Right-Angled Triangles

Lessons

Senior 3 · Mathematics

Right-Angled Triangles

5 lessons

Lesson 2 of 5

The altitude and leg theorems

25 min

Three triangles in one

Draw a right-angled triangle and drop a perpendicular from the right angle to the hypotenuse. This one line, the altitude, cuts the triangle into two smaller right-angled triangles, and all three triangles have the same shape. From that single fact come three theorems, and a second proof of Pythagoras' theorem.

The altitude CH splits the hypotenuse into the projections q = AH and p = HB.

Definition

Altitude (height) to the hypotenuse: the perpendicular segment $C H$ from the right-angle vertex $C$ to the hypotenuse $A B$ . Its length is $h = C H$ .

Projection of a leg on the hypotenuse: the part of the hypotenuse next to that leg. The projection of $a = B C$ is $p = H B$ , and the projection of $b = A C$ is $q = A H$ , with $p + q = c$ .

Example: in a 3, 4, 5 triangle the projections of the legs 3 and 4 are 1.8 and 3.2, which add up to 5.

Important

Why the three triangles are similar. Triangles $A C B$ , $A H C$ and $C H B$ each have a right angle. $A H C$ shares $\angle A$ with $A C B$ , and $C H B$ shares $\angle B$ with $A C B$ . Two equal angles are enough (AA), so

\triangle ACB \sim \triangle AHC \sim \triangle CHB.

Similar triangles have their corresponding sides in the same ratio. Each theorem below is one of these ratios.

Important

The altitude theorem. The square of the altitude equals the product of the two projections:

h^2 = p\,q.

Proof: from $\triangle AHC \sim \triangle CHB$ , $\frac{AH}{CH} = \frac{CH}{HB}$ , that is, $\frac{q}{h} = \frac{h}{p}$ , so $h^{2} = p q$ .

The leg theorem. The square of each leg equals the hypotenuse times that leg's projection:

a^2 = c\,p, \qquad b^2 = c\,q.

Proof: from $\triangle ACB \sim \triangle CHB$ , $\frac{BC}{AB} = \frac{HB}{CB}$ , that is, $\frac{a}{c} = \frac{p}{a}$ , so $a^{2} = c p$ . In the same way, $\triangle ACB \sim \triangle AHC$ gives $b^{2} = c q$ .

The area relation. The area is $\frac{1}{2}ab$ (legs as base and height) and also $\frac{1}{2}ch$ (hypotenuse as base), so $a b = c h$ .

Important

A second proof of Pythagoras' theorem. Add the two leg theorems:

a^2 + b^2 = cp + cq = c(p + q) = c \times c = c^2.

This proof uses similar triangles instead of rearranging areas. It works for every right-angled triangle, because the altitude always lands between $A$ and $B$ .

Worked example 1: from the projections

The altitude to the hypotenuse divides it into $p = 4$ cm and $q = 9$ cm. Find $h$ , $c$ , $a$ and $b$ .

Solution $h^{2} = p q = 36$ , so $h = 6$ cm, and $c = p + q = 13$ cm.

a^2 = cp = 52 \Rightarrow a = 2\sqrt{13} \approx 7.21\ \text{cm}, \qquad b^2 = cq = 117 \Rightarrow b = 3\sqrt{13} \approx 10.8\ \text{cm}.

Check: $a^{2} + b^{2} = 52 + 117 = 169 = 13^{2}$ ✓, and $ab = 6 \times 13 = 78 = ch$ ✓.

Worked example 2: from the legs

A right-angled triangle has legs 15 cm and 20 cm. Find the altitude to the hypotenuse and the two projections.

Solution $c = \sqrt{225 + 400} = 25$ cm. Area relation: $h = \frac{ab}{c} = \frac{15 \times 20}{25} = 12$ cm.

