Math by First Principles

Special Right Triangles

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Learning goals

  • Derive both special right triangle side ratios from congruence and Pythagoras
  • Identify side roles from angle markings in any orientation
  • Recover missing lengths exactly from any known side
  • Connect square diagonals and equilateral altitudes to the special triangles

Equal legs and a diagonal

Two equal legs in a right triangle force two equal acute angles. If the legs each measure 44 centimeters, their opposite angles match by the isosceles base-angle theorem. Those angles total 90∘90^\circ, so each is 45∘45^\circ.

Pythagoras gives the hypotenuse length dd:

d2=42+42.d2=32.d=42 centimeters.\begin{aligned} d^2&=4^2+4^2.\\ d^2&=32.\\ d&=4\sqrt2\text{ centimeters}. \end{aligned}
Triangle relationshipsRight triangle CAB has its marked right angle at C. The two legs have matching ticks. Side CA is labeled 4 cm. Side CB is labeled 4 cm. Hypotenuse AB is labeled 4√2 cm.4 cm4 cm4√2 cmCA45°B45°
Equal legs determine the acute angles and the diagonal.

A 45-45-90 triangle is a right triangle whose acute angles each measure 45∘45^\circ. Conversely, those equal angles force equal opposite legs by the converse of the base-angle theorem. The equal-leg condition and the angle condition describe the same triangles.

The equal-leg side relationship#

For a numerical instance inside the argument, take legs of 33 units. Pythagoras gives

h2=32+32.h2=18.h=32.\begin{aligned} h^2&=3^2+3^2.\\ h^2&=18.\\ h&=3\sqrt2. \end{aligned}

The same steps work for any positive leg length ss:

Triangle relationshipsRight triangle CAB has its marked right angle at C. The two legs have matching ticks. Side CA is labeled s. Side CB is labeled s. Hypotenuse AB is labeled h.sshCA45°B45°
The equal legs supply both terms in Pythagoras.
h2=s2+s2.h2=2s2.h=s2.\begin{aligned} h^2&=s^2+s^2.\\ h^2&=2s^2.\\ h&=s\sqrt2. \end{aligned}

We take the positive root because hh is a length. The number 33 played no special role; equal positive legs were the only length condition.

For the two legs followed by the hypotenuse, the ordered side ratio is

1:1:2.1:1:\sqrt2.

Worked example 1 One given leg

Problem. In triangle RSTRST, the givens are

m∠R=90∘.m∠S=45∘.RS=6 cm.\begin{aligned} m\angle R&=90^\circ.\\ m\angle S&=45^\circ.\\ RS&=6\text{ cm}. \end{aligned}

Find RTRT and STST exactly.

Triangle relationshipsRight triangle RST has its marked right angle at R. The two legs have matching ticks. Side RS is labeled 6 cm. Side RT is labeled RT. Hypotenuse ST is labeled ST.6 cmRTSTRS45°T45°
The marked angles and given leg identify the side roles.

Solution. The angle at TT is also 45∘45^\circ. Equal angles have equal opposite sides, so

RT=6 cm.RT=6\text{ cm}.

The side opposite the right angle is STST. Multiply a leg by 2\sqrt2:

ST=62 cm.ST=6\sqrt2\text{ cm}.

The hypotenuse is longer than either leg, as required.

Check your understanding

A right triangle has two equal legs, each 8 units long. Why is its hypotenuse not 16 units?

Work backward from the hypotenuse

Knowing the hypotenuse reverses the multiplication. For example, if an equal-leg right triangle has hypotenuse 12212\sqrt2 cm, each leg measures 1212 cm. In general, divide the hypotenuse by 2\sqrt2:

s=h2.s=\frac{h}{\sqrt2}.

Rationalizing writes the same exact length with a rational denominator:

s=h22.s=\frac{h\sqrt2}{2}.

Dividing by 22 alone would give half the hypotenuse, which belongs to a different special triangle.

Worked example 2 A diagonal gives the edge

Problem. Square PQRSPQRS has diagonal PRPR of length 1010 m. Find the exact side length and perimeter of the square. A square has four equal sides and four right angles.

Triangle relationshipsSquare PQRS has all four equal sides and four right angles. Diagonal PR is dashed. Side PQ is labeled s. Diagonal PR is labeled 10 m.PQRSs10 m
A diagonal joins opposite vertices of the square.

