Geometry · G.SRT.C.8

Special right triangles

Cut a square along its diagonal, or an equilateral triangle straight down its middle. Is one side of either half enough to find the other two exactly?

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Special right triangles

The Pythagorean theorem in three dimensions Geometric means in a right triangle

Why this exists

The Pythagorean theorem finds the third side of a right triangle from the other two. Knowing one side and the angles is not enough for it. For most right triangles, one side and an acute angle do fix the other two sides, but finding them takes trigonometry and a calculator, and the answers come out as decimals.

Two right triangles give more. Each has a shape so simple that the squaring and adding comes out the same way every time, so it can be done once, with a letter, and read off from then on.

The right triangle with two equal legs is half of a square. Its hypotenuse is a leg times \displaystyle \sqrt{2}.
The right triangle with angles of 30^\circ and 60^\circ is half of an equilateral triangle. Its hypotenuse is twice its shortest side, and its other leg is that side times \displaystyle \sqrt{3}.

For these two, one measurement gives all three sides, exactly: a whole number, or a whole number times a root. Every square’s diagonal and every equilateral triangle’s height come with them.

Definition

The special right triangles are two right triangles, each named by its three angles: the 45-45-90 triangle and the 30-60-90 triangle. In each, the sides are in a fixed exact ratio, so any one side gives the other two without trigonometry.

As in every right triangle, the two sides that form the right angle are the legs, and the side opposite the right angle is the hypotenuse (see opposite, adjacent and hypotenuse).

The 45-45-90 triangle

A 45-45-90 triangle is a triangle whose angles measure 45^\circ, 45^\circ and 90^\circ.

It is a right triangle when classified by its angles, and, as the results below show, it is the same thing as an isosceles right triangle: a right triangle with two congruent legs, which is how it is named when classified by its sides. Both names are in use. In Figure 1 the right angle is at C, the legs are \overline{CA} and \overline{CB}, and the hypotenuse is \overline{AB}.

Figure 1The 45-45-90 triangle. Two equal legs, two 45^\circ angles, and a hypotenuse \displaystyle \sqrt{2} times a leg.

Each side sits opposite one angle:

Angle	Measure	Side opposite it
\angle A	45^\circ	the leg \overline{CB}
\angle B	45^\circ	the leg \overline{CA}
\angle C	90^\circ	the hypotenuse \overline{AB}

So each leg is opposite a 45^\circ angle, and the hypotenuse is opposite the right angle. Seen from either 45^\circ angle, the opposite side and the adjacent side are the two legs, and they are equal.

The 30-60-90 triangle

A 30-60-90 triangle (read “thirty sixty ninety”) is a triangle whose three angles measure 30^\circ, 60^\circ and 90^\circ. It is a right triangle, but unlike the 45-45-90 triangle its two legs have different lengths, so they get names:

the short leg is the leg opposite the 30^\circ angle;
the long leg is the leg opposite the 60^\circ angle;
the hypotenuse is opposite the 90^\circ angle, and is the longest side.

In Figure 2 the short leg is \overline{BC}, the long leg is \overline{CA} and the hypotenuse is \overline{AB}.

Figure 2The 30-60-90 triangle, with sides in the ratio \displaystyle 1 : \sqrt{3} : 2: the short leg faces the 30^\circ angle, the long leg the 60^\circ angle, the hypotenuse the right angle.

You never need all three angles to be given. A right triangle with one acute angle of 30^\circ has a third angle of 180^\circ - 90^\circ - 30^\circ = 60^\circ by the triangle angle sum, so it is a 30-60-90 triangle. The same goes for a right triangle with a 60^\circ angle, and for any triangle with a 30^\circ angle and a 60^\circ angle, whose third angle must be 90^\circ.

Intuition

Half a square

Draw one diagonal of a square, as in Figure 3. It cuts the square into two triangles. Each has two sides of the square as its legs, so the legs are equal, and the corner between them is a corner of the square, so it is a right angle. Half a square is an isosceles right triangle.

Figure 3The diagonal cuts the square into two 45-45-90 triangles.

It works the other way round too. Take a 45-45-90 triangle and a copy of it, and put them together along the hypotenuse, one on each side of it. The new figure has four equal sides, a right angle at each of the two old right-angle corners, and 45^\circ + 45^\circ = 90^\circ at each end of the join. That is a square. So every 45-45-90 triangle is half of a square.

A bigger square gives a bigger triangle, but not a different shape. Any two 45-45-90 triangles have the same three angles, so they are similar by angle-angle similarity, and the ratio of hypotenuse to leg is the same in all of them. It only has to be found once. With legs of 1 the hypotenuse is \displaystyle \sqrt{1^2 + 1^2} = \sqrt{2}, and every other 45-45-90 triangle is that one scaled.

Since \displaystyle \sqrt{2} \approx 1.414, the hypotenuse is about 41\% longer than a leg: longer than either leg, as the hypotenuse always is, and shorter than the two legs laid end to end, as the triangle inequality requires.

Half an equilateral triangle

Take an equilateral triangle and cut it with a segment from one corner straight down to the middle of the opposite side, as in Figure 4. The two pieces are mirror images. Each piece keeps a whole 60^\circ corner at the base, gets a right angle where the cut meets the base, and gets half of the 60^\circ top corner, which is 30^\circ.

Figure 4The segment from P to the middle of the base cuts the equilateral triangle into two 30-60-90 triangles. The shaded one has hypotenuse 2x and short leg x.

Now look at the sides of the shaded piece. Its hypotenuse \overline{PQ} is a whole side of the equilateral triangle. Its short leg \overline{QR}, opposite the 30^\circ angle, is half of a side. So the hypotenuse is twice the short leg, with no calculation at all.

The third side, the cut \overline{PR}, is the long leg. It is longer than the short leg and shorter than the hypotenuse, so if the short leg is x, the long leg is somewhere between x and 2x. The Pythagorean theorem pins it down: it is \displaystyle x\sqrt{3}, about 1.73x.

As with the square, every 30-60-90 triangle has the same three angles, so all of them are similar, and what is true of one with a short leg of 1 is true of every one, scaled.

Results

The 45-45-90 triangle

The 45-45-90 Triangle Theorem. In a triangle with angles 45^\circ, 45^\circ and 90^\circ, the two legs are congruent, and the hypotenuse is \displaystyle \sqrt{2} times as long as a leg. If each leg has length x, then

\text{hypotenuse} = x\sqrt{2}

so the sides are in the ratio

\text{leg} : \text{leg} : \text{hypotenuse} = 1 : 1 : \sqrt{2}

From the hypotenuse to a leg. If the hypotenuse has length h, each leg has length

\frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2}

The two forms are the same number. The second comes from rationalizing the denominator. Multiplying the top and the bottom of a fraction by the same nonzero number does not change its value, and \displaystyle \sqrt{2} \cdot \sqrt{2} = 2, so multiplying top and bottom by \displaystyle \sqrt{2} leaves no root below the line:

\frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} = \frac{h\sqrt{2}}{2}

By convention an exact answer is written with no root in its denominator, so \displaystyle \frac{h\sqrt{2}}{2} is the usual final form (rationalizing a denominator has the general rule). When h already has a factor of \displaystyle \sqrt{2}, divide it out directly: a hypotenuse of \displaystyle 6\sqrt{2} gives legs of 6.