Leg theorem: the projection of 15 is $\frac{15^2}{25} = 9$ cm and of 20 is $\frac{20^2}{25} = 16$ cm. Check: $9 + 16 = 25$ ✓ and $9 \times 16 = 144 = 12^2$ ✓.

Worked example 3: the height of a roof

Two rafters of a roof meet at a right angle at the ridge $C$ . The span $A B$ is 10 m, and the ridge is directly above a point $H$ of the span, 3.6 m from $A$ . How high is the ridge above the span, and how long is each rafter?

Solution $q = 3.6$ m and $p = 10 - 3.6 = 6.4$ m. Altitude theorem: $h^2 = 3.6 \times 6.4 = 23.04$ , so the ridge is $h = 4.8$ m high.

Leg theorem: $b^2 = 10 \times 3.6 = 36$ , so $A C = 6$ m; $a^2 = 10 \times 6.4 = 64$ , so $B C = 8$ m. (A 6, 8, 10 triangle ✓.)

Warning

Pair each leg with its own projection. The projection of a leg is the piece of the hypotenuse next to it: $a = B C$ goes with $p = H B$ , not with $q = A H$ . A sketch with the letters on it prevents the mix-up.

The altitude theorem uses the two pieces, not the whole: $h^{2} = p q$ , not $h^{2} = p c$ .

Tip

Memory hook: "altitude squared = piece times piece; leg squared = whole times its own piece".

Khan Academy: Pythagorean theorem proof using similarity. Watch how the altitude splits the triangle into two triangles similar to the whole, and how the ratios lead to a² + b² = c², as in this lesson.

✦

Practice

Use the notation of the figure: legs $a$ and $b$ , hypotenuse $c$ , altitude $h$ , projections $p$ (of $a$ ) and $q$ (of $b$ ). The last three are harder.

$p = 2$ cm and $q = 8$ cm. Find $h$ .
$h = 6$ cm and $p = 4$ cm. Find $q$ and $c$ .
$c = 25$ cm and $p = 16$ cm. Find $a$ and $b$ .
The legs are 9 cm and 12 cm. Find $c$ , $h$ , $p$ and $q$ .
$a = 6$ cm and its projection is $p = 4$ cm. Find $c$ , $q$ , $h$ and $b$ .
Show that $\frac{a^2}{b^2} = \frac{p}{q}$ . If the legs are 3 cm and 4 cm, in what ratio does the altitude divide the hypotenuse?
A pupil writes $h^2 = a \times b$ . Test this on the 15, 20, 25 triangle of Worked example 2, and correct it.

Show answersHide answers

$h = \sqrt{16} = 4$ cm.
$q = \frac{36}{4} = 9$ cm; $c = 13$ cm.
$q = 9$ ; $a = \sqrt{25 \times 16} = 20$ cm, $b = \sqrt{25 \times 9} = 15$ cm.
$c = 15$ ; $h = \frac{108}{15} = 7.2$ ; $p = \frac{81}{15} = 5.4$ and $q = \frac{144}{15} = 9.6$ cm (if $a = 9$ ).
$c = \frac{36}{4} = 9$ ; $q = 5$ ; $h = \sqrt{20} = 2\sqrt{5} \approx 4.47$ ; $b = \sqrt{45} = 3\sqrt{5} \approx 6.71$ cm.
Divide $a^{2} = c p$ by $b^{2} = c q$ : $\frac{a^2}{b^2} = \frac{p}{q}$ . For legs 3 and 4 the ratio is $9 : 16$ (the pieces are 1.8 and 3.2 cm).
$a b = 300$ but $h^{2} = 144$ , so it is false. Correct forms: $h^{2} = p q$ , or $a b = c h$ .

Sources and credits

Written by Ganzaa.
Video: Pythagorean theorem proof using similarity, Khan Academy, embedded from YouTube.

The diagram, proofs as written, worked examples, practice items and quiz questions were written by Ganzaa; the roof measurements are illustrative.

1Three triangles in one

Three triangles in one