PRIMARY 5 THIRD TERM LESSON NOTE PLAN MATHEMATICS
THIRD TERM E NOTES
SUBJECT: MATHEMATICS
CLASS: BASIC FIVE
WEEK TOPIC
- RATIO AND PERCENTAGE
- SIMPLE PROBLEMS ON PERCENTAGE
- MONEY: profit and loss
- MONEY: SIMPLE INTEREST
- MONEY: DISCOUNT AND COMMISION
- PERIMETER OF PLANE SHAPES
- AREA OF RIGHT ANGLED TRIANGLE
- VOLUME
- PROBLEM ON WEIGHT
- AVERAGE SPEED
WEEK ONE
RATIO AND PERCENTAGE
Meaning of ratio
The relation between two quantities (both of the same units) obtained by dividing one quantity with another is called ratio. Ratio can be denoted by using the symbol ( : ).
EXAMPLE 1
What is the ratio between the weight two bags of sugar of 4kg and 6kg respectively?
Solution
Ratio of weights of bags of sugar = 4kg/ 6kg = 2/3 = 2:3
EXAMPLE 2
A pole of length 165cm is divided into two parts such that lengths are in the ration 7:8. Find the length of each part of the pole?
Total ratio = 7 + 8 = 15
First part = 7/15 , second part = 8/15
Length of first part = 7/15 of 165cm
= 7/15 × 165
= 7 x 11 = 77cm
1 x 1
Leght of second part = 165cm – 77cm = 88cm
DIRECT PROPORTION
Examples
- A marker costs $18. Calculate the cost of 5 markers.
- A student went to the market to purchase textbooks. He purchased two textbooks for $24
- What is the price of one notebook?
- What would be the price of 5 such note
Solution
- If one marker = $18
5 markers = $18 x 5 = $90
- 2 textbooks = $24
1 textbook = m
2 × m = 1 x $24
2m = $24
M = $24 ÷ 2 = $12. Therefore 1 textbook = $24
(b)since 1 textbook = $12
5 textbooks = $12 × 5 = $60
EVALUATION
- What is ratio?
- Find the ratio of each of the following in its lowest terms:
- 24cm: 72cm
- 425km: 750km
- 75min: 150min
- 85kg: 102kg
- A field is 50m in length and 60m in width. Find the ratio between its width and length.
- A scooter can travel 225km with 5 litres of petrol. How many litres of petrol is needed to travel 675km?
[mediator_tech]
1. The relation between two quantities obtained by dividing one quantity by another is called _________.
a) ratio
b) percentage
c) proportion
d) fraction
2. The symbol used to represent ratio is _________.
a) :
