Pythagoras’ Theorem

The two shorter squares equal the longest.

Look for the right angle

A right angle is a quarter turn, 90°, and a drawing marks it with a small square in the corner. A triangle with a right angle in it is called a right triangle.

Pythagoras’ theorem is a rule about the three sides of a right triangle. It works only when one corner is exactly 90°, so the first thing to do is find that corner.

34

A right triangle whose two sides at the right angle are 3 and 4 long.

The hypotenuse

The side across from the right angle, the one that does not touch it, is called the hypotenuse. It is always the longest side of a right triangle. The two sides that meet at the right angle are the shorter sides, and they are also called the legs.

In this triangle the legs are 3 and 4, and the hypotenuse is the sloping side, marked c. Its length is not given, and the theorem finds it.

34c

The hypotenuse c is the side across from the right angle.

A square on each side

Draw a square on each side of the triangle. The square on the side of 3 has an area of 3² = 3 × 3 = 9 unit squares, and the square on the side of 4 has an area of 4² = 16.

The theorem says that the two smaller squares together have the same area as the square on the hypotenuse. So the square on the hypotenuse has an area of 9 + 16 = 25. The side of a square of area 25 is 5, because 5² = 25, so the hypotenuse is 5.

θ = 60°show the unit tiles

at 60° the term −2ab cos θ is −12, so a² + b² and c² are not equal

Find the angle where a² + b² = c²

Sides of 3 and 4 meet at a corner that turns, with a square on every side. At 60° the square on the third side is smaller than 9 + 16 = 25: its side is only about 3.6. Drag the corner to 90°: the square on the third side is then exactly 9 + 16 = 25. The button under it shows the unit tiles, so that 9, 16 and 25 can be counted.

The rule for every right triangle

Call the two legs a and b and the hypotenuse c. Then Pythagoras’ theorem says that in every right triangle, a² + b² = c².

To find the hypotenuse, square the two legs, add the squares, and take the square root. If the legs are 6 cm and 8 cm, then c² = 6² + 8² = 36 + 64 = 100, so c = √100 = 10 cm. Check: 10 × 10 = 100.

The answer is not always a whole number. If the legs are 2 cm and 3 cm, then c² = 4 + 9 = 13, so c = √13, which is 3.61 cm to 2 decimal places.

abc

The legs are a and b and the hypotenuse is c, so a² + b² = c².

Only with a right angle

A triangle with sides 4, 5 and 6 has no right angle, and the rule fails: 4² + 5² = 16 + 25 = 41, but 6² = 36. The same happens to the sides 3 and 4 when the corner between them is not 90°: the square on the third side is not 9 + 16.

The rule also explains why the hypotenuse is the longest side. c² is a² with b² added on, so c² is more than a², and c is longer than a. In the same way c is longer than b. If a hypotenuse you work out comes out shorter than one of the legs, a step has gone wrong.

The usual mistakes

Adding the sides instead of their squares. 3 + 4 = 7 is the distance along the two legs, walking around the corner. The hypotenuse goes straight across, so it is shorter: 5.

Stopping at the square. c² = 25 is the area of the square on the hypotenuse, not its length. Take the square root: c = 5.

Adding the squares of the wrong sides. The two squares that are added belong to the legs, the sides that meet at the right angle. The square on the hypotenuse is their total, never one of the two that are added.

Worked example: A Footpath Along the Diagonal of a Rectangular Field

Question A rectangular field is 120 m long and 50 m wide. A footpath runs in a straight line from one corner to the opposite corner. (a) How long is the footpath? (b) Mei walks from one corner to the opposite corner along the footpath. Ravi walks between the same two corners along two sides of the field. How much shorter is Mei's walk?

  1. 1.Every corner of a rectangle is a right angle, so the length, the width and the footpath make a right-angled triangle. The footpath is opposite the right angle, so it is the hypotenuse. Let its length be c m.

    120 m50 mcthe footpath c is the hypotenuse
    120 m50 mcthe footpath c is the hypotenuse
    A corner of a rectangle is a right angle, so the footpath is a hypotenuse.
  2. 2.By Pythagoras' theorem, c2 = 1202 + 502 = 14400 + 2500 = 16900.

    120 m50 mcthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900
    120 m50 mcthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900
    c2 = 1202 + 502 = 16900.
  3. 3.Take the square root: c = √16900 = 130. (a) The footpath is 130 m long. Check: 130 m is longer than either side of the field, as a hypotenuse must be, and shorter than the two sides added together.

    120 m50 m130 mthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900c =√16900 = 130 m
    120 m50 m130 mthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900c =√16900 = 130 m
    (a) c = √16900 = 130, so the footpath is 130 m long.
  4. 4.Ravi walks one length and one width of the field: 120 + 50 = 170 m.

    120 m50 m130 mthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900c =√16900 = 130 malong two sides: 120 + 50 = 170 m
    120 m50 m130 mthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900c =√16900 = 130 malong two sides: 120 + 50 = 170 m
    Ravi walks one length and one width: 120 + 50 = 170 m.
  5. 5.(b) Mei's walk is 170 − 130 = 40 m shorter.

