📐Special Right Triangles

Master Pytha keeps two shapes chalked on the rock at Pythagoras Pass, and says everything else on the Pass is built out of them.

The first is half a square. Cut a square along its diagonal and you get a right triangle with two equal legs. Its other two angles must share the remaining 90° equally, so they are 45° each — a 45-45-90 triangle.

The second is half an equilateral triangle. Drop a height in a triangle whose three angles are all 60°, and it splits into two identical right triangles. Each one keeps a 60° angle, gains the 90° at the foot of the height, and gets half of the 60° at the top — so its angles are 30-60-90.

These two are special because you can write their sides down exactly, with no calculator anywhere. Every other right triangle needs sine and cosine; these two need only the Pythagorean theorem, once, and then you know them forever.

• ◆A 45-45-90 triangle has sides in the ratio leg : leg : hypotenuse = 1 : 1 : sqrt(2).
• ◆A 30-60-90 triangle has sides in the ratio short leg : long leg : hypotenuse = 1 : sqrt(3) : 2.

You can check both with the theorem. For the first, 1² + 1² = 2, and sqrt(2) squared is 2. For the second, 1² + (sqrt(3))^2 = 1 + 3 = 4, and 2² = 4.

Start from a square of side 1. Both legs are 1, and the diagonal is sqrt(1 + 1) = sqrt(2). Scale the whole picture by any number k and every side scales with it: the legs become k and the diagonal becomes k sqrt(2).

Going forwards (leg to hypotenuse) you multiply by sqrt(2):

leg 7 -> hypotenuse 7 sqrt(2)

Going backwards (hypotenuse to leg) you divide by sqrt(2). That is the step people get wrong, because dividing by sqrt(2) is *not* halving. sqrt(2) is about 1.41, so the leg is only a little shorter than the hypotenuse, not half as long.

A root is never left underneath, so clear it by multiplying top and bottom by sqrt(2):

hypotenuse 12 -> leg = 12 / sqrt(2) = 12 sqrt(2) / 2 = 6 sqrt(2)

If the hypotenuse already carries a sqrt(2), the division is even kinder — the roots cancel and a whole number falls out:

hypotenuse 5 sqrt(2) -> leg = 5 sqrt(2) / sqrt(2) = 5

Every question you will be asked is arranged so that the answer lands on one of these two shapes: a whole number, or a whole number times a root. You will never be left holding a fraction.

Take an equilateral triangle of side 2 and drop a height. The base is cut in half, so the bottom of each half-triangle is 1. The slanted side is still 2. The height is sqrt(2^2 - 1²) = sqrt(3).

Line the three sides up against the three angles and the pattern is easy to hold on to:

• ◆opposite the 30° angle: the short leg, the 1 in the ratio;
• ◆opposite the 60° angle: the long leg, short leg times sqrt(3);
• ◆opposite the 90° angle: the hypotenuse, exactly twice the short leg.

The short leg is the hinge of the whole triangle. Whatever you are given, the reliable move is: get the short leg first, then build whichever side you actually want.

given the short leg 6: hypotenuse = 12, long leg = 6 sqrt(3) given the hypotenuse 14: short leg = 7, long leg = 7 sqrt(3) given the long leg 9 sqrt(3): short leg = 9, hypotenuse = 18

That last line is worth staring at. Coming back from the long leg means dividing by sqrt(3), not by 3. When the long leg already carries a sqrt(3), the root cancels; when it does not, multiply top and bottom by sqrt(3) to clear it:

long leg 15 -> short leg = 15 / sqrt(3) = 15 sqrt(3) / 3 = 5 sqrt(3)

Notice which triangle owns which root. sqrt(2) belongs to the 45-45-90 and sqrt(3) to the 30-60-90. Swapping them is the single most common slip in this whole topic.

Most of the time nobody hands you a labelled right triangle. They hand you a square, or a hexagon, and the triangle is inside it waiting.

A square. Its diagonal makes two 45-45-90 triangles, so diagonal = side x sqrt(2), and side = diagonal / sqrt(2). A square of side 9 has diagonal 9 sqrt(2); a square with diagonal 10 has side 5 sqrt(2).

An equilateral triangle. Its height makes two 30-60-90 triangles whose short leg is *half* the base. A side of 12 gives a half-base of 6, so the height is 6 sqrt(3) — not 12 sqrt(3). Then the area is (1/2) x 12 x 6 sqrt(3) = 36 sqrt(3).

A regular hexagon. Join the centre to all six corners and you get six equilateral triangles. So the distance from the centre to a corner equals the side, the longest diagonal is twice the side, and the apothem (centre to the middle of a side) is the height of one of those triangles: side x sqrt(3) / 2. A hexagon of side 8 has apothem 4 sqrt(3), longest diagonal 16, and area 6 x (16 sqrt(3)) = 96 sqrt(3).

A 30-60-90 itself. The altitude drawn onto its hypotenuse is found by writing the area twice: once as half the product of the legs, and once as half the hypotenuse times that altitude. The two must agree, and the altitude drops out.

6 sqrt(2) and 8.485 are not the same thing. The first is the length; the second is a photograph of it, already slightly wrong, and getting wronger every time you build on it. In this skill you give the exact answer, which means the root stays.

Because 2 sqrt(18) and 6 sqrt(2) are the same number written two ways, an answer box needs to know which one you mean. So an exact length is typed as a pair:

a sqrt(b) is typed as (a, b)

with b as small as it can be — nothing under the root may still divide by 4, 9, 16 or 25. And a whole-number length still fills both slots, with 1 in the root slot:

• ◆6 sqrt(2) is (6, 2)
• ◆5 sqrt(3) is (5, 3)
• ◆12 is (12, 1)
• ◆sqrt(50) is not (1, 50) — simplify it first, to (5, 2)
• ◆10 sqrt(2) is not (5, 8), even though (5, 8) has the same value

Areas are typed the same way: an area of 54 sqrt(3) cm² is (54, 3), and an area of 72 cm² is (72, 1).

Just occasionally you will be asked for a length to 1 decimal place instead, and the prompt will say so plainly. Then use sqrt(2) is about 1.414 and sqrt(3) is about 1.732, and round at the very end — never in the middle.

When a question asks which triangle you are looking at, or for the ratio itself, the answer is a choice; pick the option, do not type a length.