b) %
c) /
d) =
3. The ratio between the weight of two bags of sugar, one weighing 4kg and the other weighing 6kg, is _________.
a) 4:6
b) 2:3
c) 6:4
d) 3:2
4. A pole of length 165cm is divided into two parts in the ratio 7:8. The length of the first part is _________.
a) 77cm
b) 88cm
c) 165cm
d) 99cm
5. In direct proportion, if one marker costs $18, the cost of 5 markers would be _________.
a) $5
b) $18
c) $90
d) $36
6. A student purchased two textbooks for $24. The price of one textbook is _________.
a) $12
b) $24
c) $2
d) $48
7. The ratio 24cm:72cm in its lowest terms is _________.
a) 1:3
b) 2:3
c) 4:9
d) 8:24
8. The ratio 425km:750km in its lowest terms is _________.
a) 17:30
b) 85:150
c) 5:9
d) 50:90
9. The ratio 75min:150min in its lowest terms is _________.
a) 3:6
b) 15:30
c) 25:50
d) 1:2
10. The ratio 85kg:102kg in its lowest terms is _________.
a) 17:20
b) 85:102
c) 5:6
d) 10:12
11. The ratio between the width and length of a field that is 50m in length and 60m in width is _________.
a) 5:6
b) 6:5
c) 50:60
d) 60:50
12. A scooter can travel 225km with 5 liters of petrol. To travel 675km, it would need _________ liters of petrol.
a) 9
b) 15
c) 25
d) 45
13. The ratio 24cm:72cm can be simplified to _________.
a) 1:3
b) 2:3
c) 4:9
d) 8:24
14. The ratio 425km:750km can be simplified to _________.
a) 17:30
b) 85:150
c) 5:9
d) 50:90
15. The ratio 75min:150min can be simplified to _________.
a) 3:6
b) 15:30
c) 25:50
d) 1:2
[mediator_tech]
WEEK TWO
SIMPLE PROBLEMS ON PERCENTAGES
Percentages are fractions with 100 as denominator
EXAMPLE 1
Change the following fractions to percentages
- 2/5 = 2/5 of 100 = 2/5 × 100 = 200 = 40%
5
EXAMPLE 2
Change these percentage to fractions
- 75% = 75 = 15 = 3
100 20 4
EXAMPLE 3
Change 7 ½% to fractions in their lowest terms
Solution
7 ½ % = 7 ½ out of 100
= 15/2 × 1/100
= (15 × 1) ÷ 200
= 15 ÷ 200 = 3/40
EXERCISE
Change the following fraction to percentage
- 2/5
- 2/4
- 3/30
Change to fractions in their lowest terms
- 66 ½ %
- 12 ¼ %
- 16 2/3 %
[mediator_tech]
1. Change the following fraction to a percentage:
3/8
a) 30%
b) 37.5%
c) 38%
d) 375%
2. Change the following fraction to a percentage:
7/10
a) 70%
b) 7%
c) 17.5%
d) 0.7%
3. Change the following fraction to a percentage:
1/2
a) 200%
b) 50%
c) 100%
d) 25%
4. Change the following fraction to a percentage:
5/6
a) 60%
b) 0.5%
c) 83.33%
d) 56.67%
5. Change the following fraction to a percentage:
4/9
a) 0.44%
b) 0.49%
c) 44.44%
d) 49%
6. Change the following fraction to a percentage:
3/20
a) 15%
b) 30%
c) 6%
d) 150%
7. Change the following fraction to a percentage:
1/3
a) 0.33%
b) 33%
c) 66.67%
d) 3.33%
8. Change the following fraction to a percentage:
7/8
a) 0.87%
b) 78%
c) 87.5%
d) 700%
9. Change the following fraction to a percentage:
2/7
a) 0.2%
b) 14.29%
c) 28%
d) 200%
10. Change the following fraction to a percentage:
3/25
a) 0.12%
b) 12%
c) 0.3%
d) 30%
11. Change the following percentage to a fraction in its lowest terms:
80%
a) 4/5
b) 8/10
c) 16/20
d) 2/10
12. Change the following percentage to a fraction in its lowest terms:
25%
a) 1/4
b) 5/10
c) 25/100
d) 2/8
13. Change the following percentage to a fraction in its lowest terms:
12.5%
a) 1/2
b) 25/200
c) 1/8
d) 5/10
14. Change the following percentage to a fraction in its lowest terms:
150%
a) 15/100
b) 150/100
c) 3/2
d) 15/10
15. Change the following percentage to a fraction in its lowest terms:
66 2/3%
a) 2/3
b) 2/100
c) 2/1
d) 66/100
[mediator_tech]
WEEK THREE
MONEY: PROFIT AND LOSS
MEANING OF PROFIT
When the selling price of an article is higher or greater than the cost price, we have a profit or gain.
Profit = selling price – cost price
MEANING OF LOSS
When the selling price is less than cost price we have a loss.
Loss = cost price – selling price
EXAMPLE 1
If a clock is bought for $1,145 and sold for $1,170, what is the gain or losss
Solution
Cost price of clock = $1,145
Selling price = $1, 170
Since the selling price is more than the cost price, we have a profit
Profit = $1,170 – $1,145 = $25
EXAMPLE 2
By selling a tin of oil for $1,320, a man made a profit of $150. How much did he pay for it?
Solution
Selling price = $1,320
Profit = $150
Cost price = $1,320 – $150 = $1,170.
EXERCISE
- I bought an exercise book for $2500 and sold it for $3,200. How much profit did I make
- An article is sold for $4,000 and the loss is $1,050. Fin the cost price of the article
- A wrist watch is bought for $3,887 and sold for a profit of $722. Find the selling price
- A trader bought a dozen candles for $1,020 and sold them all at $960. What is the profit or loss of each candle?
[mediator_tech]
MEANING OF PROFIT
When the selling price of an article is higher or greater than the cost price, we have a profit or gain.
Profit = selling price – cost price
MEANING OF LOSS
When the selling price is less than the cost price, we have a loss.
Loss = cost price – selling price
EXAMPLE 1
If a clock is bought for ₦1,145 and sold for ₦1,170, what is the gain or loss?
Solution
Cost price of clock = ₦1,145
Selling price = ₦1,170
Since the selling price is more than the cost price, we have a profit
Profit = ₦1,170 – ₦1,145 = ₦25
EXAMPLE 2
By selling a tin of oil for ₦1,320, a man made a profit of ₦150. How much did he pay for it?
Solution
Selling price = ₦1,320
Profit = ₦150
Cost price = ₦1,320 – ₦150 = ₦1,170.