    120 m50 m130 mthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900c =√16900 = 130 malong two sides: 120 + 50 = 170 m170 − 130 = 40 m shorter
    120 m50 m130 mthe footpath c is the hypotenusec2= 1202+ 502= 14400 + 2500 = 16900c =√16900 = 130 malong two sides: 120 + 50 = 170 m170 − 130 = 40 m shorter
    (b) Mei's walk is 170 − 130 = 40 m shorter.

Answer: (a) 130 m; (b) 40 m shorter

Common mistakes

  • Adding the sides and then squaring, (120 + 50)2, or taking √120 + 50. Pythagoras' theorem adds the squares of the sides, 1202 + 502, and the square root is taken only after the two squares are added.
  • Stopping at c2 = 16900 and giving 16900 m as the length. That number is the square of the length, so its square root still has to be taken.

More pythagoras and similar shapes problems, worked step by step →

Worked example: The Shortest Path for an Ant Over the Surface of a Box

Question A closed box is 8 cm long, 3 cm wide and 3 cm high. An ant at a bottom corner A walks over the surface of the box to the opposite top corner B. (a) The ant walks in a straight line up the front face and then across the top face. How long is this path? (b) A second ant walks from A across the front face and then across the end face to B. How long is that path, to 1 decimal place, and which of the two paths is shorter?

  1. 1.Unfold the box: lift the top face until it lies flat with the front face. The two faces make one rectangle that is 8 cm long and 3 + 3 = 6 cm high, with A at one corner and B at the opposite corner.

    fronttopAB8 cm3 + 3 = 6 cmpfront and top, unfolded: 8 cm by 6 cm
    fronttopAB8 cm3 + 3 = 6 cmpfront and top, unfolded: 8 cm by 6 cm
    Unfold the top face to lie flat with the front face: one rectangle, 8 cm by 6 cm, with A and B at opposite corners.
  2. 2.The path is the diagonal of this rectangle, which is the hypotenuse of a right-angled triangle with shorter sides of 8 cm and 6 cm. Let the path be p cm long. By Pythagoras' theorem, p2 = 82 + 62 = 64 + 36 = 100.

    fronttopAB8 cm3 + 3 = 6 cmpfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100
    fronttopAB8 cm3 + 3 = 6 cmpfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100
    The straight path is the diagonal: p2 = 82 + 62 = 100.
  3. 3.(a) The path over the top is √100 = 10 cm long.

    fronttopAB8 cm3 + 3 = 6 cm10 cmfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100p =√100 = 10 cm
    fronttopAB8 cm3 + 3 = 6 cm10 cmfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100p =√100 = 10 cm
    (a) The path over the top is √100 = 10 cm long.
  4. 4.For the second ant, unfold the end face until it lies flat beside the front face. The two faces make a rectangle that is 8 + 3 = 11 cm long and 3 cm high. Let this path be q cm long: q2 = 112 + 32 = 121 + 9 = 130.

    fronttopAB8 cm3 + 3 = 6 cm10 cmfrontendAB8 + 3 = 11 cm3 cmqfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100p =√100 = 10 cmq2= 112+ 32= 121 + 9 = 130
    fronttopAB8 cm3 + 3 = 6 cm10 cmfrontendAB8 + 3 = 11 cm3 cmqfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100p =√100 = 10 cmq2= 112+ 32= 121 + 9 = 130
    Unfold the end face beside the front face: a rectangle 11 cm by 3 cm, so q2 = 112 + 32 = 130.
  5. 5.(b) The second path is √130 = 11.4 cm long, to 1 decimal place, so the path over the top, 10 cm, is the shorter one. Check: both paths are shorter than walking along three edges, which is 8 + 3 + 3 = 14 cm.

    fronttopAB8 cm3 + 3 = 6 cm10 cmfrontendAB8 + 3 = 11 cm3 cm11.4 cmfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100p =√100 = 10 cmq2= 112+ 32= 121 + 9 = 130q =√130 = 11.4 cm (1 d.p.), so 10 cm is shorter
    fronttopAB8 cm3 + 3 = 6 cm10 cmfrontendAB8 + 3 = 11 cm3 cm11.4 cmfront and top, unfolded: 8 cm by 6 cmp2= 82+ 62= 64 + 36 = 100p =√100 = 10 cmq2= 112+ 32= 121 + 9 = 130q =√130 = 11.4 cm (1 d.p.), so 10 cm is shorter
    (b) √130 = 11.4 cm, to 1 decimal place, so the path over the top is the shorter one.

Answer: (a) 10 cm; (b) 11.4 cm, so the path over the top is shorter

Common mistakes

  • Using Pythagoras' theorem on each face separately and adding the two diagonals: √82 + 32 + 3 ≈ 11.5 cm, for example. A path that bends at a corner of a face is longer than it needs to be. The faces are unfolded first, so that the whole path is one straight line.
  • Assuming that every way of unfolding gives the same length. The two rectangles here are 8 cm by 6 cm and 11 cm by 3 cm, and their diagonals are 10 cm and about 11.4 cm, so each unfolding has to be worked out.

More pythagoras and similar shapes problems, worked step by step →

Practice Pythagoras’ Theorem in the app