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.

Pythagoras’ Theorem

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.

Problems on Pythagoras’ Theorem

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:

Application on Pythagoras’ Theorem

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











9th Grade Math

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