MathematicsGrade 7–9

Pythagorean theorem

Squares on the sides of a right triangle and a puzzle proof show: a² + b² = c².

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What is it about?

In a right triangle the sides next to the right angle are the legs aa and bb; the side opposite the right angle is the hypotenuse cc.

Pythagorean theorem: In every right triangle, the areas of the squares on the two legs add up to the area of the square on the hypotenuse: a2+b2=c2a^2 + b^2 = c^2

So c=a2+b2c = \sqrt{a^2 + b^2} and a=c2−b2a = \sqrt{c^2 - b^2}. Converse: if a2+b2=c2a^2 + b^2 = c^2 in a triangle, the triangle is right-angled, with the right angle opposite cc.

Puzzle proof

Four copies of the triangle are placed in a square with side a+ba + b. In the first arrangement the free area is c2c^2, in the second it is a2+b2a^2 + b^2. The big square and the triangles are the same, so the free areas are equal.

Try it

  1. Drag the two coloured vertices and count the unit squares. Isosceles
  2. Switch to the puzzle proof and move the triangles. Puzzle proof

Exercises

Exercise 1: Hypotenuse

Legs a=1.5 cma = 1.5\,\text{cm}, b=2 cmb = 2\,\text{cm}. Find cc.

Show solution

c=2.25+4=6.25=2.5 cmc = \sqrt{2.25 + 4} = \sqrt{6.25} = 2.5\,\text{cm}. Show in the simulation

Exercise 2: Ladder

A 5 m5\,\text{m} ladder stands 1.4 m1.4\,\text{m} from a wall. How high does it reach?

Show solution

h=25−1.96=23.04=4.8 mh = \sqrt{25 - 1.96} = \sqrt{23.04} = 4.8\,\text{m}.

Exercise 3: Right-angled or not?

Check the triangles with sides 7,24,257, 24, 25 and 5,6,85, 6, 8.

Show solution

49+576=625=25249 + 576 = 625 = 25^2: right-angled. 25+36=61≠6425 + 36 = 61 \neq 64: not right-angled.

Ideas for the classroom

  • Use: Grade 8/9. Counting squares works best with integer triples like 3-4-5.
  • Common stumbling blocks: applying the theorem to non-right triangles; choosing the wrong side as cc; a2+b2=a+b\sqrt{a^2 + b^2} = a + b.