The lengths of the sides of a right-angled triangle have a special relationship between them. This relation is widely used in many branches of mathematics, such as mensuration and trigonometry.
Pythagoras’ Theorem: In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
Given: Let XYZ be a triangle in which ∠YXZ = 90°.
YZ is the hypotenuse.
To prove: XY^{2} + XZ^{2} = YZ^{2}.
Construction: Draw XM ⊥YZ.
Therefore, ∠XMY = ∠XMZ = 90°.
Proof:
Statement |
Reason |
1. In ∆XYM and ∆XYZ, (i) ∠XMY = ∠YXZ = 90° (ii) ∠XYM = ∠XYZ |
1. (i) Given and by construction (ii) Common angle |
2. Therefore, ∆XYM ∼ ∆ZYX |
2. BY AA criterion of similarity |
3. Therefore, \(\frac{XY}{YZ}\) = \(\frac{YM}{XY}\) |
3. Corresponding sides of similar triangle are proportional |
4. Therefore, XY\(^{2}\) = YZ ∙ YM |
4. By cross multiplication in statement 3. |
5. In ∆XMZ and ∆XYZ, (i) ∠XMY = ∠YXZ = 90° (ii) ∠XZM = ∠XZY |
5. (i) Given and by construction (ii) Common angle |
6. Therefore, ∆XMZ ∼ ∆YXZ. |
6. BY AA criterion of similarity |
7. Therefore, \(\frac{XZ}{YZ}\) = \(\frac{MZ}{XZ}\) |
7. Corresponding sides of similar triangle are proportional |
8. Therefore, XZ\(^{2}\) = YZ ∙ MZ |
8. By cross multiplication in statement 7. |
9. Therefore, XY\(^{2}\) + XZ\(^{2}\) = YZ ∙ YM + YZ ∙ MZ ⟹ XY\(^{2}\) + XZ\(^{2}\) = YZ(YM+ MZ) ⟹ XY\(^{2}\) + XZ\(^{2}\) = YZ ∙ YZ ⟹ XY\(^{2}\) + XZ\(^{2}\) = YZ\(^{2}\) |
9. By adding statements 4 and 8 |
Problems on Pythagoras’ Theorem:
1. In ∆XYZ, ∠Y = 90°. If XY = 3 cm and YZ = 4 cm, find XZ.
Solution:
By Pythagoras, theorem,
XZ\(^{2}\) = XY\(^{2}\) + YZ\(^{2}\)
= (3\(^{2}\) + 4\(^{2}\)) cm\(^{2}\)
= (9 + 16) cm\(^{2}\)
= 25 cm\(^{2}\)
Therefore, XZ = \(\sqrt{25 cm^{2}}\)
Therefore, XZ = 5 cm
2. Two poles, 15 feet and 35 feet high, are 15 feet apart. Find distance between the tops of the poles.
Solution:
Let the first pole XY = 15 ft
The second pole PQ = 35 ft.
The distance between two poles YQ = 15 ft.
Draw XR ⊥ PQ.
Now, we have,
PR = PQ - RQ = PQ - XY = (35 - 15) ft = 20 ft.
Also, XR = YQ = 15 ft.
Therefore, distance between tops of the poles = XP
= \(\sqrt{XR^{2} + RP^{2}}\)
= \(\sqrt{15^{2} + 20^{2}}\) ft
= \(\sqrt{225 + 400}\) ft
= \(\sqrt{625}\) ft
= 25 ft
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