The two ratios, memorised once.
45-45-90
1 : 1 : √2
half a square
30-60-90
1 : √3 : 2
half an equilateral triangle
In both, the sides are listed in the order of the angles they face. The smallest angle always faces the smallest side.
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Special Right Triangles
Two shapes with fixed side ratios — no calculator, no trigonometry.
Find the hypotenuse
A 45-45-90 triangle with a leg of 7
In a 45-45-90 the legs match and the hypotenuse is a leg times √2: 7√2 ≈ 9.9. D uses √3, which belongs to the 30-60-90. Sanity check: the hypotenuse must beat 7 but stay under 14.
The 45-45-90: multiply the leg by √2.
legs = 7 and 7
isosceles — both legs equal
hypotenuse = 7√2
the ratio scaled by 7
≈ 9.9
√2 ≈ 1.414
Going the other way, divide by √2 — a hypotenuse of 12 gives legs of 12√2 = 6√2.
Check it with Pythagoras: 7² + 7² = 98, and (7√2)² = 49 · 2 = 98. ✓
The 30-60-90: find the short side first.
The short side faces the 30° angle and is always half the hypotenuse.
hypotenuse = 10
this is the “2”
short side = 10 ÷ 2 = 5
opposite the 30°
long side = 5√3 ≈ 8.7
opposite the 60°
Always route through the short side
Given the long side instead, divide by √3 to get the short one first, then double it. Trying to jump straight from √3 to 2 is where the errors happen.
Where the ratios come from.
cut a square in half
two sides of 1
diagonal = √1+1 = √2
cut an equilateral triangle in half
side 2, base halves to 1
height = √4−1 = √3
Nothing is being memorised that Pythagoras would not give you in thirty seconds — the ratios just save the thirty seconds.
Where they show up.
| Figure | Hidden triangle | Result |
|---|---|---|
| square, side s | 45-45-90 | diagonal = s√2 |
| equilateral triangle, side s | 30-60-90 | height = s√32 |
| regular hexagon | six equilateral triangles | area = (3√32)s² |
| 45° ramp or roof | 45-45-90 | rise = run |
These two also give the exact values of sin, cos and tan at 30°, 45° and 60°.
Working backwards from a height.
An equilateral triangle has height 5√3. Find its perimeter. The height is the long leg of a 30-60-90, and the short leg is half a side.
short leg = 5√3 ÷ √3 = 5
long leg ÷ √3
side = 2 × 5 = 10
the short leg is half the base
perimeter = 3 × 10 = 30
three equal sides
The half-side is not the side
Stopping at 5 and tripling it gives 15 — the perimeter of the wrong triangle. The 5 is only half of one side.
A rhombus has side 8 and an angle of 60°. Find its area.
short diagonal = 8
it cuts off an equilateral triangle
long diagonal = 2 × 4√3 = 8√3
two heights of that triangle
area = 8 × 8√3 ÷ 2 = 32√3
d₁d₂ ÷ 2
Two special triangles sharing one altitude.
In △ABC, AD ⊥ BC with D on BC, AD = 9, ∠B = 45° and ∠C = 60°. Find BC. Solve each half separately; AD is the side they share.
△ABD is 45-45-90 → BD = AD = 9
the legs are equal
in △ADC, AD faces the 60° at C
so AD is the long leg
DC = 9 ÷ √3 = 3√3
long leg ÷ √3, rationalised
BC = 9 + 3√3
add the two pieces
Decide which angle the shared side faces before multiplying or dividing by √3. If the angle at C had been 30°, AD would be the short leg and DC = 9√3.
On the exam
These triangles hide inside squares (diagonal), equilateral triangles (height), hexagons, 60° rhombi and figures cut by an altitude, and answers come in exact surd form. Label the side opposite each angle, route through the short leg, and add or subtract pieces only at the end. Ten-second check: the sides must rank with their angles — the longest side faces the 90°.
Check: Pythagoras on both.
45-45-90
49 + 49 = 98 = (7√2)²
✓
30-60-90
25 + 75 = 100 = 10²
(5√3)² = 75 ✓
sensible?
5 < 8.7 < 10
sides rank with their angles ✓
Now you try
Practice problem
The problem
A square has a diagonal of 8 cm. How long is each side?
The diagonal splits the square into two 45-45-90 triangles, so diagonal = side × √2 and side = 8√2 = 4√2 ≈ 5.66. C multiplies where it should divide — the side has to come out shorter than the diagonal.