Draw it, and find the right angle.
The wall meets the ground at 90°, so this is a right triangle.
Which side is which
The hypotenuse is always opposite the right angle and always the longest. Here that is the ladder, 13 m — not the wall.
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The Pythagorean Theorem
Find the hypotenuse by adding. Find a leg by subtracting.
The problem
A 13 m ladder leans against a wall. Its foot is 5 m from the base of the wall. How far up the wall does it reach?
The ladder is the hypotenuse, so the squares subtract: a² = 13² − 5² = 169 − 25 = 144, giving a = 12 m. B adds them, which is only right when the unknown is the longest side.
Put the known side in the right slot.
a² + b² = c²
c is the hypotenuse — every time
a — the wall
h
b — the ground
5
c — the ladder
13
h² + 5² = 13²
The trap
The unknown here is a leg, not the hypotenuse. Writing h² = 13² + 5² gives 13.9 m — a ladder reaching higher than its own length, which is impossible.
Subtract, then square-root.
h² + 25 = 169
5² and 13²
h² = 144
− 25 both sides
h = 12
√144, and a length is positive
The negative root is thrown away without comment here. A ladder cannot reach −12 m up a wall.
Triples worth memorising.
These come up constantly, and spotting one saves the whole calculation.
| Triple | Check | Also appears as |
|---|---|---|
| 3, 4, 5 | 9 + 16 = 25 | 6-8-10, 9-12-15 |
| 5, 12, 13 | 25 + 144 = 169 | 10-24-26 |
| 8, 15, 17 | 64 + 225 = 289 | 16-30-34 |
| 7, 24, 25 | 49 + 576 = 625 | 14-48-50 |
Our problem was the 5-12-13 triple all along.
Doubling or tripling every side of a triple gives another triple, so 10-24-26 works too.
Reading the question for the right unknown.
looking for the hypotenuse
c = √a² + b²
add
looking for a leg
a = √c² − b²
subtract
One useful sanity check: the answer for a leg must always come out smaller than the hypotenuse.
Ladder, ramp, kite string, diagonal of a rectangle, distance between two points — they are all this one triangle wearing different clothes.
A rhombus is four of them. Its diagonals cut each other in half at right angles, so each half-diagonal is a leg and the side is the hypotenuse. Diagonals 18 and 24 give legs 9 and 12, side √81 + 144 = √225 = 15, and perimeter 4 × 15 = 60. Halve the diagonals first — using 18 and 24 as the legs doubles every side.
When the sides are expressions.
A right triangle has legs 12 and x, and hypotenuse x + 8. Find its perimeter.
12² + x² = (x + 8)²
the theorem, as usual
144 + x² = x² + 16x + 64
expand the square fully
80 = 16x, x = 5
the x² terms cancel
sides 12, 5, 13 → perimeter 30
use all three sides
(x + 8)² is not x² + 64
Dropping the middle term 16x makes the x² terms cancel into nonsense. Expand with the middle term every time.
Sides in a ratio work the same way. Legs in the ratio 5 : 12 are 5k and 12k, so the hypotenuse is 13k. If the hypotenuse is 39, then k = 3 and the sides are 15, 36, 39 — perimeter 90.
Variations: no triple, decimal sides, two equal legs.
A right triangle has hypotenuse 9 and one leg 6. Find the other leg.
x² = 9² − 6² = 81 − 36 = 45
a leg, so subtract
x = √45 = √9 × 5
pull out the perfect square
x = 3√5 ≈ 6.7
still less than 9 ✓
Different: 45 is not a perfect square. The answer is a simplified root, and the choices will show it as 3√5, not 6.7.
The root of a difference is not a difference of roots
√9² − 6² is not 9 − 6 = 3. Square, subtract, then take one root at the end.
The legs of a right triangle are 0.9 m and 1.2 m. Find the hypotenuse.
0.9 : 1.2 = 3 : 4 (× 0.3)
spot the triple
c = 5 × 0.3 = 1.5 m
0.81 + 1.44 = 2.25 = 1.5² ✓
Different: triples can be scaled by a decimal. Spotting 3-4-5 avoids squaring decimals.
A square has a diagonal of 10 cm. Find its side and its area.
s² + s² = 10²
two equal legs
2s² = 100, s² = 50
÷ 2
s = 5√2 cm, area = 50 cm²
the area is s², already found
Different: one unknown appears twice. Combine the like terms first; for the area we never need the root at all.
Two triangles in a row.
∠B and ∠ACD are right angles, AB = 6, BC = 8 and CD = 24. Find AD. The side the two triangles share is the bridge.
AC² = 6² + 8² = 100, AC = 10
first triangle, right angle at B
AD² = 10² + 24² = 676
second triangle, right angle at C
AD = 26
10-24-26, a 5-12-13 doubled
The sliding ladder is the same idea: work out the first triangle completely, change the one length the story changes, and solve a second triangle with the same hypotenuse.
On the exam
Straight “find the hypotenuse” items are rare. Expect a rectangle’s diagonal, sides given as expressions or a ratio, a ladder that slides, or two right triangles sharing a side — and a final ask for a perimeter or area, not the side you solved for. Spot a triple (3-4-5, 5-12-13, 8-15-17, 7-24-25 and multiples) and skip the arithmetic. Ten-second check: the hypotenuse must be the longest side.
Check: rebuild the ladder.
the legs
12² + 5² = 144 + 25
= 169
the ladder
13² = 169
a match ✓
is it sensible
12 < 13
the leg is shorter ✓
Now you try
Practice problem
The problem
A rectangular garden is 9 m by 12 m. How long is the diagonal path across it?
The diagonal is the hypotenuse, so here the squares add: √9² + 12² = √225 = 15 m. C just adds the sides. It is the 3-4-5 triangle scaled by three — worth recognising on sight.
Hint: the diagonal cuts the rectangle into two right triangles, and it is the hypotenuse of each. This time you add — and the numbers are a familiar triple scaled up.