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
Here is a list of pythagorean triples
| (3, 4, 5) | (5, 12, 13) | (7, 24, 25) | (8, 15, 17) |
| (9, 40, 41) | (11, 60, 61) | (12, 35, 37) | (13, 84, 85) |
| (15, 112, 113) | (16, 63, 65) | (17, 144, 145) | (19, 180, 181) |
| (20, 21, 29) | (20, 99,101) | (21, 220, 221) | (23, 264, 265) |
| (24, 143, 145) | (25, 312, 313) | (27, 364, 365) | (28, 45, 53) |
| (28, 195, 197) | (29, 420, 421) | (31, 480, 481) | (32, 255, 257) |
| (33, 56, 65) | (33, 544, 545) | (35, 612, 613) | (36, 77, 85) |
| (36, 323, 325) | (37, 684, 685) | … infinitely many more … |
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 bisectBisect means dividing into two equal parts. It means to divide a geometric figure such as a line, an angle or any other shape into two congruent parts (or two parts that have the same shape and size). Point M bisects the segment. Since point M divides the segment into two parts that have the same size and shape, the segment is bisected. Since ray AB divides the angle into two parts that have the same size and shape, the... 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
\( \scriptsize \sqrt{OM^2} = \sqrt{80 \: – \: 25} \) \( \scriptsize OM = \sqrt{80 \: – \: 25} \) \( \scriptsize OM = \sqrt{55} \) \( \scriptsize OM = 7.416\) \( \scriptsize OM \approx 7.4 \: cm\)
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