The converse. A triangle is a 45-45-90 triangle if any one of these holds:

it is a right triangle with two congruent legs;
it is a right triangle with one acute angle of 45^\circ;
it has two angles of 45^\circ. No right angle needs to be given; the angle sum forces one;
it is a right triangle whose hypotenuse is \displaystyle \sqrt{2} times one of its legs;
its sides have lengths x, x and \displaystyle x\sqrt{2} for some x > 0. No right angle needs to be given; the lengths force one.

The diagonal of a square. In a square with side s, each diagonal has length

d = s\sqrt{2}

The hypotheses are the two that make half the square an isosceles right triangle: the sides are equal and the corners are right angles. It follows that a square with diagonal d has side \displaystyle \frac{d\sqrt{2}}{2} and area \displaystyle \frac{d^2}{2}.

The 30-60-90 triangle

The 30-60-90 Triangle Theorem. In a triangle with angles 30^\circ, 60^\circ and 90^\circ, if the short leg has length x, then

\begin{aligned} \text{hypotenuse} &= 2x \\ \text{long leg} &= x\sqrt{3} \end{aligned}

So the short leg, the long leg and the hypotenuse are in the ratio

\text{short leg} : \text{long leg} : \text{hypotenuse} = 1 : \sqrt{3} : 2

Any one side gives the other two. Start from the side you know and pass through the short leg. In each row below, the side you know is the bare letter:

Short leg	Long leg	Hypotenuse
x	\displaystyle x\sqrt{3}	2x
\dfrac{h}{2}	\displaystyle \dfrac{h\sqrt{3}}{2}	h
\displaystyle \dfrac{\ell\sqrt{3}}{3}	\ell	\displaystyle \dfrac{2\ell\sqrt{3}}{3}

In the last row the short leg is \displaystyle \frac{\ell}{\sqrt{3}} and the hypotenuse is \displaystyle \frac{2\ell}{\sqrt{3}}, written in the table after rationalizing the denominator, this time by multiplying top and bottom by \displaystyle \sqrt{3}.

The converse. In a right triangle, if the hypotenuse is twice one of the legs, then the angle opposite that leg is 30^\circ and the other acute angle is 60^\circ. So a right triangle whose sides are in the ratio \displaystyle 1 : \sqrt{3} : 2 is a 30-60-90 triangle.

The height of an equilateral triangle. In an equilateral triangle with side s, the height is \displaystyle \frac{s\sqrt{3}}{2} and the area is \displaystyle \frac{s^2\sqrt{3}}{4} (Example 6).

The two side by side

	45-45-90	30-60-90
Half of	a square	an equilateral triangle
Legs	equal	long leg \displaystyle = \sqrt{3} \times short leg
Hypotenuse	\displaystyle \sqrt{2} \times a leg	2 \times the short leg
Sides in the ratio	\displaystyle 1 : 1 : \sqrt{2}	\displaystyle 1 : \sqrt{3} : 2
The cut gives	a square’s diagonal	an equilateral triangle’s height

To tell which one a right triangle is, compare its sides. Equal legs, or a hypotenuse \displaystyle \sqrt{2} times a leg, make it a 45-45-90 triangle; a hypotenuse twice a leg makes it a 30-60-90 triangle. In a 45-45-90 triangle the hypotenuse is about 1.41 times the shortest side, and in a 30-60-90 triangle it is exactly 2 times, so no right triangle is both.

Never three whole numbers

Neither special right triangle has three whole-number sides, at any size. In a 45-45-90 triangle a leg and the hypotenuse are never both whole, and in a 30-60-90 triangle the two legs are never both whole, or even both fractions. So neither is ever a Pythagorean triple.

Why it is true

The 45-45-90 theorem

Let \triangle ABC have m\angle A = m\angle B = 45^\circ and m\angle C = 90^\circ, as in Figure 1.

The legs are congruent. \angle A \cong \angle B, so by the converse of the Base Angles Theorem the sides opposite them are congruent: CB = CA. These two sides meet at C, the right angle, so they are the legs. Call the length of each one x.

The hypotenuse. The triangle has a right angle at C, so the Pythagorean theorem applies to it. With c = AB,

\begin{aligned} c^2 &= x^2 + x^2 = 2x^2 \\ c &= \sqrt{2x^2} = \sqrt{2}\,\sqrt{x^2} = x\sqrt{2} \end{aligned}

Only the positive root is taken, because c is a length, and \displaystyle \sqrt{x^2} = x because x > 0 (simplifying radicals).

Half a square, the same fact another way. Let ABCD be a square with side s, as in Figure 3. In \triangle ABC, AB = BC = s and m\angle B = 90^\circ, since those are two sides and a corner of the square. By the Base Angles Theorem, \angle BAC \cong \angle BCA, and by the triangle angle sum they add to 180^\circ - 90^\circ = 90^\circ, so each is 45^\circ. So \triangle ABC is a 45-45-90 triangle, and by the theorem \displaystyle AC = s\sqrt{2}. That is the diagonal result, and the proof used exactly its two hypotheses: AB = BC and a right angle at B.

The diagonal also halves the square’s corners. The corner at A is 90^\circ and the diagonal takes m\angle BAC = 45^\circ of it, so the rest is m\angle CAD = 90^\circ - 45^\circ = 45^\circ; in the same way m\angle ACD = 90^\circ - 45^\circ = 45^\circ at C. So \triangle ADC is a 45-45-90 triangle too, the other half of the square.

The converse. Take the five conditions in turn.

A right triangle with congruent legs. By the Base Angles Theorem the angles opposite the legs are congruent. Those are the two acute angles, and they add to 180^\circ - 90^\circ = 90^\circ, so each is 45^\circ.

A right triangle with one acute angle of 45^\circ. The third angle is 180^\circ - 90^\circ - 45^\circ = 45^\circ.

Two angles of 45^\circ. The third angle is 180^\circ - 45^\circ - 45^\circ = 90^\circ, so the angles are 45^\circ, 45^\circ and 90^\circ.

A right triangle whose hypotenuse is \displaystyle \sqrt{2} times a leg. Call that leg a and the other leg b. The Pythagorean theorem gives \displaystyle a^2 + b^2 = \left(a\sqrt{2}\right)^2 = 2a^2, so b^2 = a^2, and since both are lengths, b = a. The legs are congruent, which is the first case.

Sides x, x and \displaystyle x\sqrt{2}. Here \displaystyle x^2 + x^2 = 2x^2 = \left(x\sqrt{2}\right)^2, so by the converse of the Pythagorean theorem the angle opposite the side \displaystyle x\sqrt{2} is a right angle. The two sides of length x are then congruent legs, which is the first case.

