Geometry

The Pythagorean theorem

In a right triangle, the legs a and b and the hypotenuse c (the longest side, across from the right angle) always satisfy a2+b2=c2a^2 + b^2 = c^2. Use it to find a missing side. The converse tells you a triangle's type from its sides: if a2+b2=c2a^2 + b^2 = c^2 it is right, if greater it is acute, and if less it is obtuse. The distance formula is the same rule on a graph.

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The key idea

Label the hypotenuse c first. It is always across from the right angle and always the longest side. The other two sides are the legs, a and b.

a2+b2=c2c=a2+b2a=c2−b2a^2 + b^2 = c^2 \qquad c = \sqrt{a^2 + b^2} \qquad a = \sqrt{c^2 - b^2}

To find the hypotenuse, add the squares. To find a leg, subtract. Choosing the wrong one of those causes most wrong answers.

Pythagorean triples

Some sets of whole numbers fit exactly. Any multiple of a triple is also a triple, so 6, 8, 10 and 9, 12, 15 work because 3, 4, 5 does.

TripleCheckSome multiples
3, 4, 59+16=259 + 16 = 256, 8, 10 and 9, 12, 15
5, 12, 1325+144=16925 + 144 = 16910, 24, 26
8, 15, 1764+225=28964 + 225 = 28916, 30, 34
7, 24, 2549+576=62549 + 576 = 62514, 48, 50

The converse: what kind of triangle is it?

Let c be the longest side and compare a2+b2a^2 + b^2 with c2c^2.

IfThe triangle is
a2+b2=c2a^2 + b^2 = c^2Right
a2+b2>c2a^2 + b^2 > c^2Acute (every angle is less than 90°)
a2+b2<c2a^2 + b^2 < c^2Obtuse (one angle is more than 90°)

On a graph, the horizontal gap and the vertical gap between two points are the legs of a right triangle, so the distance between the points is its hypotenuse:

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Worked examples

Example 1: find the hypotenuse

Problem A right triangle has legs of 9 and 12. Find the hypotenuse.

  1. The hypotenuse is the unknown, so add the squares of the legs.
    92+122=c29^2 + 12^2 = c^2
  2. Square each leg.
    81+144=c281 + 144 = c^2
  3. Add.
    225=c2225 = c^2
  4. Take the square root. A length is positive.
    c=15c = 15
  5. This is the 3, 4, 5 triple times 3, so you could have spotted it without the work.

Answer c=15c = 15

Example 2: find a leg

Problem A 20-foot ladder leans against a wall. Its foot is 6 feet from the base of the wall. How high up the wall does the ladder reach?

  1. The wall and the ground meet at a right angle. The ladder is across from it, so the ladder is the hypotenuse. The height h is a leg.
    62+h2=2026^2 + h^2 = 20^2
  2. Square the numbers.
    36+h2=40036 + h^2 = 400
  3. Subtract 36 from both sides. Finding a leg means subtracting.
    h2=364h^2 = 364
  4. Take the square root and simplify, since 364=4⋅91364 = 4 \cdot 91.
    h=364=291h = \sqrt{364} = 2\sqrt{91}
  5. On a calculator, 291≈19.082\sqrt{91} \approx 19.08. That is a bit less than 20, which makes sense for a ladder standing almost straight up.

Answer 2912\sqrt{91} feet, about 19.08 feet.

Example 3: use the converse

Problem A triangle has sides of 7, 9 and 12. Is it acute, right or obtuse?

  1. The longest side is 12, so it plays the role of c.
  2. Add the squares of the two shorter sides.
    72+92=49+81=1307^2 + 9^2 = 49 + 81 = 130
  3. Square the longest side.
    122=14412^2 = 144
  4. Compare.
    130<144130 < 144
  5. The two short sides are too short to meet at a right angle, so the angle across from the side of 12 is wider than 90°.

Answer Obtuse.

Example 4 (test-hard): the diagonal of a box

Problem A box is 6 inches wide, 8 inches long and 24 inches tall. How long is the longest straight rod that fits inside, from a bottom corner to the opposite top corner?

  1. First find the diagonal across the bottom of the box. It is the hypotenuse of a right triangle with legs 6 and 8.
    62+82=100=10\sqrt{6^2 + 8^2} = \sqrt{100} = 10
  2. That bottom diagonal and the 24-inch height form a second right triangle, standing up inside the box. The rod is its hypotenuse.
    102+242=676=26\sqrt{10^2 + 24^2} = \sqrt{676} = 26
  3. Shortcut for any box: d=l2+w2+h2=36+64+576=26d = \sqrt{l^2 + w^2 + h^2} = \sqrt{36 + 64 + 576} = 26.

Answer 26 inches.

