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Pythagorean theorem

In a right triangle, there is a special relationship between the hypotenuse and the two legs. This relationship is known as the Pythagorean theorem.

3.6 Explain and prove the Pythagorean theorem and Pythagorean triples.

4.6 Solve problems related to the Pythagorean theorem and Pythagorean triples.

After studying this material, students are expected to be able to:

  1. Verify the Pythagorean theorem.
  2. Determine an unknown side of a right triangle when two sides are known.
  3. Determine the type of a triangle from its side lengths.
  4. Test whether three numbers form a Pythagorean triple.
  5. Apply the theorem to real-life problems.

Pythagorean theorem

Triangle ABCABC is right-angled at BB. The side opposite the right angle is called the hypotenuse, while the other two sides are called the legs.

Proof idea

From the square arrangements in the figure:

area of the large square=a2+2bcarea of the large square=b2+c2+2bc \begin{aligned} \text{area of the large square} &= a^2 + 2bc \\ \text{area of the large square} &= b^2 + c^2 + 2bc \end{aligned}

Since both large squares have the same area:

a2+2bc=b2+c2+2bca2=b2+c2 \begin{aligned} a^2 + 2bc &= b^2 + c^2 + 2bc \\ a^2 &= b^2 + c^2 \end{aligned}

Area interpretation 1

Area interpretation 2

This means that in a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs.

Pythagorean formula

If the hypotenuse is aa and the legs are bb and cc, then:

a2=b2+c2 a^2 = b^2 + c^2

or

a=b2+c2 a = \sqrt{b^2 + c^2}

Rearranged forms:

b=a2c2c=a2b2 b = \sqrt{a^2 - c^2} \qquad c = \sqrt{a^2 - b^2}

The theorem can be used to determine an unknown side of a right triangle when the other two sides are known.

Examples of side calculation

  1. For figure (a):
a2=132122=169144=25a=5 \begin{aligned} a^2 &= 13^2 - 12^2 \\ &= 169 - 144 \\ &= 25 \\ a &= 5 \end{aligned}
  1. For figure (b):
b2=172152=289225=64b=8 \begin{aligned} b^2 &= 17^2 - 15^2 \\ &= 289 - 225 \\ &= 64 \\ b &= 8 \end{aligned}
  1. For figure (c):
c2=242+72=576+49=625c=25 \begin{aligned} c^2 &= 24^2 + 7^2 \\ &= 576 + 49 \\ &= 625 \\ c &= 25 \end{aligned}

If a triangle satisfies:

a2=b2+c2 a^2 = b^2 + c^2

then the triangle is a right triangle.

Converse of Pythagoras

In general, if cc is the longest side:

  1. If c2=a2+b2c^2 = a^2 + b^2, the triangle is right.
  2. If c2>a2+b2c^2 > a^2 + b^2, the triangle is obtuse.
  3. If c2<a2+b2c^2 < a^2 + b^2, the triangle is acute.
  1. A triangle has side lengths 55, 1212, and 1313.
132=169,122+52=144+25=169 13^2 = 169,\qquad 12^2 + 5^2 = 144 + 25 = 169

So it is a right triangle.

  1. A triangle has side lengths 99, 77, and 55.
92=81,72+52=49+25=74 9^2 = 81,\qquad 7^2 + 5^2 = 49 + 25 = 74

Because 81>7481 > 74, the triangle is obtuse.

A Pythagorean triple is a set of three natural numbers that can be the side lengths of a right triangle.

Examples:

  1. 1515, 2020, and 2525
152+202=225+400=625=252 15^2 + 20^2 = 225 + 400 = 625 = 25^2

So they form a Pythagorean triple.

  1. 2020, 2424, and 2929
202+242=400+576=976292 20^2 + 24^2 = 400 + 576 = 976 \neq 29^2

So they do not form a Pythagorean triple.

Isosceles right triangle

If AB=BC=aAB = BC = a, then:

AC2=AB2+BC2=a2+a2=2a2AC=a2 \begin{aligned} AC^2 &= AB^2 + BC^2 \\ &= a^2 + a^2 \\ &= 2a^2 \\ AC &= a\sqrt{2} \end{aligned}

Therefore:

AB:BC:AC=1:1:2 AB : BC : AC = 1 : 1 : \sqrt{2}

45 degree example

If AC=82AC = 8\sqrt{2}, then:

AB=BC=8 AB = BC = 8

30 degree triangle

60 degree triangle

From the construction:

BC:AC:AB=1:2:3 BC : AC : AB = 1 : 2 : \sqrt{3}

Meaning:

  • the side opposite 3030^\circ is the shortest side,
  • the hypotenuse is twice that side,
  • the remaining leg is 3\sqrt{3} times that side.

30 degree example

If AB=33AB = 3\sqrt{3}, then:

AC=6,BC=3 AC = 6,\qquad BC = 3

Cube diagonals

If the edge length of a cube is aa, then the face diagonal is:

BG2=BC2+CG2=a2+a2BG=a2 \begin{aligned} BG^2 &= BC^2 + CG^2 \\ &= a^2 + a^2 \\ BG &= a\sqrt{2} \end{aligned}

The space diagonal is:

AG2=AB2+BG2=a2+2a2AG=a3 \begin{aligned} AG^2 &= AB^2 + BG^2 \\ &= a^2 + 2a^2 \\ AG &= a\sqrt{3} \end{aligned}

So for a cube with edge length aa:

  • face diagonal = a2a\sqrt{2}
  • space diagonal = a3a\sqrt{3}

General steps for word problems using the Pythagorean theorem:

  1. Read the problem carefully.
  2. Draw a sketch.
  3. Identify the known and unknown sides.
  4. Use the appropriate formula.
  5. Check the result.

Andi leans a ladder of length 55 m against a tree. The bottom of the ladder is 33 m from the base of the tree. How high is the top of the ladder above the ground?

Ladder problem

52=32+h225=9+h2h2=16h=4 \begin{aligned} 5^2 &= 3^2 + h^2 \\ 25 &= 9 + h^2 \\ h^2 &= 16 \\ h &= 4 \end{aligned}

So the height is 44 m.

A ship sails west for 66 km and then south for 88 km. How far is it now from its starting point?

Ship problem

x2=62+82=36+64=100x=10 \begin{aligned} x^2 &= 6^2 + 8^2 \\ &= 36 + 64 \\ &= 100 \\ x &= 10 \end{aligned}

So the ship is 1010 km from its starting point.

The old interactive worksheet in this article depended on a third-party embed that is no longer stable with the current blog theme. It has been removed so the page stays intact and readable.

It can be replaced later with:

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This completes the Pythagorean theorem lesson. Corrections and suggestions can be shared through the available feedback channel.

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