The 30-60-90 theorem

The hypotenuse is twice the short leg. Start with any triangle PQR with the right angle at R, m\angle P = 30^\circ and m\angle Q = 60^\circ, and call its short leg QR = x. The argument builds Figure 4 around it.

Extend \overline{QR} past R to a point Q' with RQ' = x, and join P to Q'. Because Q, R and Q' lie on one line with R between them, the angles \angle PRQ and \angle PRQ' form a linear pair and add to 180^\circ, so \angle PRQ' is 180^\circ - 90^\circ = 90^\circ, a right angle too. The new triangle PRQ' is the mirror image of PRQ across the long leg \overline{PR}.

Triangles PRQ and PRQ' have RQ = RQ', the right angles at R equal, and the side \overline{PR} in common. They are congruent by SAS, so their matching parts are equal:

\begin{aligned} PQ' &= PQ \\ m\angle Q' &= m\angle Q = 60^\circ \\ m\angle RPQ' &= m\angle RPQ = 30^\circ \end{aligned}

Now look at the big triangle PQQ'. Since R lies between Q and Q', the segment \overline{PR} runs inside its angle at P and splits it in two, so that angle is 30^\circ + 30^\circ = 60^\circ. Its angles at Q and Q' are 60^\circ each, so all three angles are equal and the triangle is equilateral. The one step of that needed here: the angles at P and Q' are equal, so by the converse of the Base Angles Theorem the sides opposite them are equal. The side opposite P is \overline{QQ'} and the side opposite Q' is \overline{PQ}, so

PQ = QQ' = x + x = 2x

The long leg is \displaystyle x\sqrt{3}. Now use the Pythagorean theorem in the right triangle PQR, with legs x and PR and hypotenuse 2x:

\begin{aligned} x^2 + PR^2 &= (2x)^2 \\ PR^2 &= 4x^2 - x^2 = 3x^2 \\ PR &= \sqrt{3x^2} = x\sqrt{3} \end{aligned}

A length is positive, so only the positive square root is kept, and \displaystyle \sqrt{3x^2} = \sqrt{3}\,\sqrt{x^2} = x\sqrt{3} because x > 0.

Nothing in the argument depended on the size of x, so the ratio \displaystyle 1 : \sqrt{3} : 2 holds in every 30-60-90 triangle.

The converse. Now start with a right triangle PQR, right angle at R, whose hypotenuse is twice a leg: QR = x and PQ = 2x. Build the same picture: extend \overline{QR} past R to Q' with RQ' = x. As before, triangles PRQ and PRQ' are congruent by SAS, so PQ' = PQ = 2x. Now triangle PQQ' has sides PQ = 2x, PQ' = 2x and QQ' = x + x = 2x. It is equilateral, so by the Base Angles Theorem, used on each pair of equal sides, its three angles are equal, and as they add to 180^\circ each is 60^\circ. In particular m\angle Q = 60^\circ. The angle sum in triangle PQR then gives

m\angle QPR = 180^\circ - 90^\circ - 60^\circ = 30^\circ

and \angle QPR is the angle opposite the leg \overline{QR} of length x. The right angle at R was used twice, to make \angle PRQ' a right angle and to make the two small triangles congruent, which is why the converse has to say right triangle.

Never three whole numbers

The 45-45-90 triangle. Suppose a leg x and the hypotenuse \displaystyle x\sqrt{2} were both whole numbers. Then

\sqrt{2} = \frac{x\sqrt{2}}{x}

would be a fraction with whole numbers on top and bottom. But \displaystyle \sqrt{2} is not any such fraction. So the leg and the hypotenuse are never both whole, which is all this argument proves, and it is enough. One of them can be: legs of 5 give a hypotenuse of \displaystyle 5\sqrt{2}, and a hypotenuse of 4 gives legs of \displaystyle \frac{4\sqrt{2}}{2} = 2\sqrt{2}.

The 30-60-90 triangle. Suppose some 30-60-90 triangle had a short leg x and a long leg \displaystyle x\sqrt{3} that were both fractions. Dividing one fraction by another nonzero fraction gives a fraction, so \displaystyle {\dfrac{x\sqrt{3}}{x} = \sqrt{3}} would be a fraction. It is not. So at least one of the two legs is always irrational.

That \displaystyle \sqrt{2} and \displaystyle \sqrt{3} are not fractions is used here, not proved. The proof for \displaystyle \sqrt{2} is in the square root of 2 is irrational, and it carries over to \displaystyle \sqrt{3} with divisibility by 3 in place of evenness; see also irrational numbers.

Worked examples

Example 1

From one side of a 45-45-90 triangle

(a) A 45-45-90 triangle has a leg of length 7. Find the other two sides.

(b) Do the same for a 45-45-90 triangle with a leg of \displaystyle 3\sqrt{2}.

Solution

(a) The legs of a 45-45-90 triangle are congruent, so the other leg is also 7. The hypotenuse is a leg times \displaystyle \sqrt{2}:

\text{hypotenuse} = 7\sqrt{2} \approx 9.90

Check. 7^2 + 7^2 = 98 and \displaystyle \left(7\sqrt{2}\right)^2 = 49 \cdot 2 = 98.

(b) With a leg of \displaystyle 3\sqrt{2}, the other leg is \displaystyle 3\sqrt{2} as well, and

\text{hypotenuse} = 3\sqrt{2} \cdot \sqrt{2} = 3 \cdot 2 = 6

The \displaystyle \sqrt{2} in the leg and the \displaystyle \sqrt{2} of the ratio multiply to the whole number 2, which is why a leg that is a whole number times \displaystyle \sqrt{2} gives a whole-number hypotenuse.

(c) An isosceles right triangle is a 45-45-90 triangle (the first condition of the converse), so its hypotenuse is a leg times \displaystyle \sqrt{2}. Call each leg x:

x\sqrt{2} = 10 \qquad\Longrightarrow\qquad x = \frac{10}{\sqrt{2}}

Multiply top and bottom by \displaystyle \sqrt{2}, which does not change the value, to clear the root from the denominator:

x = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07

Check. \displaystyle 5\sqrt{2} \cdot \sqrt{2} = 5 \cdot 2 = 10.

The Pythagorean theorem gives the same answer the long way: x^2 + x^2 = 100, so x^2 = 50 and \displaystyle x = \sqrt{50} = 5\sqrt{2}. The ratio is that calculation done once in advance.

Example 2

From the short leg or the hypotenuse of a 30-60-90 triangle

(a) A 30-60-90 triangle has a short leg of 5. Find the other two sides.

(b) The hypotenuse of a 30-60-90 triangle is 14. Find the two legs.

Solution

(a) The hypotenuse is twice the short leg: 2 \cdot 5 = 10. The long leg is the short leg times \displaystyle \sqrt{3}: \displaystyle 5\sqrt{3}, which is about 8.66.

Check. \displaystyle 5^2 + (5\sqrt{3})^2 = 25 + 25 \cdot 3 = 25 + 75 = 100, and 10^2 = 100, so the Pythagorean theorem holds.

The exact answer is \displaystyle 5\sqrt{3}. The decimal 8.66 is a rounding of it, useful for a measurement, but it is not the length.

