer forms. If the root works out exactly (a whole number, or a decimal that stops, such as \(\sqrt{110.25} = 10.5\)) that is the exact answer. If the decimal never stops, the exact answer keeps the root sign, a surd such as \(\sqrt{56}\), and a rounded decimal is given to the places the question asks for.
Scaling. A hypotenuse of \(45\) with a shorter side of \(27\) is \(9\) times \(5\) and \(3\), so the third side is \(9 \times 4 = 36\). Same units: change every length to one unit before you square.
A height inside a triangle. A line from a vertex at right angles to the opposite side splits the triangle into two right-angled triangles. Find the shared height in one, then use it in the other.
Pythagoras' theorem rearranged (c is the hypotenuse):
How to find a shorter side
- Find the right angle; the side opposite it is the hypotenuse \(c\).
- Change all lengths to the same unit.
- Write \(b^2 = c^2 - a^2\), hypotenuse squared first, and substitute.
- Subtract, then take the square root (exact, surd or rounded).
- Check: the answer is shorter than the hypotenuse.
How to use a height inside a triangle
- Find the height in the triangle where you know two sides.
- Use the height (or its square) in the second triangle.
- Add the two pieces of the base if the whole base is asked for.
| \(x^2 + 24^2\) | \(=\) | \(25^2\) |
| \(x^2\) | \(=\) | \(25^2 - 24^2\) |
| \(\) | \(=\) | \(625 - 576 = 49\) |
| \(x\) | \(=\) | \(\sqrt{49} = 7\) |
\(x = 7\) cm. Check: \(7 < 25\).
\(50\) m is opposite the right angle, so it is the hypotenuse. Subtract, hypotenuse first.
| \(x^2\) | \(=\) | \(50^2 - 14^2\) |
| \(\) | \(=\) | \(2500 - 196 = 2304\) |
| \(x\) | \(=\) | \(\sqrt{2304} = 48\) |
\(x = 48\) m. Check: \(48 < 50\).
| \(y^2\) | \(=\) | \(9^2 - 5^2\) |
| \(\) | \(=\) | \(81 - 25 = 56\) |
| \(y\) | \(=\) | \(\sqrt{56} = 7.4833\ldots\) |
\(y \approx 7.5\) cm
| \(w^2\) | \(=\) | \(8.5^2 - 4^2\) |
| \(\) | \(=\) | \(72.25 - 16 = 56.25\) |
| \(w\) | \(=\) | \(\sqrt{56.25} = 7.5\) |
\(56.25\) has two decimal places, so its root has one: \(w = 7.5\) m. Check: \(7.5 < 8.5\).
(i) exactly
(ii) correct to \(1\) decimal place.
| \(t^2\) | \(=\) | \(19^2 - 11^2\) |
| \(\) | \(=\) | \(361 - 121 = 240\) |
(i) \(15^2 = 225 < 240 < 256 = 16^2\), so \(240\) is not a square number: \(t = \sqrt{240}\) cm, a surd.
(ii) \(\sqrt{240} = 15.4919\ldots \approx 15.5\) cm
(i) How high does it reach?
(ii) The foot slips out to \(3.3\) m. How far does the top drop?
The ladder is opposite the right angle, so it is the hypotenuse both times.
| \(h^2\) | \(=\) | \(6.5^2 - 2.5^2\) |
| \(\) | \(=\) | \(42.25 - 6.25 = 36\) |
| \(h^2\) | \(=\) | \(6.5^2 - 3.3^2\) |
| \(\) | \(=\) | \(42.25 - 10.89 = 31.36\) |
(i) \(h = \sqrt{36} = 6\) m
(ii) Now \(h = \sqrt{31.36} = 5.6\) m, so the top drops \(6 - 5.6 = 0.4\) m.
(i) Find \(AD\).
(ii) Find \(DC\) and \(BC\).
| \(AD^2\) | \(=\) | \(13^2 - 5^2\) |
| \(\) | \(=\) | \(169 - 25 = 144\) |
| \(DC^2\) | \(=\) | \(37^2 - 12^2\) |
| \(\) | \(=\) | \(1369 - 144 = 1225\) |
(i) \(AD = \sqrt{144} = 12\) cm
(ii) \(DC = \sqrt{1225} = 35\) cm, so \(BC = 5 + 35 = 40\) cm
Common pitfalls
Frequently asked questions
How do you find a shorter side of a right-angled triangle?
Square the hypotenuse, subtract the square of the other side, then take the square root: \(\sqrt{13^2 - 5^2} = 12\).
Why do you subtract to find a shorter side?
Because \(c^2 = a^2 + b^2\), taking \(a^2\) from both sides gives \(b^2 = c^2 - a^2\).
Which number is squared first?
The hypotenuse, the longest side. Its square is the biggest, so the answer stays positive.
How do I know my answer is sensible?
A shorter side must be shorter than the hypotenuse.
When is the answer left as a surd?
When the number under the root is not a square number and an exact answer is asked for, such as \(\sqrt{56}\).