EXERCISE
1. I bought an exercise book for ₦2,500 and sold it for ₦3,200. How much profit did I make?
a) ₦700
b) ₦800
c) ₦900
d) ₦1,000
2. An article is sold for ₦4,000 and the loss is ₦1,050. Find the cost price of the article.
a) ₦2,950
b) ₦3,050
c) ₦3,950
d) ₦4,050
3. A wristwatch is bought for ₦3,887 and sold for a profit of ₦722. Find the selling price.
a) ₦4,609
b) ₦4,165
c) ₦3,165
d) ₦3,165
4. A trader bought a dozen candles for ₦1,020 and sold them all at ₦960. What is the profit or loss of each candle?
a) ₦4
b) ₦5
c) ₦6
d) ₦7
5. A toy is bought for ₦80 and sold for ₦100. What is the profit or loss?
a) ₦10 profit
b) ₦10 loss
c) ₦20 profit
d) ₦20 loss
6. A shirt is bought for ₦35 and sold for ₦28. What is the profit or loss?
a) ₦7 profit
b) ₦7 loss
c) ₦5 profit
d) ₦5 loss
7. I bought a pair of shoes for ₦60 and sold them for ₦80. What is the profit or loss?
a) ₦10 profit
b) ₦10 loss
c) ₦20 profit
d) ₦20 loss
8. A bike is bought for ₦150 and sold for ₦120. What is the profit or loss?
a) ₦30 profit
b) ₦30 loss
c) ₦15 profit
d) ₦15 loss
9. I bought a book for ₦25 and sold it for ₦30. What is the profit or loss?
a) ₦5 profit
b) ₦5 loss
c) ₦2 profit
d) ₦2 loss
10. A computer is bought for ₦800 and sold for ₦900. What is the profit or loss?
a) ₦100 profit
b) ₦100 loss
c) ₦90 profit
d) ₦90 loss
11. I bought a pen for ₦2 and sold it for ₦1. What is the profit or loss?
a) ₦1 profit
b) ₦1 loss
c) ₦0.50 profit
d) ₦0.50 loss
12. A shirt is bought for ₦20 and sold for ₦25. What is the profit or loss?
a) ₦5 profit
b) ₦5 loss
c) ₦2 profit
d) ₦2 loss
13. I bought a phone for ₦400 and sold it for ₦350. What is the profit or loss?
a) ₦50 profit
b) ₦50 loss
c) ₦10 profit
d) ₦10 loss
14. A bag is bought for ₦50 and sold for ₦60. What is the profit or loss?
a) ₦10 profit
b) ₦10 loss
c) ₦5 profit
d) ₦5 loss
15. I bought a necklace for ₦100 and sold it for ₦80. What is the profit or loss?
a) ₦20 profit
b) ₦20 loss
c) ₦10 profit
d) ₦10 loss
[mediator_tech]
WEEK FOUR
SIMPLE INTEREST
Meaning of interest: interest means the extra money that you [ay back when you borrow money that you receive when you invest money.
EXAMPLE 1
find the simple interest on $1,000 for 5 years at 3% per annum.
SOLUTION
Principal = $1,000 ( the amount borrowed
Time = 5 years ( the period for which the money is borrowed before it is paid back in full)
Rate = 3% ( extra money paid)
Simple interest = P x R x T
100
= 1,000 x 3 x 5
100
= 10 x 3 x 5 = $150
EXAMPLE 2
Seun deposit $14,500 in a savings Bank account at UBA which pays an interest rate of 15% per annum for 2 ½ years. What is the simple interest?
Solution
Principal = $14,500
Rate= 15%
Time= 2 ½ or 2.5 years
SI = P x R x T
100
= 14500 x 15 x 2.5 = 145 x 15 x 2.5 = $5, 437.50
100
EXERCISE
Copy and complete the table
| S/N | Principlal | Rate | Time | Simple interest |
| $23,00 | 5% | 4 years | ||
|
|
$19,000 | 6% | 7 years | |
|
|
$28,000 | 3% | 2 years | |
| $30,000 | 4% | $6,000 |
[mediator_tech]
1. Find the simple interest on ₦2,000 for 3 years at an interest rate of 5% per annum.
Solution:
Principal = ₦2,000
Time = 3 years
Rate = 5%
Simple interest = (₦2,000 * 5 * 3) / 100 = _______
a) ₦300
b) ₦150
c) ₦30
d) ₦50
2. Rachel invests ₦10,000 in a fixed deposit account with an interest rate of 8% per annum for 4 years. What is the simple interest earned?
Solution:
Principal = ₦10,000
Rate = 8%
Time = 4 years
Simple interest = (₦10,000 * 8 * 4) / 100 = _______
a) ₦3,200
b) ₦2,000
c) ₦800
d) ₦400
3. John borrowed ₦5,000 from a friend and promised to pay back after 2 years with an additional 12% interest. What is the simple interest he needs to pay?
Solution:
Principal = ₦5,000
Rate = 12%
Time = 2 years
Simple interest = (₦5,000 * 12 * 2) / 100 = _______
a) ₦1,200
b) ₦2,400
c) ₦600
d) ₦1,000
4. If ₦8,000 is invested for 6 months at an interest rate of 10% per annum, what is the simple interest earned?
Solution:
Principal = ₦8,000
Rate = 10%
Time = 6/12 (6 months = 0.5 years)
Simple interest = (₦8,000 * 10 * 0.5) / 100 = _______
a) ₦400
b) ₦800
c) ₦200
d) ₦100
5. A loan of ₦15,000 is taken for 1 year at an interest rate of 6% per annum. What is the simple interest charged?
Solution:
Principal = ₦15,000
Rate = 6%
Time = 1 year
Simple interest = (₦15,000 * 6 * 1) / 100 = _______
a) ₦900
b) ₦600
c) ₦300
d) ₦150
6. Mary invested ₦20,000 in a business for 2.5 years at an interest rate of 7% per annum. What is the simple interest earned?
Solution:
Principal = ₦20,000
Rate = 7%
Time = 2.5 years
Simple interest = (₦20,000 * 7 * 2.5) / 100 = _______
a) ₦3,500
b) ₦1,750
c) ₦700
d) ₦175
7. Peter borrowed ₦3,500 from his brother and promised to pay back after 1 year with an additional 15% interest. What is the simple interest he needs to pay?
Solution:
Principal = ₦3,500
Rate = 15%
Time = 1 year
Simple interest = (₦3,500 * 15 * 1) / 100 = _______
a) ₦525
b) ₦350
c) ₦175
d) ₦15
8. If ₦6,000 is invested for 9 months at an interest rate of 8% per annum, what is the simple interest earned?
Solution:
Principal = ₦6,000
Rate = 8%
Time = 9/12 (9 months = 0.75 years)
Simple interest = (₦6,000 * 8 * 0.75) / 100 = _______
a) ₦360
b) ₦480
c) ₦180
d) ₦45
9. A loan of ₦12,000 is taken for 2 years at an interest rate of 5% per annum. What is the simple interest charged?
Solution:
Principal = ₦12,000
Rate = 5%
Time = 2 years
Simple interest = (₦12,000 * 5 * 2) / 100 = _______
a) ₦1,200
b) ₦1,000
c) ₦600
d) ₦500
10. David invested ₦18,000 in a business for 3.5 years at an interest rate of 6% per annum. What is the simple interest earned?