(b) The hypotenuse is twice the short leg, so the short leg is half the hypotenuse: 14 \div 2 = 7. Now the long leg is the short leg times \displaystyle \sqrt{3}: \displaystyle 7\sqrt{3}, about 12.12.

Check. \displaystyle 7^2 + (7\sqrt{3})^2 = 49 + 147 = 196, and 14^2 = 196.

Example 3

From the long leg of a 30-60-90 triangle

The long leg of a 30-60-90 triangle is 9. Find the short leg and the hypotenuse exactly.

Solution

Call the short leg x. The long leg is \displaystyle x\sqrt{3}, so

x\sqrt{3} = 9 \qquad\Longrightarrow\qquad x = \frac{9}{\sqrt{3}}

That is exact, and it can be tidied by multiplying top and bottom by \displaystyle \sqrt{3}, which does not change the value because it multiplies by \displaystyle \frac{\sqrt{3}}{\sqrt{3}} = 1:

x = \frac{9}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}

The hypotenuse is twice the short leg: \displaystyle 2 \cdot 3\sqrt{3} = 6\sqrt{3}.

So the short leg is \displaystyle 3\sqrt{3}, about 5.20, and the hypotenuse is \displaystyle 6\sqrt{3}, about 10.39.

Check. \displaystyle (3\sqrt{3})^2 + 9^2 = 27 + 81 = 108, and \displaystyle (6\sqrt{3})^2 = 36 \cdot 3 = 108.

Going from the long leg is the one direction in this triangle that divides by \displaystyle \sqrt{3}, so it is the one where a radical lands in a denominator. In the 45-45-90 triangle the same thing happens going from the hypotenuse, as in Example 1(c).

Example 4

Naming the sides from the angles

(a) In \triangle JKL, m\angle L = 90^\circ, m\angle J = 45^\circ and JK = 12. Find m\angle K, KL, JL, the perimeter and the area.

(b) In Figure 5, triangle PQR has its right angle at R, m\angle P = 60^\circ and \displaystyle PQ = 4\sqrt{3}. Find PR and QR.

Figure 5The hypotenuse is along the bottom here, for Example 4.

Solution

(a) The angles add to 180^\circ, so m\angle K = 180^\circ - 90^\circ - 45^\circ = 45^\circ. The triangle is a 45-45-90 triangle.

Now name each side by the angle it is opposite. \overline{JK} is opposite the right angle at L, so it is the hypotenuse. \overline{KL} is opposite \angle J and \overline{JL} is opposite \angle K; those are the legs, and they are equal. From the hypotenuse:

KL = JL = \frac{12}{\sqrt{2}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2} \approx 8.49

The perimeter is the sum of the three sides:

6\sqrt{2} + 6\sqrt{2} + 12 = 12 + 12\sqrt{2} \approx 28.97

The legs are perpendicular, so one is a base and the other the height to it:

\text{area} = \tfrac{1}{2} \cdot 6\sqrt{2} \cdot 6\sqrt{2} = \tfrac{1}{2} \cdot 72 = 36

Here \displaystyle 6\sqrt{2} \cdot 6\sqrt{2} = 36 \cdot 2 = 72. The area is a whole number even though the legs are not.

(b) The triangle is drawn resting on its hypotenuse, so read the sides from the angles, not from the picture. The third angle is m\angle Q = 180^\circ - 90^\circ - 60^\circ = 30^\circ, so this is a 30-60-90 triangle.

\overline{PQ} is opposite the right angle at R: it is the hypotenuse, \displaystyle 4\sqrt{3}.
\overline{PR} is opposite the 30^\circ angle at Q: it is the short leg.
\overline{QR} is opposite the 60^\circ angle at P: it is the long leg.

The short leg is half the hypotenuse:

PR = \frac{4\sqrt{3}}{2} = 2\sqrt{3}

The long leg is the short leg times \displaystyle \sqrt{3}:

QR = 2\sqrt{3} \cdot \sqrt{3} = 2 \cdot 3 = 6

So \displaystyle PR = 2\sqrt{3}, about 3.46, and QR = 6.

Check. \displaystyle (2\sqrt{3})^2 + 6^2 = 12 + 36 = 48, and \displaystyle (4\sqrt{3})^2 = 16 \cdot 3 = 48.

Here the hypotenuse is irrational and the long leg is a whole number.

Example 5

Which special right triangle is it?

Decide whether each triangle is a 45-45-90 triangle, a 30-60-90 triangle, or neither.

(a) A triangle with sides 5, 5 and \displaystyle 5\sqrt{2}.

(b) A right triangle with a leg of 6 and a hypotenuse of \displaystyle 6\sqrt{2}.

(d) A right triangle with legs 3 and \displaystyle 3\sqrt{2}.

(e) A triangle with sides 4, \displaystyle 4\sqrt{3} and 8. If it is special, say which angle is opposite each side.

Solution

(a) A 45-45-90 triangle. The sides are x, x and \displaystyle x\sqrt{2} with x = 5. Indeed \displaystyle 5^2 + 5^2 = 50 = \left(5\sqrt{2}\right)^2, so by the converse of the Pythagorean theorem the angle opposite \displaystyle 5\sqrt{2} is right, and the legs 5 and 5 are equal.

(b) A 45-45-90 triangle. The hypotenuse is \displaystyle \sqrt{2} times a leg. To see the legs agree, call the other leg b: \displaystyle 36 + b^2 = \left(6\sqrt{2}\right)^2 = 72, so b^2 = 36 and b = 6.

(c) Neither. The triangle is isosceles, but it has no right angle: 4^2 + 4^2 = 32, while 6^2 = 36. For a 45-45-90 triangle with legs 4, the third side would have to be \displaystyle 4\sqrt{2} \approx 5.66. And every special right triangle has a right angle.

(d) Neither. It is a right triangle, but its legs are unequal, so it is not a 45-45-90 triangle: if one acute angle were 45^\circ, the other would be too, and then the converse of the Base Angles Theorem would make the legs equal. Nor is it a 30-60-90 triangle: the long leg divided by the short leg is \displaystyle \sqrt{2}, not \displaystyle \sqrt{3}.

(e) A 30-60-90 triangle. First it has to be a right triangle, since the converse is about right triangles only. The longest side is 8, because \displaystyle 4\sqrt{3} is about 6.93. Test the converse of the Pythagorean theorem:

4^2 + (4\sqrt{3})^2 = 16 + 48 = 64 = 8^2

So the triangle is right, with the right angle opposite the side 8. Now the hypotenuse 8 is twice the leg 4. By the converse of the 30-60-90 theorem, the angle opposite the side 4 is 30^\circ, and so the angle opposite \displaystyle 4\sqrt{3} is 60^\circ.

Example 6

The diagonal of a square and the height of an equilateral triangle

(a) The diagonal of a square is 14 centimeters. Find its area without finding the side first. Then find the side, and use it to check the area.

(b) An equilateral triangle has side s. Find its height and its area in terms of s.

