## Pythagoras Theorem

Consider the triangle \(ΔPMK, P = 90^{∘}\)

A triangle is called a right-angled triangle if one of its angles is \(90^{∘}\). So, ΔPMK is a right-angled triangle (or rectangular triangle) with right-angle \(\angle P\). The side of the rectangular triangle, which lies opposite the angle \(90^{∘}\), is called the hypotenuse.

One of the most important elements in the course of geometry is **Pythagoras’ theorem**:

In the right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides. Therefore, for the triangle ΔPMK:

\(MK^{2}=PK^{2}+MP^{2}\)

Let \(MK=c,PK=a,MP=b\) Then:

\(c^{2}=a^{2}+b^{2}\)

Using Pythagoras’ theorem, knowing two sides of a right-angled triangle, we can find the third one. The following scheme will help us:

The other two sides of the rectangular triangle are called opposite and adjacent. In any rectangular triangle, they are smaller than the hypotenuse. Separate cases of Pythagorean theorems, in particular, so-called Egyptian or “sacred”, triangles with sides 3, 4 and 5, were known prior to Pythagoras in ancient Egypt, Babylon, India and China. Perhaps, Pythagoras was the first who proved this theorem.

**Example 1.** Two sides of a right-angled triangle are 6 cm and 8 cm. Find the hypotenuse.

**Solution 1.** We have a=6 cm; b=8 cm. Then, if

\(c^{2}=a^{2}+b^{2}\)

\(c^{2}=6^{2}+8^{2}\)

\(c^{2}=36+64\)

\(c^{2}=100\)

\(c=\sqrt{100}=10\)

Consequently, the length of the hypotenuse is 10 cm.

**Example 2.** The basis of the triangle with two equal sides is 10 cm, and its perimeter is 36 cm. Find the height of the triangle held to the basis.

**Solution 2. **According to the condition of the task, AC=AB, ∠AKB = ∠AKC = 90°, because AK is height of the triangle ∆ABC, CB=10 cm, \(P_{∆ABC}=\)36 cm.

Knowing that the perimeter is the sum of all sides of the triangle, we have:

\(P_{∆ABC}=\)AC+AB+CB,

36=AC+AB+10

AC+AB=36-10

AC+AB=26

But AC=AB, so, AC=AB=13 cm.

As AK is height of the triangle ∆ABC, then CK=KB=\(\frac{10}{2}\)=5 cm.

Consider the triangle ∆AKB: ∠AKB = 90°. Therefore, ∆AKB is a right-angled triangle, where AB is a hypotenuse. Let’s use the Pythagorean theorem:

\(AB^{2}=AK^{2}+KB^{2}\)

\(132=AK^{2}+52\)

\(AK^{2}=169-25\)

\(AK^{2}=144\)

\(AK=12\)

So, the height AK of the triangle ∆ABC is equal 12 cm.

**Example 3.** One of the rectangular triangle sides is 8 cm, and the second one is 2 cm smaller than the hypotenuse. Find the perimeter of the triangle.

**Solution 3.
**Let a=8 cm be the opposite side of a rectangular triangle. We denote the adjacent side b=(x) cm, then the hypotenuse c=(x+2) cm.

By the Pythagorean theorem:

\(c^{2}=a^{2}+b^{2}\)

\((x+2)^{2}=8^{2}+x^{2}\)

\(x^{2}+4x+4=64+x^{2}\)

\(4x+4=64\)

\(4x=60\)

\(x=15\)

So, b=15 cm; c=x+2=15+2=17 (cm).

Then the perimeter of the triangle is P = 8 + 15 + 17 = 40 (cm).

Also, the inverse statement (also called inverse to the Pythagorean theorem) is proved:

For any three positive numbers a, b and c, for which the equation \(c^{2}=a^{2}+b^{2}\) is executed, there is a rectangular triangle with the opposite and adjacent sides a and b and the hypotenuse c.

**Example 4.** Is a triangle with sides:

1) 7 cm; 8 cm; 9 cm;

2) 7 cm; 24 cm; 25 cm

Is it a right-angled triangle?

**Solution 4.
**1) Since 81 ≠ 49 + 64, 81 ≠ 113, the triangle is not a right-angled triangle.

2) Since 625 = 576 + 49, 625 = 625, then the triangle is a right-angled triangle.

**Example 5.** The sides of the triangle are 12 cm; 16 cm and 20 cm. Find the median taken to the largest side of the triangle.

**Solution 5.** Since 144 + 256 = 400, then the triangle is a right-angled triangle with a hypotenuse equal to 20 cm.

2) The median CM, taken to the hypotenuse, is equal to hypotenuse’s half, and therefore is equal to CM=AB/2=20/2 = 10 (cm).