Finding a Shorter Side with Pythagoras’ Theorem

Take the known square away instead of adding.

This time the hypotenuse is known

Pythagoras’ theorem says that in a right triangle, a² + b² = c², where a and b are the legs, the two sides that meet at the right angle, and c is the hypotenuse, the side across from the right angle.

The theorem is often used to find the hypotenuse from the two legs. Here it is the other way round: the hypotenuse is 13, one leg is 12, and the other leg is missing. Before anything else, find the right angle and the side across from it, so that you know which of the given lengths is the hypotenuse.

?1213

The hypotenuse is 13 and one leg is 12. The other leg is missing.

Subtract the known square

The two squares on the legs still add up to the square on the hypotenuse. Call the missing leg a and write the theorem with the lengths you know: a² + 12² = 13², which is a² + 144 = 169.

Subtract 144 from both sides: a² = 169 − 144 = 25. In words, the square on the hypotenuse is made of the two smaller squares, so taking the known one away leaves the missing one.

The rule for a missing leg is: leg² = hypotenuse² − other leg². To find the hypotenuse you add the two squares; to find a leg you subtract.

Then take the square root

a² = 25 is the area of the square on the missing leg, not the leg itself. The leg is the side of that square: the number whose square is 25, which is √25 = 5.

Check the answer in the theorem: 5² + 12² = 25 + 144 = 169 = 13². The leg, 5, is also shorter than the hypotenuse, 13, as every leg must be.

333² = 3 × 3 = 99 < 25

3² = 9 < 25: a square of side 3 holds too few tiles, so √25 is more than 3

Grow the square until it holds 25 tiles: which side? √25 = ?

The missing leg is the side of a square of 25 tiles. Grow the square from its corner: side 4 holds 16 tiles, too few, and side 6 holds 36, too many. Side 5 holds exactly 25.

51213

The missing leg is 5, and 5² + 12² = 13².

Add or subtract?

Look at which side is missing. If it is the hypotenuse, add the squares of the two legs. If it is a leg, subtract the square of the other leg from the square of the hypotenuse.

The hypotenuse is 10 cm and one leg is 6 cm. A leg is missing, so subtract: 10² − 6² = 100 − 36 = 64, and the leg is √64 = 8 cm.

The answer need not be a whole number. The hypotenuse is 7 cm and one leg is 4 cm: 7² − 4² = 49 − 16 = 33, so the other leg is √33 cm, which is 5.74 cm to 2 decimal places. It is shorter than 7 cm, as it should be.

The usual mistakes

Adding when a leg is missing. 13² + 12² = 169 + 144 = 313, and √313 is about 17.7, which is longer than the hypotenuse. No leg can be longer than the hypotenuse, so a sum was used where a difference belonged.

Subtracting the lengths instead of their squares. 13 − 12 = 1 is wrong: 1² + 12² = 1 + 144 = 145, not 169. The theorem is about the squares, so subtract 169 − 144.

Stopping at the square. 25 is the square of the missing leg. The leg is √25 = 5.

Worked example: A Ladder Against a Wall That Slips Down

Question A ladder 2.5 m long leans against a vertical wall on level ground. Its foot is 0.7 m from the wall. (a) How high up the wall does the ladder reach? (b) The ladder slips until its top is 2.0 m above the ground. How much further from the wall is its foot now?

  1. 1.The wall is vertical and the ground is level, so the wall, the ground and the ladder make a right-angled triangle. The ladder is opposite the right angle, so it is the hypotenuse, 2.5 m. Let the height that the ladder reaches be h m.

    2.5 m0.7 mhthe ladder is the hypotenuse: 2.5 m
    2.5 m0.7 mhthe ladder is the hypotenuse: 2.5 m
    The wall and the ground meet at a right angle, so the ladder is the hypotenuse.

  2. 2.By Pythagoras' theorem, h2 + 0.72 = 2.52. To find a shorter side, subtract the known square from the square of the hypotenuse: h2 = 2.52 − 0.72 = 6.25 − 0.49 = 5.76.

    2.5 m0.7 mhthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76
    2.5 m0.7 mhthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76
    The height is a shorter side: h2 = 2.52 − 0.72 = 5.76.

  3. 3.Take the square root: h = √5.76 = 2.4. (a) The ladder reaches 2.4 m up the wall. Check: 2.4 m is less than the length of the ladder, as a shorter side must be.

    2.5 m0.7 m2.4 mthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76h =√5.76= 2.4 m
    2.5 m0.7 m2.4 mthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76h =√5.76= 2.4 m
    (a) h = √5.76 = 2.4, so the ladder reaches 2.4 m up the wall.

  4. 4.After the slip the ladder is still 2.5 m long, and its top is 2.0 m high. Let the foot be d m from the wall. Then d2 = 2.52 − 2.02 = 6.25 − 4 = 2.25, so d = √2.25 = 1.5.

    2.5 m0.7 m2.4 m2.0 m1.5 mthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76h =√5.76= 2.4 md2= 2.52− 2.02= 6.25 − 4 = 2.25d =√2.25= 1.5 m
    2.5 m0.7 m2.4 m2.0 m1.5 mthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76h =√5.76= 2.4 md2= 2.52− 2.02= 6.25 − 4 = 2.25d =√2.25= 1.5 m
    After the slip the same ladder has its top at 2.0 m: d2 = 2.52 − 2.02 = 2.25, so d = 1.5.

  5. 5.(b) The foot has moved from 0.7 m to 1.5 m from the wall, so it is 1.5 − 0.7 = 0.8 m further away. Check: 1.52 + 2.02 = 2.25 + 4 = 6.25 = 2.52.

    2.5 m0.7 m2.4 m2.0 m1.5 m0.8 mthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76h =√5.76= 2.4 md2= 2.52− 2.02= 6.25 − 4 = 2.25d =√2.25= 1.5 m1.5 − 0.7 = 0.8 m further from the wall
    2.5 m0.7 m2.4 m2.0 m1.5 m0.8 mthe ladder is the hypotenuse: 2.5 mh2= 2.52− 0.72= 6.25 − 0.49 = 5.76h =√5.76= 2.4 md2= 2.52− 2.02= 6.25 − 4 = 2.25d =√2.25= 1.5 m1.5 − 0.7 = 0.8 m further from the wall
    (b) The foot has moved 1.5 − 0.7 = 0.8 m further from the wall.

Answer: (a) 2.4 m; (b) 0.8 m further

Common mistakes

  • Adding the squares, 2.52 + 0.72, to find the height. The squares are added only to find the hypotenuse. The ladder is the hypotenuse here, so the height is a shorter side and the known square is subtracted.
  • Saying that the foot moves out by 0.4 m because the top moves down by 0.4 m. The two distances are linked through their squares, not directly, so each position has to be worked out with Pythagoras' theorem.

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