Right-Angled Triangles

Lessons

Senior 3 · Mathematics

Right-Angled Triangles

5 lessons

Lesson 1 of 5

Pythagoras' theorem and the median theorem

25 min

One tap for three houses

Three families live at the corners of a plot shaped like a right-angled triangle. They want to put a shared water tap where it is the same distance from all three houses. Where should it go? Surprisingly, there is always such a point, and for a right-angled triangle it lies exactly in the middle of the longest side. This lesson proves it, after a quick return to the most famous fact about right-angled triangles.

Definition

Hypotenuse and legs: in a right-angled triangle, the side opposite the right angle is the hypotenuse; it is always the longest side. The two sides that form the right angle are the legs.

Notation: a side is often named with the small letter of the opposite vertex: $a = B C$ , $b = C A$ , $c = A B$ .

Important

Pythagoras' theorem. In a triangle with a right angle at $C$ ,

c^2 = a^2 + b^2 \qquad (\text{hypotenuse}^2 = \text{leg}^2 + \text{leg}^2).

Converse: if $c^{2} = a^{2} + b^{2}$ , then the angle opposite $c$ is a right angle.

In S2 you proved the theorem by rearranging four copies of the triangle inside a square. Lesson 2 of this unit gives a second proof, from similar triangles.

Worked example 1: a side and a check

(a) A right-angled triangle has hypotenuse 17 cm and one leg 8 cm. Find the other leg. (b) Is a triangle with sides 9 cm, 40 cm and 41 cm right-angled?

Solution (a) $b^{2} = 17^{2} - 8^{2} = 289 - 64 = 225$ , so $b = 15$ cm.

(b) The longest side is 41: $9^{2} + 40^{2} = 81 + 1600 = 1681 = 41^{2}$ . By the converse, yes: the right angle is opposite the 41 cm side.

Worked example 2: Pythagoras twice, in a classroom

A classroom is a cuboid 8 m long, 6 m wide and 3 m high. How long is the straight line from a corner of the floor to the opposite corner of the ceiling?

First find the floor diagonal AF, then use the right angle at F.

Solution Floor diagonal: $A F^{2} = 8^{2} + 6^{2} = 100$ , so $A F = 10$ m. The edge $F G$ is vertical, so the angle $A F G$ is a right angle:

AG^2 = AF^2 + FG^2 = 100 + 9 = 109, \qquad AG = \sqrt{109} \approx 10.4\ \text{m}.

In one step: the space diagonal of an $l \times w \times h$ cuboid is $\sqrt{l^2 + w^2 + h^2}$ .

Definition

Median of a triangle: the segment from a vertex to the midpoint of the opposite side. Every triangle has three medians.

Example: in triangle $A B C$ , if $M$ is the midpoint of $B C$ , then $A M$ is the median from $A$ .

Important

The median theorem. In a right-angled triangle, the median to the hypotenuse is half the hypotenuse:

\angle A = 90^\circ \text{ and } M \text{ the midpoint of } BC \quad\Longrightarrow\quad AM = BM = MC = \frac{1}{2}BC.

Proof: complete the triangle to a rectangle $A B D C$ by drawing a line through $B$ parallel to $A C$ and a line through $C$ parallel to $A B$ . A rectangle's diagonals are equal and bisect each other. So the diagonals $B C$ and $A D$ cross at their common midpoint $M$ , and $AM = \frac{1}{2}AD = \frac{1}{2}BC$ .

M is the midpoint of both diagonals of the rectangle, so it is the same distance from A, B, C and D.

Converse: if the median from $A$ is half the side $B C$ , then $\angle A = 90^\circ$ . So $M$ is the centre of a circle through $A$ , $B$ and $C$ : the circumscribed circle of a right-angled triangle has the hypotenuse as a diameter.

Worked example 3: the shared tap

The plot has legs of 30 m and 40 m. Where should the tap go, and how far is it from each house?