Solution

(a) Call the side s. A square has equal sides and right-angled corners, so its diagonal is the hypotenuse of an isosceles right triangle with legs s, and \displaystyle d = s\sqrt{2}. Squaring both sides,

d^2 = 2s^2 \qquad\Longrightarrow\qquad s^2 = \frac{d^2}{2}

The area of a square is s^2, so

\text{area} = \frac{14^2}{2} = \frac{196}{2} = 98 \text{ cm}^2

The side is

s = \frac{14}{\sqrt{2}} = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \approx 9.90 \text{ cm}

Check. \displaystyle \left(7\sqrt{2}\right)^2 = 49 \cdot 2 = 98.

The formula \displaystyle \text{area} = \frac{d^2}{2} used nothing about this square except \displaystyle d = s\sqrt{2}, so it holds for every square.

(b) The altitude from one corner cuts the triangle into two 30-60-90 triangles, as in Figure 4. In each one the hypotenuse is a whole side, s, so the short leg is half a side, \displaystyle \frac{s}{2}. The height is the long leg:

\text{height} = \frac{s}{2} \cdot \sqrt{3} = \frac{s\sqrt{3}}{2}

The area is half the base times the height (area of triangles):

\text{area} = \frac{1}{2} \cdot s \cdot \frac{s\sqrt{3}}{2} = \frac{s^2\sqrt{3}}{4}

Check. With s = 6 these give a height of \displaystyle 3\sqrt{3} and an area of \displaystyle 9\sqrt{3}, the same values the Pythagorean theorem gives when the height is found directly: \displaystyle \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3}.

The square’s diagonal and the equilateral triangle’s height are the two cuts that make the two special right triangles, so these two formulas are the theorems seen from the outside.

Example 7

The altitude to the hypotenuse

In Figure 6, \triangle ABC is a 45-45-90 triangle with its right angle at C and hypotenuse AB = h. The segment \overline{CD} is perpendicular to \overline{AB}. Show that \displaystyle CD = \frac{h}{2} and that D is the midpoint of \overline{AB}.

Figure 6The altitude from the right angle to the hypotenuse, for Example 7.

Solution

Look at \triangle ACD first. m\angle ADC = 90^\circ because \overline{CD} \perp \overline{AB}, and m\angle A = 45^\circ. The angles add to 180^\circ, so

m\angle ACD = 180^\circ - 90^\circ - 45^\circ = 45^\circ

So \triangle ACD is a 45-45-90 triangle too, with its right angle at D. Its hypotenuse is \overline{AC}, which is a leg of the big triangle, so \displaystyle AC = \frac{h}{\sqrt{2}}. The legs of \triangle ACD are \overline{AD} and \overline{CD}, and each is its hypotenuse divided by \displaystyle \sqrt{2}:

AD = CD = \frac{h}{\sqrt{2}} \div \sqrt{2} = \frac{h}{2}

The same argument in \triangle BCD gives \displaystyle DB = CD = \frac{h}{2}. So AD = DB, which makes D the midpoint of \overline{AB}, and \displaystyle CD = \frac{h}{2}.

So in a 45-45-90 triangle the altitude from the right angle is half the hypotenuse, and it cuts the triangle into two smaller 45-45-90 triangles. This is a fact about this triangle only: the proof used the 45^\circ angles at A and B. In a right triangle with unequal legs the altitude to the hypotenuse is shorter than half of it, and its foot is not the midpoint (Exercise 31 finds where it lands in a 30-60-90 triangle). With h = 10, as in Example 1(c), the altitude is 5.

Check. Find the area two ways, from the hypotenuse and its altitude, and from the two legs:

\begin{aligned} \tfrac{1}{2} \cdot 10 \cdot 5 &= 25 \\ \tfrac{1}{2} \cdot 5\sqrt{2} \cdot 5\sqrt{2} &= \tfrac{1}{2} \cdot 50 = 25 \end{aligned}

Example 8

An isosceles triangle with a 120^\circ angle

In Figure 7, AB = AC = 10 and m\angle A = 120^\circ. Find the base BC and the area of the triangle, exactly.

Figure 7An isosceles triangle with a 120^\circ angle between its equal sides, for Example 8.

Solution

There is no right angle in the figure, so make one. Draw the altitude \overline{AD} from A perpendicular to \overline{BC}, meeting it at D.

The two base angles are equal by the Base Angles Theorem, and the angles add to 180^\circ, so each is

\frac{180^\circ - 120^\circ}{2} = 30^\circ

Triangles ABD and ACD are right triangles with equal hypotenuses AB = AC and a common leg \overline{AD}, so they are congruent by HL. That gives BD = DC.

Triangle ABD has a right angle at D and a 30^\circ angle at B, so it is a 30-60-90 triangle with hypotenuse AB = 10.

The short leg is opposite the 30^\circ angle at B: it is AD = 10 \div 2 = 5.
The long leg is opposite the 60^\circ angle \angle BAD: it is the short leg times \displaystyle \sqrt{3}, \displaystyle BD = 5\sqrt{3}.

So \displaystyle BC = BD + DC = 5\sqrt{3} + 5\sqrt{3} = 10\sqrt{3}, about 17.32. The altitude AD = 5 is the height to that base, so the area is

\frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot 10\sqrt{3} \cdot 5 = 25\sqrt{3}

which is about 43.30.

Example 9

A regular octagon from a square

A regular octagon is made from a square of side 12 by cutting off four congruent corners. Each cut-off corner is a right triangle whose two legs, of length t, lie along the sides of the square (Figure 8). Find t and the side of the octagon, exactly and to the nearest hundredth.

Figure 8Each cut-off corner is a 45-45-90 triangle, for Example 9.

Solution

Each cut-off corner has a right angle, the corner of the square, and two equal legs t. By the converse it is a 45-45-90 triangle, so its third side, which is a side of the octagon, has length \displaystyle t\sqrt{2}.

The other four sides of the octagon are what is left of each side of the square once a length t is cut from each end, so each is 12 - 2t. For the octagon to be regular these must equal the cut sides:

\begin{aligned} 12 - 2t &= t\sqrt{2} \\ 12 &= 2t + t\sqrt{2} = t\left(2 + \sqrt{2}\right) \\ t &= \frac{12}{2 + \sqrt{2}} \end{aligned}

Multiplying top and bottom by \displaystyle \sqrt{2} would not clear this denominator, because it is a sum. Multiply instead by \displaystyle 2 - \sqrt{2}, which does, because \displaystyle \left(2 + \sqrt{2}\right)\left(2 - \sqrt{2}\right) = 4 - 2 = 2 by the difference of squares:

t = \frac{12\left(2 - \sqrt{2}\right)}{2} = 12 - 6\sqrt{2} \approx 3.51

The side of the octagon is

t\sqrt{2} = \left(12 - 6\sqrt{2}\right)\sqrt{2} = 12\sqrt{2} - 12 \approx 4.97

Check. The sides along the square are \displaystyle 12 - 2t = 12 - 24 + 12\sqrt{2} = 12\sqrt{2} - 12, the same.