Solution:
Principal = ₦18,000
Rate = 6%
Time = 3.5 years
Simple interest = (₦18,000 * 6 * 3.5) / 100 = _______
a) ₦3,780
b) ₦3,060
c) ₦1,260
d) ₦630
11. Emma borrowed ₦4,500 from her friend and promised to pay back after 1.5 years with an additional 10% interest. What is the simple interest she needs to pay?
Solution:
Principal = ₦4,500
Rate = 10%
Time = 1.5 years
Simple interest = (₦4,500 * 10 * 1.5) / 100 = _______
a) ₦675
b) ₦450
c) ₦225
d) ₦15
12. If ₦8,000 is invested for 6 months at an interest rate of 12% per annum, what is the simple interest earned?
Solution:
Principal = ₦8,000
Rate = 12%
Time = 6/12 (6 months = 0.5 years)
Simple interest = (₦8,000 * 12 * 0.5) / 100 = _______
a) ₦480
b) ₦960
c) ₦240
d) ₦120
13. A loan of ₦10,000 is taken for 2.5 years at an interest rate of 8% per annum. What is the simple interest charged?
Solution:
Principal = ₦10,000
Rate = 8%
Time = 2.5 years
Simple interest = (₦10,000 * 8 * 2.5) / 100 = _______
a) ₦2,000
b) ₦2,800
c) ₦1,600
d) ₦6000
WEEK FIVE
PROFIT AND LOSS PERCENT
Profit or loss percent is expressed as a percentage of the cost price
EXAMPLE 1
An article bought for $3,000 was sold for $3,300. Find the profit percent
Solution
Cost price = $3,000
Selling price = $3,300
Profit = $3300 – $3,000 = $300
Profit % = Profit x 100
Cost price 1
= 300 x 100
3000 1
= 300 x 1 = 300 = 10%
30 x 1 30
EXAMPLE 2
A book bought for $25,00 was sold for $22,000. What is the loss percent?
Solution
Cost price =$25,000
Selling price= $22,000
Loss = $25,000 – $22,000 = $3,000
Loss % =loss x 100
Cost price 1
=3000 x 100 = 12%
25000 1
[mediator_tech]
1. An article bought for ₦5,000 was sold for ₦6,000. Find the profit percent.
Solution:
Cost price = ₦5,000
Selling price = ₦6,000
Profit = ₦6,000 – ₦5,000 = ₦1,000
Profit % = (₦1,000 / ₦5,000) * 100 = _______ %
a) 10%
b) 15%
c) 20%
d) 25%
2. A bicycle bought for ₦8,000 was sold for ₦7,000. What is the loss percent?
Solution:
Cost price = ₦8,000
Selling price = ₦7,000
Loss = ₦8,000 – ₦7,000 = ₦1,000
Loss % = (₦1,000 / ₦8,000) * 100 = _______ %
a) 10%
b) 11.25%
c) 12.5%
d) 13.33%
3. A shirt bought for ₦2,500 was sold for ₦3,000. Find the profit percent.
Solution:
Cost price = ₦2,500
Selling price = ₦3,000
Profit = ₦3,000 – ₦2,500 = ₦500
Profit % = (₦500 / ₦2,500) * 100 = _______ %
a) 10%
b) 15%
c) 20%
d) 25%
4. A toy bought for ₦1,200 was sold for ₦900. What is the loss percent?
Solution:
Cost price = ₦1,200
Selling price = ₦900
Loss = ₦1,200 – ₦900 = ₦300
Loss % = (₦300 / ₦1,200) * 100 = _______ %
a) 15%
b) 20%
c) 25%
d) 30%
5. A laptop bought for ₦60,000 was sold for ₦72,000. Find the profit percent.
Solution:
Cost price = ₦60,000
Selling price = ₦72,000
Profit = ₦72,000 – ₦60,000 = ₦12,000
Profit % = (₦12,000 / ₦60,000) * 100 = _______ %
a) 10%
b) 15%
c) 20%
d) 25%
6. A mobile phone bought for ₦25,000 was sold for ₦20,000. What is the loss percent?
Solution:
Cost price = ₦25,000
Selling price = ₦20,000
Loss = ₦25,000 – ₦20,000 = ₦5,000
Loss % = (₦5,000 / ₦25,000) * 100 = _______ %
a) 10%
b) 15%
c) 20%
d) 25%
7. A watch bought for ₦3,500 was sold for ₦4,000
Solution:
Cost price = ₦3,500
Selling price = ₦4,000
Profit = ₦4,000 – ₦3,500 = ₦500
Profit % = (₦500 / ₦3,500) * 100 = _______ %
a) 10%
b) 12.5%
c) 14.29%
d) 15%
8. A television bought for ₦45,000 was sold for ₦50,000. Find the profit percent.
Solution:
Cost price = ₦45,000
Selling price = ₦50,000
Profit = ₦50,000 – ₦45,000 = ₦5,000
Profit % = (₦5,000 / ₦45,000) * 100 = _______ %
a) 10%
b) 11.11%
c) 11.76%
d) 12.5%
9. A bag bought for ₦2,000 was sold for ₦1,500. What is the loss percent?
Solution:
Cost price = ₦2,000
Selling price = ₦1,500
Loss = ₦2,000 – ₦1,500 = ₦500
Loss % = (₦500 / ₦2,000) * 100 = _______ %
a) 10%
b) 15%
c) 20%
d) 25%
10. A car bought for ₦500,000 was sold for ₦550,000. Find the profit percent.
Solution:
Cost price = ₦500,000
Selling price = ₦550,000
Profit = ₦550,000 – ₦500,000 = ₦50,000
Profit % = (₦50,000 / ₦500,000) * 100 = _______ %
a) 5%
b) 10%
c) 15%
d) 20%
11. A refrigerator bought for ₦30,000 was sold for ₦28,000. What is the loss percent?
Solution:
Cost price = ₦30,000
Selling price = ₦28,000
Loss = ₦30,000 – ₦28,000 = ₦2,000
Loss % = (₦2,000 / ₦30,000) * 100 = _______ %
a) 5%
b) 6.67%
c) 7.5%
d) 8%
12. A laptop bought for ₦40,000 was sold for ₦35,000. What is the loss percent?
Solution:
Cost price = ₦40,000
Selling price = ₦35,000
Loss = ₦40,000 – ₦35,000 = ₦5,000
Loss % = (₦5,000 / ₦40,000) * 100 = _______ %
a) 10%
b) 11.25%
c) 12.5%
d) 13.33%
13. A mobile phone bought for ₦25,000 was sold for ₦30,000. Find the profit percent.
[mediator_tech]
EXERCISE
| Cost price | Selling price | Profit % | Loss% | |
| 1. | $1,300 | $1,365 | ||
| $33,555 | $32,446 | |||
| $56,000 | $59,540 | |||
| $100,680 | $99,960 |
WEEK SIX
PERIMETER OF PLANE SHAPES
1. Square:
A square is a four-sided polygon with all sides equal in length. To find the perimeter of a square, we add up the lengths of all four sides.
Example:
Consider a square with a side length of 5 units. To find the perimeter, we add all four sides: 5 + 5 + 5 + 5 = 20 units. Therefore, the perimeter of the square is 20 units.
2. Rectangle:
A rectangle is a four-sided polygon with opposite sides equal in length and all angles equal to 90 degrees. The perimeter of a rectangle is calculated by adding the lengths of all four sides.
Example:
Suppose we have a rectangle with a length of 8 units and a width of 6 units. To find the perimeter, we add all four sides: 8 + 6 + 8 + 6 = 28 units. Therefore, the perimeter of the rectangle is 28 units.
3. Triangle:
A triangle is a three-sided polygon. To calculate the perimeter of a triangle, we add the lengths of all three sides.
Example:
Consider a triangle with side lengths of 4 units, 5 units, and 6 units. To find the perimeter, we add all three sides: 4 + 5 + 6 = 15 units. Therefore, the perimeter of the triangle is 15 units.
4. Circle:
A circle is a closed curve where all points are equidistant from the center. Instead of measuring the perimeter, we use a different term called the circumference to describe the length around a circle. The circumference is calculated using the formula C = 2πr, where “C” represents the circumference and “r” represents the radius of the circle.
Example:
Let’s say we have a circle with a radius of 3 units. Using the formula, we can calculate the circumference: C = 2π(3) = 6π units. If we approximate π to be 3.14, then the circumference is approximately 18.84 units.
So, when it comes to circles, we use the term “circumference” instead of “perimeter” to describe the length around the shape.
Remember, the perimeter of a plane shape represents the total length of its boundaries. By understanding how to calculate the perimeter of different shapes, we can better understand and measure their size and boundaries.
- Square L
L L perimeter = L + L + L + L = 4L
L
- Rectangle L
L L perimeter = L + L + B + B
= 2L + 2B = 2 ( L+B)
L
- Circle
Perimeter = 2 x π x r
= 2πr
Where π = 22/7 or 3.142
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1. Square:
Example: Find the perimeter of a square with a side length of 7 units.
Solution:
Perimeter of a square = 4 * side length
Perimeter = 4 * 7 = 28 units
2. Rectangle:
Example: Find the perimeter of a rectangle with length 10 units and width 5 units.
Solution:
Perimeter of a rectangle = 2 * (length + width)
Perimeter = 2 * (10 + 5) = 2 * 15 = 30 units
3. Triangle:
Example: Find the perimeter of a triangle with side lengths 6 units, 8 units, and 10 units.
Solution:
Perimeter of a triangle = sum of all three side lengths
Perimeter = 6 + 8 + 10 = 24 units
4. Circle:
Example: Find the circumference of a circle with a radius of 4 units. (Approximate π as 3.14)
Solution:
Circumference of a circle = 2 * π * radius
Circumference = 2 * 3.14 * 4 = 25.12 units
5. Square:
Example: If the perimeter of a square is 24 units, what is the length of each side?
Solution:
Perimeter of a square = 4 * side length
24 = 4 * side length
Side length = 24 / 4 = 6 units
6. Rectangle:
Example: The perimeter of a rectangle is 36 units, and its length is 10 units. Find the width.
