# Pythagoras’ Theorem

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: XY2 + XZ2 = YZ2.

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