Equal sides alone do not make a polygon regular, so the angles need checking too. Every vertex of the octagon is an end of a cut, on a side of the square. There the straight side, 180^\circ, is split into the cut-off corner’s 45^\circ angle and the octagon’s angle, so each angle of the octagon is 180^\circ - 45^\circ = 135^\circ. With this t the octagon has eight equal sides and eight equal angles, so it really is regular.

Boundaries

The angles must be exact. The ratios come from the two legs being equal, or from the triangle being exactly half of an equilateral one. A right triangle with acute angles of 44^\circ and 46^\circ has unequal legs, so its sides are close to the ratio \displaystyle {1 : 1 : \sqrt{2}} but not in it; one with acute angles of 35^\circ and 55^\circ is close in shape to a 30-60-90 triangle, but its sides are not in the ratio \displaystyle 1 : \sqrt{3} : 2. Such triangles need trigonometry. A triangle with a 30^\circ angle and no right angle, such as one with angles 30^\circ, 75^\circ and 75^\circ, is not covered either. Equal legs, a 45^\circ angle or a 30^\circ angle must be given or proved; a figure that looks right proves nothing.

An isosceles triangle with a 45^\circ angle need not be a 45-45-90 triangle. If the 45^\circ angle is the vertex angle, the base angles are each \displaystyle \frac{180^\circ - 45^\circ}{2} = 67.5^\circ and there is no right angle. The conditions that do suffice are the ones in the converse.

The 30-60-90 converse needs the right angle, and the hypotenuse. The triangle with sides 6, 10 and 12 has one side twice another, but 6^2 + 10^2 = 136 and 12^2 = 144, so it is not a right triangle, and so it is not a 30-60-90 triangle. And it must be the hypotenuse that is twice a leg. A right triangle with legs 1 and 2 has one leg twice the other, but its hypotenuse is \displaystyle \sqrt{5}, and \displaystyle 1 : 2 : \sqrt{5} is not \displaystyle 1 : \sqrt{3} : 2.

The diagonal rule needs a square. \displaystyle d = s\sqrt{2} used both equal sides and a right-angled corner. A 3 by 4 rectangle has right angles but unequal sides, and its diagonal is 5, which is neither \displaystyle 3\sqrt{2} nor \displaystyle 4\sqrt{2}. A rhombus has equal sides, but without right angles the half it cuts off is not a right triangle, and the rule does not apply.

They need a flat plane. The Pythagorean step of each proof, and the angle sums used throughout, rest on the parallel postulate. On a curved surface such as a globe, a triangle’s angles add to more than 180^\circ, so no triangle there has angles 45^\circ, 45^\circ and 90^\circ, or 30^\circ, 60^\circ and 90^\circ, at all.

Where this goes next

The ratio of two sides of a special right triangle is the same in every triangle of that shape, so fractions such as \displaystyle \frac{\text{short leg}}{\text{hypotenuse}} = \frac{1}{2} depend only on the angle, not on the size of the triangle. A ratio that depends only on the angle is exactly what a trigonometric ratio is, and these two triangles supply the exact values at 30^\circ, 45^\circ and 60^\circ: see the trig values of the special right triangles. Placed with the hypotenuse as a radius of length 1, they give the exact points of the unit circle at those angles, and the exact values on the unit circle that follow.

On a grid, a segment that runs as far across as it runs up, such as the one from (0, 0) to (a, a) with a > 0, is the hypotenuse of a 45-45-90 triangle, so its length is \displaystyle a\sqrt{2}; the distance formula gives the same answer. A face diagonal of a cube with edge s is \displaystyle s\sqrt{2} for the same reason, and it is the first step toward the cube’s long diagonal in distance in space.

A regular hexagon is six equilateral triangles around a point, so every length in it, and its area, come from the 30-60-90 triangle: see the area of a regular polygon and constructing a regular hexagon.

Exercises

A side of a 45-45-90 triangle is given

Each triangle is a 45-45-90 triangle. Find the other two sides exactly.

A leg of \displaystyle \sqrt{6}.
A hypotenuse of \displaystyle 5\sqrt{2}.
A hypotenuse of 7.
A hypotenuse of \displaystyle 3\sqrt{6}.

A side of a 30-60-90 triangle is given

Each exercise gives one side of a 30-60-90 triangle. Find the other two exactly.

Short leg \displaystyle \sqrt{3}.
Hypotenuse 9.
Hypotenuse \displaystyle 6\sqrt{3}.
Long leg \displaystyle 3\sqrt{3}.
Long leg 12.
Long leg 5.

Naming the sides

In \triangle XYZ, m\angle Y = 90^\circ, m\angle X = 45^\circ and XY = 11. Which side is opposite \angle X? Find YZ and XZ.
In \triangle KLM, m\angle K = m\angle M = 45^\circ and KM = 20. No right angle is given. Find KL, LM, the perimeter and the area.
In \triangle ABC, \angle C is a right angle and m\angle A = 30^\circ. Name the short leg, the long leg and the hypotenuse.

Which triangle is it?

Say whether each triangle is a 45-45-90 triangle, a 30-60-90 triangle, or neither. (a) Sides 6, 6, \displaystyle 6\sqrt{2}. (b) Sides 2, 2, 3. (c) A right triangle with a hypotenuse of 10 and a leg of \displaystyle 5\sqrt{2}. (d) Sides \displaystyle \sqrt{3}, \displaystyle \sqrt{3}, \displaystyle \sqrt{6}. (e) Sides 6, 10, 12. (f) A right triangle with legs 7 and \displaystyle 7\sqrt{3}.
A right triangle has a leg of 8 and a hypotenuse of 16. Find its two acute angles and its other leg.
A right triangle has a hypotenuse of 12. Find its legs and its area if (a) it is a 45-45-90 triangle, (b) it is a 30-60-90 triangle. Which of the two has the larger area?

Areas, squares and equilateral triangles

An isosceles right triangle has an area of 72. Find its legs and its hypotenuse.
A square has a diagonal of 18. Find its side, its perimeter and its area.
The diagonal of a square is 4 longer than its side. Find the side and the diagonal, exactly and to the nearest hundredth.
An equilateral triangle has a height of 6. Find its side and its area exactly.

Inside other figures

Find the distance between (-3, 1) and (4, 8) by spotting a 45-45-90 triangle.
A regular octagon with sides of 6 is cut from a square by removing four congruent corners, as in Example 9. Find the side of the square, exactly and to the nearest hundredth.
The hypotenuse of a 45-45-90 triangle runs from (1, 2) to (7, 2), and the vertex at the right angle lies above it. Find that vertex.
A ladder 16 feet long leans against a vertical wall and makes a 60^\circ angle with the level ground. How far from the wall is its foot, and how high up the wall does it reach? Give the height exactly and to two decimal places.
A rectangle’s diagonal is 18 long and makes a 30^\circ angle with the longer side. Find the two sides and the area.
A regular hexagon with side 4 is cut from its center into six equilateral triangles. Find the distance between two opposite sides, and the area of the hexagon.
An isosceles triangle has base angles of 30^\circ and a base of 12. Find its height and the length of each equal side.

