Finding the Shorter Side
When the hypotenuse is known, a shorter side is found by subtracting: \(a^2 = c^2 - b^2\). This NSW Year 9 Advanced maths subtopic revises that Stage 4 skill (NESA outcome MA4‑PYT‑C‑01) and uses it inside other shapes.
You will find shorter sides in triangles drawn in any position, give answers rounded or as simplified surds such as \(4\sqrt{6}\), and solve problems about a ladder, a gate brace and a TV screen. Harder questions find the area of an isosceles triangle, a diagonal of a kite, the side of an equilateral triangle from its height, and an unknown \(x\) when the hypotenuse is \(x + 8\). It is part of Year 9 Advanced Pythagoras Theorem.
Every question with a fully worked solution.
- Finding the Shorter Side - Video - Pythagoras Theorem Finding Shorter Sides Watch
Theory
To find a shorter side of a right-angled triangle, subtract: \(a^2 = c^2 - b^2\), where \(c\) is the hypotenuse. This Year 9 Advanced maths page revises the earlier method and extends it to exact surd answers, the height of an isosceles triangle, kites, and equations where a side is written with a pronumeral.
A right-angled triangle has one hypotenuse, the side opposite the right angle, and two shorter sides. When the hypotenuse is known and one shorter side is missing, Pythagoras' theorem is used backwards.
Start from \(c^2 = a^2 + b^2\) and subtract \(b^2\) from both sides: \(a^2 = c^2 - b^2\). The hypotenuse squared always comes first, so the answer is positive and smaller than the hypotenuse.
Many shapes contain a right-angled triangle with a known hypotenuse: a ladder against a wall, the diagonal of a rectangle, the equal sides of an isosceles triangle once its height is drawn, and the sides of a kite cut by its diagonals.
At this level the sides can be surds, such as \(4\sqrt{3}\), or expressions, such as \(x + 9\). The theorem then becomes an equation to solve.
With \(c\) the hypotenuse, subtract to find a shorter side:
Surd sides square to whole numbers, and a bracket squares in full:
How to find a shorter side
- Find the hypotenuse: the side opposite the right angle, even when the triangle is turned. In a word problem it is the slanting ladder, brace, wire or diagonal.
- Subtract the square of the known shorter side from the square of the hypotenuse: \(a^2 = c^2 - b^2\).
- Square root and give the form asked: whole number, simplest surd or rounded decimal.
- Use the side if the question goes on: a perimeter, an area, or the other part of a kite diagonal.
- With pronumerals, write the theorem as an equation, expand any brackets and solve. When two triangles share a side, write that side squared both ways and make the two expressions equal.
(i) as a surd in simplest form
(ii) correct to \(2\) decimal places.
The right angle is at \(V\), so the hypotenuse is \(UW = 11\). \(VW\) is a shorter side, so subtract.
| \(VW^2\) | \(=\) | \(UW^2 - UV^2\) |
| \(\) | \(=\) | \(11^2 - 5^2\) |
| \(\) | \(=\) | \(121 - 25\) |
| \(\) | \(=\) | \(96\) |
(i) The largest square factor of \(96\) is \(16\).
| \(VW\) | \(=\) | \(\sqrt{96}\) |
| \(\) | \(=\) | \(\sqrt{16 \times 6}\) |
| \(\) | \(=\) | \(\mathbf{4\sqrt{6}}\) |
(ii) \(\sqrt{96} = 9.7979\ldots\), so \(VW \approx \mathbf{9.80}\). Keep the zero: \(2\) decimal places means two digits after the point.
The area needs the height. The height from the top vertex meets the base at right angles and cuts it in half, so each half of the base is \(7\) cm.
| \(h^2\) | \(=\) | \(25^2 - 7^2\) |
| \(\) | \(=\) | \(625 - 49\) |
| \(\) | \(=\) | \(576\) |
| \(h\) | \(=\) | \(24\ \text{cm}\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 14 \times 24\) |
| \(\) | \(=\) | \(\mathbf{168}\ \text{cm}^2\) |
The equal side \(25\) cm is the hypotenuse of each half, not the base.
Write the theorem with the hypotenuse on its own, then expand \((x + 9)^2\) in full.
| \((x + 9)^2\) | \(=\) | \(x^2 + 15^2\) |
| \(x^2 + 18x + 81\) | \(=\) | \(x^2 + 225\) |
| \(18x + 81\) | \(=\) | \(225\) |
| \(18x\) | \(=\) | \(144\) |
| \(x\) | \(=\) | \(\mathbf{8}\) |
The \(x^2\) terms cancel, leaving a linear equation. Check: the sides are \(8\), \(15\) and \(17\), and \(64 + 225 = 289 = 17^2\).
\(AD\) splits the triangle into two right-angled triangles. Let \(BD = x\), so \(DC = 21 - x\). Write \(h^2\) from each triangle and make them equal.
| \(h^2\) | \(=\) | \(17^2 - x^2\) |
| \(h^2\) | \(=\) | \(10^2 - (21 - x)^2\) |
| \(289 - x^2\) | \(=\) | \(100 - (441 - 42x + x^2)\) |
| \(289 - x^2\) | \(=\) | \(-341 + 42x - x^2\) |
| \(630\) | \(=\) | \(42x\) |
| \(x\) | \(=\) | \(15\) |
| \(h^2\) | \(=\) | \(17^2 - 15^2\) |
| \(\) | \(=\) | \(289 - 225 = 64\) |
| \(h\) | \(=\) | \(\mathbf{8}\) |
Common pitfalls
Frequently asked questions
How do you find the shorter side of a right-angled triangle?
Square the hypotenuse, subtract the square of the other shorter side, then take the square root. For a hypotenuse of \(11\) and a side of \(5\): \(121 - 25 = 96\), and \(\sqrt{96} = 4\sqrt{6} \approx 9.80\).
Why do you subtract to find a shorter side?
Because \(c^2 = a^2 + b^2\). Taking \(b^2\) from both sides leaves \(a^2 = c^2 - b^2\).
How do you find the height of an isosceles triangle?
Draw the height from the top vertex to the base. It cuts the base in half at right angles. Use Pythagoras with the equal side as the hypotenuse and half the base as the other side.
How do you use Pythagoras when a side is x + 9?
Write the theorem with the hypotenuse on its own, expand the brackets, and solve. The \(x^2\) terms often cancel, leaving a linear equation.
Can a shorter side be longer than the hypotenuse?
No. The hypotenuse is always the longest side. If your answer for a shorter side is longer than the hypotenuse, check whether you added the squares instead of subtracting.