Solution Hypotenuse: $\sqrt{30^2 + 40^2} = \sqrt{2500} = 50$ m. By the median theorem, the midpoint of the hypotenuse is $\frac{50}{2} = 25$ m from all three corners. The tap goes at the middle of the 50 m side, 25 m from each house.

Worked example 4: proving a right angle with the converse

Triangle $A B C$ has $A (2, 5)$ , $B (- 1, 1)$ and $C (6, 2)$ . Show that $\angle A = 90^\circ$ by using the median from $A$ .

Solution Midpoint of $B C$ : $M = \left(\frac{-1 + 6}{2}, \frac{1 + 2}{2}\right) = (2.5, 1.5)$ .

BC = \sqrt{7^2 + 1^2} = \sqrt{50}, \qquad AM = \sqrt{0.5^2 + 3.5^2} = \sqrt{12.5} = \frac{\sqrt{50}}{2}.

So $AM = \frac{1}{2}BC$ , and by the converse of the median theorem, $\angle A = 90^\circ$ . (Check with Pythagoras: $A B^{2} + A C^{2} = 25 + 25 = 50 = B C^{2}$ ✓.)

Warning

Find the hypotenuse first. Pythagoras subtracts only when you want a leg: $\text{leg}^2 = \text{hyp}^2 - \text{other leg}^2$ . Adding when a leg is missing ( $17^{2} + 8^{2}$ ) is the most common error.

The median theorem is about the median to the hypotenuse only. The medians to the legs are not half of anything in particular.

Tip

Learn the common Pythagorean triples: 3, 4, 5; 5, 12, 13; 8, 15, 17; 7, 24, 25; 9, 40, 41, and their multiples such as 6, 8, 10 or 30, 40, 50. They save time in exams.

Khan Academy: Pythagorean theorem in 3D. Watch how a right-angled triangle is found inside a solid, and how Pythagoras is used twice to find a diagonal, as in Worked example 2.

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Practice

Give lengths to 3 significant figures where they are not exact. The last three are harder.

Find the hypotenuse of a right-angled triangle with legs 20 cm and 21 cm.
A ladder 6.5 m long reaches 6 m up a vertical wall. How far is its foot from the wall?
Is a triangle with sides 12 cm, 35 cm and 37 cm right-angled? Explain.
A right-angled triangle has legs 10 cm and 24 cm. Find the length of the median to the hypotenuse.
The median to the hypotenuse of a right-angled triangle is 6.5 cm and one leg is 5 cm. Find the other leg.
Find the space diagonal of a box 12 cm by 4 cm by 3 cm.
$P (0, 0)$ , $Q (8, 0)$ and $R (2, k)$ with $k > 0$ . Find $k$ so that $\angle PRQ = 90^\circ$ , using the median from $R$ .

Show answersHide answers

$\sqrt{400 + 441} = \sqrt{841} = 29$ cm.
$\sqrt{6.5^2 - 6^2} = \sqrt{6.25} = 2.5$ m.
$12^{2} + 35^{2} = 144 + 1225 = 1369 = 37^{2}$ : yes, by the converse of Pythagoras.
Hypotenuse $\sqrt{100 + 576} = 26$ cm, so the median is 13 cm.
The hypotenuse is $2 \times 6.5 = 13$ cm, so the other leg is $\sqrt{169 - 25} = 12$ cm.
$\sqrt{144 + 16 + 9} = \sqrt{169} = 13$ cm.
The midpoint of $P Q$ is $(4, 0)$ , and $R$ must be $\frac{8}{2} = 4$ from it: $(2 - 4)^{2} + k^{2} = 16$ , so $k^{2} = 12$ and $k = 2\sqrt{3} \approx 3.46$ .

Sources and credits

Written by Ganzaa.
Video: Pythagorean theorem in 3D, Khan Academy, embedded from YouTube.

The diagrams, worked examples, practice items and quiz questions were written by Ganzaa; the plot and classroom measurements are illustrative.

1One tap for three houses

One tap for three houses