Reasoning

Explain why neither special right triangle ever has three whole-number sides, and so why neither is a Pythagorean triple, even though 3, 4, 5 is one.
An isosceles triangle has an angle of 45^\circ. Must it be a 45-45-90 triangle? Explain.
The diagonal of a rectangle is \displaystyle \sqrt{2} times one of its sides. Show that the rectangle is a square.
In a 30-60-90 triangle ABC with the right angle at C and m\angle A = 30^\circ, the altitude from C meets the hypotenuse at D. With short leg BC = x, show that \displaystyle BD = \frac{x}{2} and \displaystyle AD = \frac{3x}{2}, so the altitude cuts the hypotenuse in the ratio 1 : 3.
A right triangle has legs a and \displaystyle a\sqrt{3}. Prove that its angles are 30^\circ, 60^\circ and 90^\circ, and say which angle is opposite which leg.
Using only the numbers x, \displaystyle x\sqrt{3} and 2x, explain why the long leg of a 30-60-90 triangle is longer than the short leg and shorter than the hypotenuse, and say how that matches the angles opposite them.
Explain why it is enough to work out the sides of a single 45-45-90 triangle, the one with legs of 1, and a single 30-60-90 triangle, the one with a short leg of 1, to know the ratio of the sides in every special right triangle.