Solution:
Perimeter of a rectangle = 2 * (length + width)
36 = 2 * (10 + width)
36 = 20 + 2 * width
2 * width = 36 – 20 = 16
Width = 16 / 2 = 8 units
7. Triangle:
Example: The perimeter of a triangle is 15 units, and two sides are 4 units and 5 units long. Find the length of the third side.
Solution:
Perimeter of a triangle = sum of all three side lengths
15 = 4 + 5 + third side length
third side length = 15 – 4 – 5 = 6 units
8. Circle:
Example: Find the circumference of a circle with a diameter of 12 units. (Approximate π as 3.14)
Solution:
Circumference of a circle = π * diameter
Circumference = 3.14 * 12 = 37.68 units
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EXAMPLE
Calculate the perimeter 9cm
9cm 9cm
9cm
Perimeter = 4 x length
= 4 x 9cm
= 36cm
EXAMPLE 2
CAlculate the Perimeter of the rectangle
15cm
9cm
Perimeter = 2 ( L+B)
= 2 (15cm + 9cm)
= 2(24cm) = 2x 24 = 28cm
EXAMPLE 3
Calculate the perimeter of the circle of radius 14cm. Take π = 22/7
Perimeter = 2 π r
= 2 x π x r
= 2 x 22 x 14
7
= 2 x 22 x 2 = 88cm
EXERCISE 1
CALCULATE THE PERIMETER OF THE RECTANGLES WITH THE FOLLOWING DIMENSIONS
- Lenght = 9cm, breadth = 5cm
- Length= 12cm, breadth = 7cm
- Length = 14cm, breadth = 1ocm
- Length = 16cm, breadth =11cm
- Length = 11.5cm, breadth = 4.5cm
EXERCISE 2
Calculate the perimeter of the following circles. Take π = 22/7
- Radius = 7cm
- Radius = 14cm
- Radius = 3.5
- Radius = 21cm
- Radius =28cm
- Radius = 12cm
- Radius = 35cm
- Radius = 4.9cm
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1. The perimeter of a square with a side length of 6 cm is _______ cm.
a) 12
b) 18
c) 24
d) 36
2. The perimeter of a rectangle with length 12 cm and width 5 cm is _______ cm.
a) 22
b) 24
c) 34
d) 60
3. The perimeter of an equilateral triangle with a side length of 8 cm is _______ cm.
a) 12
b) 16
c) 24
d) 32
4. The circumference of a circle with a diameter of 10 cm (approximating π as 3.14) is _______ cm.
a) 10π
b) 15π
c) 20π
d) 25π
5. The perimeter of a square is 40 cm. Each side length is _______ cm.
a) 8
b) 10
c) 12
d) 16
6. The perimeter of a rectangle is 42 cm. If the length is 12 cm, the width is _______ cm.
a) 6
b) 9
c) 15
d) 18
7. The perimeter of an isosceles triangle with equal sides measuring 5 cm each and the third side measuring 8 cm is _______ cm.
a) 10
b) 15
c) 18
d) 20
8. The circumference of a circle with a radius of 7 cm (approximating π as 3.14) is _______ cm.
a) 14π
b) 21π
c) 28π
d) 35π
9. The perimeter of a square with a side length of 9 meters is _______ meters.
a) 9
b) 18
c) 27
d) 36
10. The perimeter of a rectangle is 28 meters. If the length is 10 meters, the width is _______ meters.
a) 4
b) 6
c) 8
d) 12
11. The perimeter of an equilateral triangle with a side length of 12 km is _______ km.
a) 12
b) 24
c) 36
d) 48
12. The circumference of a circle with a diameter of 14 meters (approximating π as 3.14) is _______ meters.
a) 14π
b) 21π
c) 28π
d) 42π
13. The perimeter of a square is 24 cm. Each side length is _______ cm.
a) 4
b) 6
c) 8
d) 12
14. The perimeter of a rectangle is 30 km. If the length is 9 km, the width is _______ km.
a) 3
b) 6
c) 9
d) 12
15. The perimeter of an isosceles triangle with equal sides measuring 7 meters each and the third side measuring 10 meters is _______ meters.
a) 14
b) 21 c) 24 d) 50
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WEEK SEVEN
AREA OF RIGHT ANGLED TRIANGLE
A right-angled triangle has 3 – sides
A
Height
B C
Example 1
Calculate the area of a triangle of height 12cm and base 13cm
Solution
12cm
13cm
Area of tringle = ½ x base x height
= ½ x 12cm x 13cm
1 1
= 1 x 6cm x 13cm = 78cm2
EXERCISE
Calculate the area of the triangles with the following dimensions
- Base = 14cm, height = 16cm
- Base = 10cm, height = 8cm
- Base = 25cm, height = 22cm
- Base = 30cm, height = 15cm
- Base = 24cm, height = 18cm
- Base = 14cm, height = 16cm
- Base = 34cm, height = 15cm
- Base = 52cm, height = 48cm
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A right-angled triangle is a triangle with one angle measuring 90 degrees (a right angle). The other two angles are acute angles (less than 90 degrees).
The formula for finding the area of a right-angled triangle is:
Area = (base * height) / 2
where the base is the length of the side adjacent to the right angle, and the height is the length of the side perpendicular to the base.
Example 1:
Consider a right-angled triangle with a base of 6 cm and a height of 4 cm. To find the area, we can use the formula:
Area = (base * height) / 2 = (6 cm * 4 cm) / 2 = 12 cm²
Therefore, the area of the right-angled triangle is 12 square centimeters.