Solution. In triangle PQRPQR, the equal sides PQPQ and QRQR meet at a right angle. Thus PRPR is its hypotenuse, and the side length ss satisfies

s=102.s=52 m.\begin{aligned} s&=\frac{10}{\sqrt2}.\\ s&=5\sqrt2\text{ m}. \end{aligned}

The perimeter is four side lengths:

P=202 m.P=20\sqrt2\text{ m}.

The diagonal is a hypotenuse, even when the square is turned.

Check your understanding

An equal-leg right triangle has hypotenuse 969\sqrt6 cm. What is each leg?

Split an equilateral triangle

Halving an equilateral triangle gives a second fixed side pattern. Start with an equilateral triangle ABCABC of side length 1212 cm. Let MM be the midpoint of BCBC, and join AA to MM.

The triangles ABMABM and ACMACM have equal outer sides, equal half-base lengths of 66 cm, and shared side AMAM. Side-side-side (SSS) makes them congruent. Corresponding angles at MM are equal and form a straight angle, so each is 90∘90^\circ. Corresponding angles at AA are equal and split 60∘60^\circ, so each is 30∘30^\circ. Each angle at BB or CC remains 60∘60^\circ.

Only after establishing the right angle can we apply Pythagoras to the altitude AMAM:

AM2=122−62.AM2=108.AM=63 cm.\begin{aligned} AM^2&=12^2-6^2.\\ AM^2&=108.\\ AM&=6\sqrt3\text{ cm}. \end{aligned}
Triangle relationshipsEquilateral triangle ABC has midpoint M on BC and segment AM. Matching single ticks mark the equal outer sides AB and AC. The proved right angle at M and equal half-base double ticks are shown. The sloping sides are labeled 12 cm. Each half of BC is labeled 6 cm. AM is labeled 6√3 cm.ABCM12 cm12 cm6 cm6 cmAM: 6√3 cm30°30°
The perpendicular and bisection markings follow from SSS.

In either half, the short leg is half the original equilateral side. The altitude is the long leg; the original sloping side is the hypotenuse.

The half-equilateral side relationship#

Take an equilateral triangle of side 88 units and join a vertex to the opposite midpoint. SSS gives congruent halves: the outer sides are 88, the half-base lengths are 44, and the dividing side is shared. The equal adjacent angles at the midpoint are each 90∘90^\circ, and the equal halves of the vertex angle are each 30∘30^\circ. The other acute angle in each half is 60∘60^\circ.

For the altitude ll, Pythagoras gives

l2=82−42.l2=48.l=43.\begin{aligned} l^2&=8^2-4^2.\\ l^2&=48.\\ l&=4\sqrt3. \end{aligned}

For a general positive short-leg length ss, start instead with an equilateral triangle of side 2s2s. The midpoint join again gives congruent halves by SSS, right angles at the midpoint, and 30∘30^\circ vertex halves. Now

l2=(2s)2−s2.l2=3s2.l=s3.\begin{aligned} l^2&=(2s)^2-s^2.\\ l^2&=3s^2.\\ l&=s\sqrt3. \end{aligned}

Nothing depended on the particular side length 88. Every 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle is similar to one of these halves by AA, so it has the same ordered side ratio.

Triangle relationshipsEquilateral triangle ABC has midpoint M on BC and segment AM. Matching single ticks mark the equal outer sides AB and AC. The proved right angle at M and equal half-base double ticks are shown. The sloping sides are labeled 2s. Each half of BC is labeled s. AM is labeled s√3.ABCM2s2sssAM: s√330°30°
Blue and dashed red distinguish the congruent halves after their properties have been proved.

A 30-60-90 triangle is a right triangle whose acute angles measure 30∘30^\circ and 60∘60^\circ. Its side lengths, in order opposite those angles and the right angle, are

short leg=s.long leg=s3.hypotenuse=2s,\begin{aligned} \text{short leg}&=s.\\ \text{long leg}&=s\sqrt3.\\ \text{hypotenuse}&=2s, \end{aligned}

where ss is positive. Thus the ordered ratio of short leg to long leg to hypotenuse is

1:3:2.1:\sqrt3:2.

Check your understanding

M is the midpoint of BC in equilateral triangle ABC. Why is AM perpendicular to BC?

Match the side before choosing an operation

Rotating a triangle changes no angle or length. Locate the right angle first, then identify the short leg as the side opposite 30∘30^\circ. In particular, a vertical side can be either leg or the hypotenuse.