Answers

The other leg is \displaystyle \sqrt{6} and the hypotenuse is \displaystyle 2\sqrt{3}. The legs are equal, and \displaystyle \sqrt{6} \cdot \sqrt{2} = \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}.
Each leg is 5. The hypotenuse already has the factor \displaystyle \sqrt{2}: \displaystyle \frac{5\sqrt{2}}{\sqrt{2}} = 5.
Each leg is \displaystyle \frac{7\sqrt{2}}{2} \approx 4.95. Each leg is \displaystyle \frac{7}{\sqrt{2}}; multiplying top and bottom by \displaystyle \sqrt{2} gives \displaystyle \frac{7\sqrt{2}}{2}.
Each leg is \displaystyle 3\sqrt{3} \approx 5.20. Each leg is \displaystyle \frac{3\sqrt{6}}{\sqrt{2}} = 3\sqrt{\frac{6}{2}} = 3\sqrt{3}.
Long leg 3, hypotenuse \displaystyle 2\sqrt{3}. The long leg is \displaystyle \sqrt{3} \cdot \sqrt{3} = 3, and the hypotenuse is twice the short leg.
Short leg \displaystyle \frac{9}{2}, long leg \displaystyle \frac{9\sqrt{3}}{2}. Half the hypotenuse is \displaystyle \frac{9}{2}, then multiply by \displaystyle \sqrt{3}.
Short leg \displaystyle 3\sqrt{3}, long leg 9. Half of \displaystyle 6\sqrt{3} is \displaystyle 3\sqrt{3}, and \displaystyle 3\sqrt{3} \cdot \sqrt{3} = 9.
Short leg 3, hypotenuse 6. The long leg already has the factor \displaystyle \sqrt{3}: from \displaystyle x\sqrt{3} = 3\sqrt{3}, x = 3, with nothing to rationalize.
Short leg \displaystyle 4\sqrt{3}, hypotenuse \displaystyle 8\sqrt{3}. \displaystyle x = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}, and the hypotenuse is 2x.
Short leg \displaystyle \frac{5\sqrt{3}}{3}, hypotenuse \displaystyle \frac{10\sqrt{3}}{3}. The short leg is \displaystyle \frac{5}{\sqrt{3}}, rationalized, and the hypotenuse is twice it.
\overline{YZ} is opposite \angle X. YZ = 11 and \displaystyle XZ = 11\sqrt{2} \approx 15.56. The third angle is m\angle Z = 45^\circ, so the legs \overline{XY} and \overline{YZ} are equal, and \overline{XZ}, opposite the right angle, is the hypotenuse.
\displaystyle KL = LM = 10\sqrt{2} \approx 14.14, the perimeter is \displaystyle 20 + 20\sqrt{2} \approx 48.28 and the area is 100. The angle sum gives m\angle L = 180^\circ - 45^\circ - 45^\circ = 90^\circ, so \overline{KM}, opposite it, is the hypotenuse and each leg is \displaystyle \frac{20}{\sqrt{2}} = 10\sqrt{2}. The area is \displaystyle {\frac{1}{2} \cdot 10\sqrt{2} \cdot 10\sqrt{2} = \frac{1}{2} \cdot 200 = 100}.
Short leg \overline{BC}, long leg \overline{AC}, hypotenuse \overline{AB}. The third angle is m\angle B = 60^\circ. \overline{BC} is opposite the 30^\circ angle at A, \overline{AC} is opposite the 60^\circ angle at B, and \overline{AB} is opposite the right angle.
(a) 45-45-90: the sides are x, x, \displaystyle x\sqrt{2} with x = 6. (b) Neither: 2^2 + 2^2 = 8, not 3^2 = 9, so there is no right angle. (c) 45-45-90: the other leg b has b^2 = 100 - 50 = 50, so \displaystyle b = 5\sqrt{2}, and the legs are equal. (d) 45-45-90: \displaystyle \sqrt{3} \cdot \sqrt{2} = \sqrt{6}, so the sides are x, x, \displaystyle x\sqrt{2} with \displaystyle x = \sqrt{3}. (e) Neither: 6^2 + 10^2 = 136 but 12^2 = 144, so it is not a right triangle, even though 12 is twice 6. (f) 30-60-90: the hypotenuse is \displaystyle \sqrt{49 + 147} = \sqrt{196} = 14, twice the leg 7, so 30^\circ is opposite the leg 7 and 60^\circ is opposite \displaystyle 7\sqrt{3}.
30^\circ opposite the leg 8, 60^\circ opposite the other leg, which is \displaystyle 8\sqrt{3}. The hypotenuse is twice the leg 8, so the converse applies; the other leg is \displaystyle \sqrt{16^2 - 8^2} = \sqrt{192} = 8\sqrt{3}.
(a) Legs \displaystyle 6\sqrt{2} and \displaystyle 6\sqrt{2}, area 36. (b) Legs 6 and \displaystyle 6\sqrt{3}, area \displaystyle 18\sqrt{3} \approx 31.18. The 45-45-90 triangle has the larger area. In (a) each leg is \displaystyle \frac{12}{\sqrt{2}} = 6\sqrt{2} and the area is \displaystyle \frac{1}{2} \cdot 72; in (b) the short leg is 12 \div 2 = 6 and the area is \displaystyle \frac{1}{2} \cdot 6 \cdot 6\sqrt{3}.
The legs are 12 and the hypotenuse is \displaystyle 12\sqrt{2}. The legs are equal and perpendicular, so with legs x the area is \displaystyle \frac{1}{2}x^2 = 72; then x^2 = 144 and x = 12.
The side is \displaystyle 9\sqrt{2}, the perimeter is \displaystyle 36\sqrt{2} \approx 50.91 and the area is 162. The side is \displaystyle \frac{18}{\sqrt{2}} = 9\sqrt{2}, and the area is \displaystyle \frac{18^2}{2} = \frac{324}{2}.
The side is \displaystyle 4 + 4\sqrt{2} \approx 9.66 and the diagonal is \displaystyle 8 + 4\sqrt{2} \approx 13.66. From \displaystyle s\sqrt{2} = s + 4, \displaystyle s\left(\sqrt{2} - 1\right) = 4, so \displaystyle s = \frac{4}{\sqrt{2} - 1}. Multiply top and bottom by \displaystyle \sqrt{2} + 1, since \displaystyle \left(\sqrt{2} - 1\right)\left(\sqrt{2} + 1\right) = 2 - 1 = 1: \displaystyle s = 4\sqrt{2} + 4. The diagonal is s + 4.
Side \displaystyle 4\sqrt{3}, area \displaystyle 12\sqrt{3}. The height is the long leg of each half, so the short leg is \displaystyle \frac{6}{\sqrt{3}} = 2\sqrt{3} and the side, a hypotenuse, is \displaystyle 4\sqrt{3}. The area is \displaystyle \frac{1}{2} \cdot 4\sqrt{3} \cdot 6 = 12\sqrt{3}.
\displaystyle 7\sqrt{2} \approx 9.90. The horizontal step is {4 - (-3) = 7} and the vertical step is {8 - 1 = 7}. The two steps meet at a right angle and are equal, so the segment is the hypotenuse of a 45-45-90 triangle with legs 7.
\displaystyle 6 + 6\sqrt{2} \approx 14.49. Each cut-off corner is a 45-45-90 triangle whose hypotenuse is a side of the octagon, so \displaystyle t\sqrt{2} = 6 and \displaystyle t = \frac{6}{\sqrt{2}} = 3\sqrt{2}. Each side of the square is \displaystyle t + 6 + t = 6 + 6\sqrt{2}.
(4, 5). By Example 7 the altitude from the right angle meets the hypotenuse at its midpoint, (4, 2), and is half the hypotenuse, \displaystyle \frac{6}{2} = 3. The hypotenuse is horizontal, so the altitude is vertical, and the vertex is 3 above (4, 2). Check: from (4, 5) each end of the hypotenuse is 3 across and 3 down, a distance of \displaystyle 3\sqrt{2}, and \displaystyle 3\sqrt{2} \cdot \sqrt{2} = 6.
The foot is 8 feet from the wall and the top is \displaystyle 8\sqrt{3} feet up, about 13.86 feet. The ladder is the hypotenuse. The ground angle is 60^\circ, so the wall side is opposite it and is the long leg; the ground side is opposite the 30^\circ angle at the top and is 16 \div 2 = 8.
Sides 9 and \displaystyle 9\sqrt{3}, area \displaystyle 81\sqrt{3}. The diagonal is the hypotenuse of a right triangle whose legs are two sides of the rectangle. The shorter side is opposite the 30^\circ angle, so it is 18 \div 2 = 9; the longer is \displaystyle 9\sqrt{3}; the area is \displaystyle 9 \cdot 9\sqrt{3}.
Distance \displaystyle 4\sqrt{3}, area \displaystyle 24\sqrt{3}. Each triangle has side 4 and height \displaystyle \frac{4\sqrt{3}}{2} = 2\sqrt{3} (Example 6(b)). Opposite sides are two heights apart, and the area is \displaystyle 6 \cdot \frac{1}{2} \cdot 4 \cdot 2\sqrt{3} = 24\sqrt{3}.
Height \displaystyle 2\sqrt{3}, each equal side \displaystyle 4\sqrt{3}. The altitude to the base cuts it into two 30-60-90 triangles with long leg 6, half the base, opposite the 60^\circ half of the 120^\circ apex angle. The short leg, the height, is \displaystyle \frac{6}{\sqrt{3}} = 2\sqrt{3}, and the hypotenuse is twice that.
In a 45-45-90 triangle the hypotenuse divided by a leg is \displaystyle \sqrt{2}, and in a 30-60-90 triangle the long leg divided by the short leg is \displaystyle \sqrt{3}. If those two sides were both whole numbers, their quotient would be a fraction, but \displaystyle \sqrt{2} and \displaystyle \sqrt{3} are irrational. So at least one side is not whole, and a Pythagorean triple needs all three. The 3, 4, 5 triangle is right, but its legs are unequal and its hypotenuse is not twice a leg, so it is neither special triangle.
No. If the 45^\circ angle is the vertex angle, each base angle is \displaystyle \frac{180^\circ - 45^\circ}{2} = 67.5^\circ, and the triangle has no right angle. If the 45^\circ angle is a base angle, the other base angle is also 45^\circ, the third angle is 90^\circ, and the triangle is a 45-45-90 triangle. So it depends on where the 45^\circ angle is.
Call the sides a and b and the diagonal \displaystyle d = a\sqrt{2}. The diagonal and two sides form a right triangle, because a rectangle’s corners are right angles, so a^2 + b^2 = d^2 = 2a^2. Then b^2 = a^2, and since both are lengths, b = a. A rectangle with two neighboring sides equal has all four sides equal, so it is a square.
Triangle BCD has a right angle at D and m\angle B = 60^\circ, so it is a 30-60-90 triangle with hypotenuse BC = x; its angle at C is 180^\circ - 90^\circ - 60^\circ = 30^\circ. Its short leg BD is opposite that angle, so BD = \tfrac{x}{2}. The whole hypotenuse is AB = 2x, so AD = 2x - \tfrac{x}{2} = \tfrac{3x}{2}. So BD : AD = \tfrac{x}{2} : \tfrac{3x}{2} = 1 : 3.
The hypotenuse is \displaystyle \sqrt{a^2 + 3a^2} = \sqrt{4a^2} = 2a, twice the leg a. By the converse, the angle opposite a is 30^\circ and the other acute angle, opposite \displaystyle a\sqrt{3}, is 60^\circ. The right angle is opposite the hypotenuse.
Since 1 < 3 < 4, taking square roots gives \displaystyle 1 < \sqrt{3} < 2, and multiplying by x > 0 gives \displaystyle x < x\sqrt{3} < 2x. The sides are in the same order as the angles opposite them, 30^\circ < 60^\circ < 90^\circ: the larger angle faces the longer side, as it does in every triangle (the side and angle inequality).
Every 45-45-90 triangle has the same three angles, so any two are similar by angle-angle similarity, and similar triangles have their sides in the same ratio. So the ratio {\text{leg} : \text{leg} : \text{hypotenuse}} is the same in all of them, and the one with legs 1 shows it is \displaystyle {1 : 1 : \sqrt{2}}, since \displaystyle {\sqrt{1^2 + 1^2} = \sqrt{2}}. In the same way every 30-60-90 triangle is similar to the one with short leg 1, which has hypotenuse 2 and long leg \displaystyle \sqrt{2^2 - 1^2} = \sqrt{3}, so the ratio is \displaystyle 1 : \sqrt{3} : 2 in all of them.