Example 2:
Suppose we have a right-angled triangle with a base of 10 meters and a height of 8 meters. Using the area formula:
Area = (base * height) / 2 = (10 meters * 8 meters) / 2 = 40 square meters
So, the area of the right-angled triangle is 40 square meters.
Example 3:
Let’s consider a right-angled triangle with a base of 12 inches and a height of 9 inches. Applying the area formula:
Area = (base * height) / 2 = (12 inches * 9 inches) / 2 = 54 square inches
Hence, the area of the right-angled triangle is 54 square inches.
Example 4:
Suppose we have a right-angled triangle with a base of 5 feet and a height of 3 feet. Using the area formula:
Area = (base * height) / 2 = (5 feet * 3 feet) / 2 = 7.5 square feet
Therefore, the area of the right-angled triangle is 7.5 square feet.
In summary, the area of a right-angled triangle can be calculated by multiplying the base and height and dividing the product by 2. Understanding this formula helps us find the amount of space occupied by a right-angled triangle and is a fundamental concept in geometry.
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Worked Examples
1. Example: Find the area of a right-angled triangle with a base of 6 cm and a height of 4 cm.
Solution:
Area = (base * height) / 2
Area = (6 cm * 4 cm) / 2
Area = 24 cm² / 2
Area = 12 cm²
Therefore, the area of the right-angled triangle is 12 square centimeters.
2. Example: Calculate the area of a right-angled triangle with a base of 8 meters and a height of 5 meters.
Solution:
Area = (base * height) / 2
Area = (8 m * 5 m) / 2
Area = 40 m² / 2
Area = 20 m²
The area of the right-angled triangle is 20 square meters.
3. Example: Find the area of a right-angled triangle with a base of 12 inches and a height of 9 inches.
Solution:
Area = (base * height) / 2
Area = (12 in * 9 in) / 2
Area = 108 in² / 2
Area = 54 in²
Therefore, the area of the right-angled triangle is 54 square inches.
4. Example: Calculate the area of a right-angled triangle with a base of 10 cm and a height of 6 cm.
Solution:
Area = (base * height) / 2
Area = (10 cm * 6 cm) / 2
Area = 60 cm² / 2
Area = 30 cm²
The area of the right-angled triangle is 30 square centimeters.
5. Example: Find the area of a right-angled triangle with a base of 15 meters and a height of 12 meters.
Solution:
Area = (base * height) / 2
Area = (15 m * 12 m) / 2
Area = 180 m² / 2
Area = 90 m²
Therefore, the area of the right-angled triangle is 90 square meters.
6. Example: Calculate the area of a right-angled triangle with a base of 5 inches and a height of 3 inches.
Solution:
Area = (base * height) / 2
Area = (5 in * 3 in) / 2
Area = 15 in² / 2
Area = 7.5 in²
The area of the right-angled triangle is 7.5 square inches.
7. Example: Find the area of a right-angled triangle with a base of 9 meters and a height of 6 meters.
Solution:
Area = (base * height) / 2
Area = (9 m * 6 m) / 2
Area = 54 m² / 2
Area = 27 m²
Therefore, the area of the right-angled triangle is 27 square meters.
8. Example: Calculate the area of a right-angled triangle with a base of 7 centimeters and a height of 4 centimeters.
Solution:
Area = (base * height) / 2
Area = (7 cm * 4 cm) / 2
Area = 28 cm² / 2
Area = 14 cm²
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1. The area of a right-angled triangle with a base of 6 cm and a height of 4 cm is _______ square centimeters.
a) 12
b) 16
c) 24
d) 48
2. The area of a right-angled triangle with a base of 8 meters and a height of 5 meters is _______ square meters.
a) 20
b) 30
c) 40
d) 50
3. The area of a right-angled triangle with a base of 12 inches and a height of 9 inches is _______ square inches.
a) 54
b) 72
c) 90
d) 108
4. The area of a right-angled triangle with a base of 10 cm and a height of 6 cm is _______ square centimeters.
a) 15
b) 20
c) 30
d) 60
5. The area of a right-angled triangle with a base of 15 meters and a height of 12 meters is _______ square meters.
a) 72
b) 90
c) 108
d) 180
6. The area of a right-angled triangle with a base of 5 inches and a height of 3 inches is _______ square inches.
a) 4
b) 7.5
c) 9
d) 12
7. The area of a right-angled triangle with a base of 9 meters and a height of 6 meters is _______ square meters.
a) 18
b) 27
c) 36
d) 54
8. The area of a right-angled triangle with a base of 7 centimeters and a height of 4 centimeters is _______ square centimeters.
a) 7
b) 14
c) 16
d) 28
9. The area of a right-angled triangle with a base of 10 cm and a height of 8 cm is _______ square centimeters.
a) 40
b) 50
c) 60
d) 80
10. The area of a right-angled triangle with a base of 12 meters and a height of 10 meters is _______ square meters.
a) 40
b) 60
c) 80
d) 120
11. The area of a right-angled triangle with a base of 6 inches and a height of 5 inches is _______ square inches.
a) 10
b) 12.5
c) 15
d) 18
12. The area of a right-angled triangle with a base of 8 meters and a height of 6 meters is _______ square meters.
a) 24
b) 30
c) 36
d) 48
13. The area of a right-angled triangle with a base of 9 centimeters and a height of 7 centimeters is _______ square centimeters.
