Back to: MATHEMATICS JSS 2

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*In today’s Mathematics class, We will be discussing Pythagoras Theorem. We hope you enjoy the class!*

**CONTENT: **

** **i. Pythagoras triple

ii Pythagoras theorem

iii. Using Pythagoras theorem to solve other related problems.

**PYTHAGORAS TRIPLE**

The sides of a right-angled triangle can be related to the proof of Pythagoras Triple. A Pythagoras triple is a set of three whole numbers which gives lengths of the sides of a right-angled triangle.

Examples of some common Pythagoras triple are (3, 4, 5), (6, 8, 10). (5, 12, 13), etc.

** **

**Worked Example**

Which of the following is a Pythagoras triple?

- a) (15, 30, 35) b) (33, 56, 65)

** **

**Solution **

15^{2} + 30^{2} = 225 + 900

= 1125

But 35^{2} = 1225

(15, 30, 35) is not a Pythagoras triple

b) 33^{2} + 56^{2} = 1089 + 3136 = 4225

65^{2} = 4225

Thus, 33^{2} + 56^{2} = 65^{2}

(33, 56, 65) is a Pythagoras triple.

** **

**Evaluation: Class Work**

Find out which of the following are Pythagoras triples.

a) (12, 16, 20) b) (27, 36, 45) c) (14, 24, 28)

**Answer to the evaluation question**

A) (12, 16, 20)

12^{2} + 16^{2} = 144 + 256 = 400

20^{2} = 400

Thus, 12^{2} + 16^{2} = 20

(12, 16, 29) is a Pythagoras triple.

B) (27, 36, 45)

27^{2} + 36^{2} = 729 + 1296 = 2025

45^{2} = 2025

Thus, 27^{2} + 36^{2} = 45^{2}

(27, 36, 45) is a Pythagoras triple

C) (14, 24, 28)

14^{2} + 24^{2} = 196 + 576 = 772

28^{2} = 784

Thus, 14, 24, 28 is not a Pythagoras triple.

**Reference: **New General Mathematics Book 2, Chapter 7, Pages 150 – 151

Essential Mathematics for JSS Book 2, Chapter, 21, pages 218 and 219

**PYTHAGORAS THEOREM**

The Pythagoras’ Theorem states that in any right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the two sides.

A) Using Pythagoras rule

C^{2} = a^{2} + b^{2}

a = 3, b = 4

c^{2} = 3^{2} + 4^{2} = 9 + 16

c^{2} = 25

c = 5m, the length of the third side is 5m.

B) Using Pythagoras rule

c^{2} = a^{2} + b^{2}

C = 13, a = a b = 5

13^{2} = a^{2} + 5^{2}

a^{2} = 169 – 25 = 144

a =

a = 12cm

**Evaluations**

**Answer to the evaluations.**

**a)** /AC/^{2} = /AB/^{2} + /BC/^{2}

AC = ?, AB = 8cm, BC = 6cm

AC^{2} = 8^{2} + 6^{2}

AC = = 10cm

**b) **/AC/^{2} = /AB/^{2}– /BC/^{2}

= 100m, = 80m, BC =?

100^{2} = 80^{2} + /BC/^{2}

1000 = 6400 + /BC/^{2}

/BC/^{2} = 1000 – 6400

/BC/ = = 60M

**c) **/AC/^{2} = /AB/^{2} + 7^{2}

AC = 25, /AB/ = 7^{2}

25^{2} = /AB/^{2} + 49

/AB/^{2} = 625 – 49 = 576

/AB/ = = 24cm

** **

**Reference **

NGM BK 2, chapter 17, pages 147 – 148

Essential mathematics for JSS BK 2, chapter 21, pages 215 – 218

** **

**USING PYTHAGORAS THEOREM TO SOLVE OTHER RELATED PROBLEM INVOLVING TRIANGLES**

In some cases, we may have more than one right-angled triangle.

**Worked examples**

- Calculate the length of the unknown in the following triangle:

**Solution**

**A)** PRS is a right-angled triangle, PQR is also a right-angled triangle

Let PR be ycm

In triangle PQR; y^{2} = 3^{2} + 2^{2}

= 9 + 4 = 13

y^{2} = 13

Let PS be xcm

In triangle PRS, x^{2} = y^{2} + 6^{2}

Substitute 13 for y^{2} in the formula

x^{2} = 13 + 62

x^{2} = 13 + 36

x^{2}= = 7

PS = 7cm

**B)** AD is the right-angled ABD. Let AB be ycm.

In triangle ABC, x^{2} = y^{2} + (8 + 12)^{2}

Substitute 225 for y^{2} in the formula

X^{2} = 225 + 20^{2}

= 225 + 400 = 625

X = = 25cm

Therefore, AD = 25cm

When solving a triangle relating to decimal fraction and whole numbers, it is advisable to find the squares and square root from tables or multiplying the decimal by itself.

**Evaluation**

- A ladder is 7.3m long and the foot of the ladder is 1.8m from the wall. How far up the wall is the ladder?
- The distances between the opposite corner of a rectangular lawn is 30m, of the lawn, is 24m. Calculate the breadth of the lawn.

**GENERAL EVALUATION**

- The distance between the opposite corners of a rectangular plot is 30m. The length of the plot is 24m. Calculate the breadth of the plot.
- A student cycles from home to school, first eastwards to a road junction 12km from home, then southwards to school. If the school is 19km from home, how far is it from the road junction?

**REVISION QUESTION:**

- A square top lid of a container has a diagonal 150cm. Find the length of one side of the lid.
- ABCD is a rectangle. AB = xcm, BC = 9cm and the diagonal AC = 19cm. Calculate the value of x.

**READING ASSIGNMENT**

Essential Mathematics for JSS 2 Chapter 21 pages 268 – 271

Exercise 21.1 1a – b, 2a – d, 3a – b, page 270

**WEEKEND ASSIGNMENT**

- The longest side of a right-angled triangle is called A. hypotenuse B. hypostasis C. base D. adjacent
- Calculate the length of the diagonal of a 15m by 12m room. A. 9m B. 81m C. 19m D. 12m
- Which of the following are Pythagorean triples? A. 6, 8. 10 B. 12, 28, 32 C. 9, 12, 20 D. 13, 15, 17
- Calculate the value of x in the diagram below.

**THEORY**

- A flagpole 5m tall is supported by a wire that is fixed at point 3m from the base of the pole. Calculate to 1 d.p the length of the wire.
- A square top lid of a container has a diagonal of 150cm. Find the length of one side of the lid.

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*The Journey still continues though, we are moving on to JSS3. we hope to meet you there. *

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