Lesson 4, Topic 2
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# Pythagoras Theorem

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From the geometrical point of view, Pythagoras theorem gives the relation between areas of squares on the sides of a right-angled triangle. Also, from an algebraic point of view, the formula below can be used to solve right-angled triangles.

⇒ $$\scriptsize a^2 = b^2 \: + \: c^2$$

which is $$\scriptsize Hyp^2 = Opp^2 \: + \: Adj^2$$

### Pythagorean Triple:

In ΔABC above, the sides are 3cm, 4cm and 5cm. This is an example of a Pythagorean triple, i.e. a set of 3 whole numbers which can be taken as the lengths of the sides

$$\scriptsize 5^2 = 4^2 \: + \: 3^2$$

$$\scriptsize 25 = 16 \: + \: 9$$

$$\scriptsize 25 = 25$$

When a triangle’s sides are a Pythagorean Triple it is a right-angled triangle.

Another example is 5, 12, 13

$$\scriptsize 13^2 = 5^2 \: + \: 12^2$$

$$\scriptsize 169 = 25 \: + \: 144$$

$$\scriptsize 169 = 169$$

Here is a list of pythagorean triples

### Example 1:

If the diagonal of a square is 8cm, what is the area of the square? (WAEC)

Solution:

Recall, that the diagonals of a square bisect at right angles.

Thus, $$\scriptsize \overline{EB} = \scriptsize \overline{EA} = 4cm$$

We can then find x using ΔAEB (Figure 3)

Using Pythagoras theorem

$$\scriptsize x^2 = \overline{EB}^2 \: + \: \overline{EA}^2$$

⇒ $$\scriptsize x^2 = 4^2 \: + \: 4^2$$

$$\scriptsize x^2 = 16 \: + \: 16$$

$$\scriptsize x^2 = 32$$

Take the square root of both sides

$$\scriptsize \sqrt{x^2} = \sqrt{32}$$

$$\scriptsize x = \sqrt{32} \: cm$$

Using Figure 1

Area of square = $$\scriptsize x \: \times \: x$$

Area of square = $$\scriptsize \sqrt{32} \: cm \: \times \: \sqrt{32} \: cm$$

Area of square = $$\scriptsize 32 cm^2$$

OR (2nd Method)

Using Figure 1, Consider ΔABC

Using Pythagoras Theorem

⇒ $$\scriptsize AC^2 = BC^2 \: + \: AB^2$$

$$\scriptsize 8^2 = x^2 \: + \: x^2$$

$$\scriptsize 64 = 2x^2$$

$$\scriptsize 2x^2 = 64$$

$$\scriptsize x^2 = \frac{64}{2}$$

$$\scriptsize x^2 = 32 cm^2$$

Same as above

### Example 2

In the diagram below, O is the centre of circle HKL, |HK| = 16cm,  |HL| = 10cm, and the perpendicular from O to |HK| is 4cm. What is the length of the perpendicular from O to |HL|? (WAEC)

Solution

Redraw the diagram using the information provided

r is the radius and is the distance from the centre O to H

the perpendicular from O to |HK| = |OP| = 4cm

If we find r we can find |OM| which is the length of the perpendicular from O to |HL|

Consider ΔHPO

Using Pythagoras theorem

$$\scriptsize r^2 = opp^2 \: + \: adj^2$$

$$\scriptsize r^2 = 8^2 \: + \: 4^2$$

$$\scriptsize r = \sqrt{64 \: + \: 16}$$

$$\scriptsize r = \sqrt{64 \: + \: 16}$$

$$\scriptsize r = \sqrt{80}$$

$$\scriptsize r = 4\sqrt{5}\: cm$$

Consider ΔHMO

Using Pythagoras theorem

$$\scriptsize r^2 = HM^2 \: + \: OM^2$$

$$\scriptsize r^2 = 5^2 \: + \: OM^2$$

$$\scriptsize OM^2 = r^2 \: – \: 5^2$$

$$\scriptsize r = \sqrt{80}$$

Therefore $$\scriptsize OM^2 = \sqrt{80}^2 \: – \: 5^2$$

$$\scriptsize OM^2 = 8.9443^2 \: – \: 5^2$$

$$\scriptsize OM^2 = 80 \: – \: 25$$

Take the square root of both sides

$$\scriptsize \sqrt{OM^2} = \sqrt{80 \: – \: 25}$$

$$\scriptsize OM = \sqrt{80 \: – \: 25}$$

$$\scriptsize OM = \sqrt{55}$$

$$\scriptsize OM = 7.416$$

$$\scriptsize OM \approx 7.4 \: cm$$