Pythagoras theorem and elementary applications

Learning outcome

Find missing lengths in right triangles and use the converse to check whether a triangle is right-angled.

Concepts and assumptions

In triangle ABC with ∠B = 90°, AB and BC are the perpendicular legs; AC is the hypotenuse, opposite the right angle. Pythagoras’ theorem states:

AC² = AB² + BC².

The hypotenuse is the longest side. The equality compares square areas, not the unsquared side lengths. Place four identical right triangles, legs along the edges, inside a square of side a + b; their hypotenuses enclose a square of side c. Thus (a + b)² = 4 × (a × b/2) + c², simplifying to a² + b² = c².

For the hypotenuse, add the leg squares; for a leg, subtract the other leg’s square from the hypotenuse’s square. Take the positive square root: lengths are positive.

The converse states: if a triangle’s longest side c satisfies c² = a² + b², the angle opposite c is 90°.

Work in a Euclidean plane with exact straight-segment lengths. Verify right angles, align units before squaring, and retain radicals unless an approximation is requested. Use ≈ when rounding to two decimal places.

Worked examples

Example 1 — Hypotenuse. Triangle ABC is right at B, with AB = 9 cm and BC = 12 cm. AC² = 9² + 12² = 81 + 144 = 225. Hence AC = 15 cm, longer than either leg.

Example 2 — Ladder height. A straight 17 m ladder rests against a wall perpendicular to level ground. Its foot is 8 m from the wall; ignore thickness. If the vertical contact height is h, then h² + 8² = 17². Thus h² = 289 - 64 = 225 and h = 15 m.

Example 3 — Construct a right triangle. In isosceles triangle ABC, AB = AC = 13 cm and BC = 10 cm. Let D be BC’s midpoint. Triangles ABD and ACD are congruent by SSS, so equal adjacent angles at D are each 90°. Thus BD = 5 cm and AD² = 13² - 5² = 169 - 25 = 144. Height AD = 12 cm; area of ABC = 10 × 12/2 = 60 cm².

Common mistakes

A sketch does not establish a right angle. A sloping side is not automatically the hypotenuse. Do not add lengths instead of their squares. An isosceles triangle’s base midpoint gives a useful perpendicular here; this is not true for an arbitrary triangle.

Practice questions

1. Triangle PQR is right at Q, with PQ = 12 cm and QR = 35 cm. Find PR.
2. Triangle ABC has AB = 10 cm, BC = 24 cm and AC = 26 cm. Is it right-angled? Identify the vertex.
3. Rectangle ABCD has consecutive sides AB = 7 cm and BC = 9 cm. Find diagonal AC exactly and to two decimal places.
4. In triangle PQR, PQ = PR = 17 cm and QR = 16 cm. Find the perpendicular height from P and the area.

Worked solutions

1. PR² = 12² + 35² = 144 + 1225 = 1369. Thus PR = 37 cm.
2. Longest side AC satisfies 26² = 676 = 10² + 24². The converse gives ∠B = 90°.
3. Rectangle corners are right angles, so AC² = 7² + 9² = 130. AC = √130 cm ≈ 11.40 cm.
4. The altitude from P bisects QR into 8 cm halves. Height² = 17² - 8² = 289 - 64 = 225, so height = 15 cm. Area = 16 × 15/2 = 120 cm².