Common mistakes

  • Adding when you need a leg. For a leg, subtract: a2=c2−b2a^2 = c^2 - b^2. Fix: if the hypotenuse is given, you subtract.
  • Calling the wrong side c. c is always across from the right angle. Fix: find the right-angle mark first and look straight across from it.
  • Thinking a2+b2=a+b\sqrt{a^2 + b^2} = a + b. Legs 8 and 15 give a hypotenuse of 17, not 23. Square roots do not split over addition.
  • Using it on a triangle with no right angle. a2+b2=c2a^2 + b^2 = c^2 only holds for right triangles. For other triangles, use the converse to classify them.
  • Forgetting the last square root. c2=225c^2 = 225 means c=15c = 15, not 225. Fix: ask whether your answer is a length or a squared length.

Quick methods

Practice

5 practice questions

  1. A right triangle has legs of 8 and 15. What is the hypotenuse?

    1. 1717
    2. 2323
    3. 161\sqrt{161}
    4. 289289
    Show answer

    Answer: 1717

    82+152=64+225=2898^2 + 15^2 = 64 + 225 = 289, and 289=17\sqrt{289} = 17. 23 adds the legs, 161\sqrt{161} subtracts instead of adding, and 289 forgets the square root.

  2. A right triangle has a hypotenuse of 10 and one leg of 4. What is the other leg?

    1. 66
    2. 2212\sqrt{21}
    3. 2292\sqrt{29}
    4. 14\sqrt{14}
    Show answer

    Answer: 2212\sqrt{21}

    102−42=100−16=8410^2 - 4^2 = 100 - 16 = 84, and 84=221\sqrt{84} = 2\sqrt{21}, about 9.17. 6 subtracts the sides without squaring, 2292\sqrt{29} adds the squares instead of subtracting, and 14\sqrt{14} adds the sides without squaring.

  3. Which set of side lengths makes a right triangle?

    1. 5, 11, 12
    2. 9, 40, 41
    3. 6, 7, 9
    4. 10, 20, 25
    Show answer

    Answer: 9, 40, 41

    81+1600=1681=41281 + 1600 = 1681 = 41^2. For the others, 25+121=14625 + 121 = 146 is not 144, 36+49=8536 + 49 = 85 is not 81, and 100+400=500100 + 400 = 500 is not 625.

  4. What is the distance between the points (−1,2)(-1, 2) and (5,10)(5, 10)?

    1. 1010
    2. 1414
    3. 454\sqrt{5}
    4. 100100
    Show answer

    Answer: 1010

    The horizontal gap is 5−(−1)=65 - (-1) = 6 and the vertical gap is 10−2=810 - 2 = 8, so the distance is 36+64=10\sqrt{36 + 64} = 10. 14 adds the gaps, 454\sqrt{5} uses 5−1=45 - 1 = 4 and drops the negative sign, and 100 forgets the square root.

  5. A triangle has sides of 8, 10 and 13. What kind of triangle is it?

    1. Acute
    2. Right
    3. Obtuse
    4. It cannot be a triangle
    Show answer

    Answer: Obtuse

    82+102=1648^2 + 10^2 = 164 and 132=16913^2 = 169. Since 164<169164 < 169, the triangle is obtuse. It is a real triangle because 8+10=188 + 10 = 18 is more than 13.

Frequently asked questions

Does the Pythagorean theorem work for every triangle?

No. a2+b2=c2a^2 + b^2 = c^2 is only true for right triangles. For other triangles, comparing a2+b2a^2 + b^2 with c2c^2 tells you whether the triangle is acute or obtuse. To find missing sides in those triangles, you will later learn the law of cosines.

How do I know which side is the hypotenuse?

It is the side across from the right angle, the one that does not touch the small square mark. It is always the longest side. In a word problem it is usually the slanted thing: the ladder, the ramp, the wire or the diagonal.

Should I memorize Pythagorean triples?

Memorize 3, 4, 5 and 5, 12, 13 at least, and 8, 15, 17 and 7, 24, 25 if you can. Then watch for multiples like 6, 8, 10. Spotting a triple skips the squaring and square root steps and catches arithmetic slips.

How is the distance formula related to the Pythagorean theorem?

Draw a right triangle with the two points at the ends of the hypotenuse. The legs are the horizontal gap x2−x1x_2 - x_1 and the vertical gap y2−y1y_2 - y_1. Put those into a2+b2=c2a^2 + b^2 = c^2, solve for c, and you have the distance formula.

Try asking Ducky

  • “I got 23 for the hypotenuse with legs 8 and 15. What did I do wrong?”
  • “How do I tell which side is c when the triangle is turned sideways?”
  • “Give me a ladder problem to try, then check my work.”

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