Worked example 3 A tilted triangle

Problem. In triangle JKLJKL, the givens are

m∠J=90∘.m∠K=30∘.JL=5 cm.\begin{aligned} m\angle J&=90^\circ.\\ m\angle K&=30^\circ.\\ JL&=5\text{ cm}. \end{aligned}

Find JKJK and KLKL exactly.

Triangle relationshipsRight triangle JKL has its marked right angle at J. The angle at K is 30 degrees and the angle at L is 60 degrees. Side JK is labeled JK. Side JL is labeled 5 cm. Hypotenuse KL is labeled KL.JK5 cmKLJK30°L60°
The short leg lies opposite the marked 30-degree angle.

Solution. Side JLJL is opposite 30∘30^\circ, so it is the short leg. The long leg is JKJK:

JK=53 cm.JK=5\sqrt3\text{ cm}.

The side opposite the right angle is KLKL:

KL=10 cm.KL=10\text{ cm}.

The side roles come from the angles rather than the page orientation.

Worked example 4 A given longer leg

Problem. A 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle has long leg 767\sqrt6 meters. Find its short leg and hypotenuse in simplest exact form.

Triangle relationshipsRight triangle CAB has its marked right angle at C. The angle at A is 30 degrees and the angle at B is 60 degrees. Side CA is labeled 7√6 m. Side CB is labeled s. Hypotenuse AB is labeled h.7√6 mshCA30°B60°
Identify the given longer leg before recovering the short leg.

Solution. The long leg is 3\sqrt3 times the short leg. Divide to recover the short leg ss:

s=763.s=72 m.\begin{aligned} s&=\frac{7\sqrt6}{\sqrt3}.\\ s&=7\sqrt2\text{ m}. \end{aligned}

Double the short leg to get the hypotenuse hh:

h=142 m.h=14\sqrt2\text{ m}.

Check by squares, keeping the values exact:

(72)2=98.(76)2=294.(142)2=392.\begin{aligned} (7\sqrt2)^2&=98.\\ (7\sqrt6)^2&=294.\\ (14\sqrt2)^2&=392. \end{aligned}

The first two squares sum to the third.

Worked example 5 An equilateral altitude

Problem. An equilateral triangle has side length 1818 mm. Find the exact length of its altitude.

Triangle relationshipsEquilateral triangle ABC has midpoint M on BC and segment AM. Matching single ticks mark the equal outer sides AB and AC. The proved right angle at M and equal half-base double ticks are shown. The sloping sides are labeled 18 mm. Each half of BC is labeled 9 mm. AM is labeled h.ABCM18 mm18 mm9 mm9 mmAM: h30°30°
The midpoint join forms the two right triangles used in the solution.

Solution. The vertex-to-midpoint segment is perpendicular to the opposite side by the SSS argument above. Each half-base is 99 mm. In either right triangle the short leg is 99 mm, and the altitude is the long leg:

h=93 mm.h=9\sqrt3\text{ mm}.

For any equilateral side length ee, the same reasoning gives

h=e32.h=\frac{e\sqrt3}{2}.

The altitude is shorter than the sloping side, which is the hypotenuse of each half.

The ratios also identify angles when all three positive sides are given. Construct a special triangle with the same short side, then compare the three sides by SSS. For instance, sides 22, 232\sqrt3, and 44 match a half-equilateral triangle, so their opposite angles are 30∘30^\circ, 60∘60^\circ, and 90∘90^\circ respectively. Knowing only one pair of lengths in that ratio, without a right angle or third side, would not justify that conclusion.

Check your understanding

In a 30-60-90 triangle the hypotenuse is 22 cm. Which is the long leg?

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Other explanations of this lesson, if you want a second take.

A bit of history (optional)

In Plato’s dialogue Meno, Socrates asks an enslaved boy how to make a square with twice the area of a given square. The boy first proposes doubling its side. Questions about the drawing reveal the problem: that produces four times the area, not twice.

Socrates arranges four equal squares and draws their corner-to-corner lines. Those lines enclose a new square made from four half-squares. The boy recognizes that its area is twice the original square’s area. The side they needed was already present as the original square’s diagonal.

This is a scene in a philosophical dialogue, not a transcript documenting an actual lesson. Plato uses the boy’s changing answers in a larger discussion about learning. You can read the exchange in the translated Meno.

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