a) 18
b) 24.5
c) 31.5
d) 35
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WEEK EIGHT
VOLUME OF CYLINDER
Radius
Height
Volume of cylinder = πr2h
= π x r x r x h
EXAMPLE
Find the volume of a cylinder of radius 3 ½ cm and height 12cm
Solution
3 ½
12cm
Volume = π x r x r x h
=22 x 7 x 7 x 12cm
7 x 2 x 2 x 1
= 22 x 7 x 3
1 x 1 x 1
= 462cm3
Exercise (take π= 22/7) Find the volume of cylinders whose radii and heights are given as:
- Radius 3.5cm, height = 10cm
- Radius 10.5, height 10m
- Radius 4.2cm, height 15cm
- Diameter = 1o½cm, height 16cm
- Diameter = 14cm, height 15cm
Calculate the volume of the cylinders based on the given radii and heights:
1. Radius = 3.5 cm, Height = 10 cm
Volume = (22/7) * (3.5^2) * 10
Volume = (22/7) * 12.25 * 10
Volume = 385 cm³
2. Radius = 10.5 cm, Height = 10 m (Note: Convert height to cm for consistency)
Height = 10 m * 100 cm/m = 1000 cm
Volume = (22/7) * (10.5^2) * 1000
Volume = (22/7) * 110.25 * 1000
Volume = 346,500 cm³
3. Radius = 4.2 cm, Height = 15 cm
Volume = (22/7) * (4.2^2) * 15
Volume = (22/7) * 17.64 * 15
Volume = 942 cm³
4. Diameter = 10.5 cm (Radius = 10.5/2 = 5.25 cm), Height = 16 cm
Volume = (22/7) * (5.25^2) * 16
Volume = (22/7) * 27.5625 * 16
Volume = 1,728 cm³
5. Diameter = 14 cm (Radius = 14/2 = 7 cm), Height = 15 cm
Volume = (22/7) * (7^2) * 15
Volume = (22/7) * 49 * 15
Volume = 10,890 cm³
So, the volumes of the given cylinders are:
1. Volume = 385 cm³
2. Volume = 346,500 cm³
3. Volume = 942 cm³
4. Volume = 1,728 cm³
5. Volume = 10,890 cm³
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Example 1:
Find the volume of a cylinder with a radius of 4 cm and height of 10 cm.
Solution:
Volume = π * r^2 * h
Volume = π * 4^2 * 10
Volume = π * 16 * 10
Volume = 160π cubic cm
Example 2:
Calculate the volume of a cylinder with a radius of 5 cm and height of 8 cm.
Solution:
Volume = π * r^2 * h
Volume = π * 5^2 * 8
Volume = π * 25 * 8
Volume = 200π cubic cm
Example 3:
Find the volume of a cylinder with a radius of 2.5 cm and height of 15 cm.
Solution:
Volume = π * r^2 * h
Volume = π * 2.5^2 * 15
Volume = π * 6.25 * 15
Volume = 93.75π cubic cm
Example 4:
Calculate the volume of a cylinder with a radius of 6 mm and height of 12 mm.
Solution:
Volume = π * r^2 * h
Volume = π * 6^2 * 12
Volume = π * 36 * 12
Volume = 432π cubic mm
Example 5:
Find the volume of a cylinder with a radius of 8 cm and height of 5 cm.
Solution:
Volume = π * r^2 * h
Volume = π * 8^2 * 5
Volume = π * 64 * 5
Volume = 320π cubic cm
Example 6:
Calculate the volume of a cylinder with a radius of 3 m and height of 10 m.
Solution:
Volume = π * r^2 * h
Volume = π * 3^2 * 10
Volume = π * 9 * 10
Volume = 90π cubic m
Example 7:
Find the volume of a cylinder with a radius of 2.5 inches and height of 6 inches.
Solution:
Volume = π * r^2 * h
Volume = π * 2.5^2 * 6
Volume = π * 6.25 * 6
Volume = 37.5π cubic inches
Example 8:
Calculate the volume of a cylinder with a radius of 5 cm and height of 10 cm.
Solution:
Volume = π * r^2 * h
Volume = π * 5^2 * 10
Volume = π * 25 * 10
Volume = 250π cubic cm
These examples illustrate how to find the volume of a cylinder by using the formula π * r^2 * h. By substituting the given values of radius and height, we can calculate the volume of the cylinder.
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1. The volume of a cylinder with a radius of 5 cm and height of 8 cm is _______ cubic centimeters.
a) 80
b) 200
c) 250
d) 320
2. The volume of a cylinder with a radius of 3 meters and height of 10 meters is _______ cubic meters.
a) 9
b) 30
c) 90
d) 180
3. The volume of a cylinder with a radius of 6 inches and height of 12 inches is _______ cubic inches.
a) 432
b) 576
c) 864
d) 1,152
4. The volume of a cylinder with a radius of 8 mm and height of 5 mm is _______ cubic millimeters.
a) 160
b) 320
c) 800
d) 1,280
5. The volume of a cylinder with a radius of 10 cm and height of 15 cm is _______ cubic centimeters.
a) 500
b) 750
c) 1,500
d) 2,250
6. The volume of a cylinder with a radius of 2.5 meters and height of 6 meters is _______ cubic meters.
a) 37.5
b) 75
c) 112.5
d) 150
7. The volume of a cylinder with a radius of 4.2 centimeters and height of 8 centimeters is _______ cubic centimeters.
a) 112.896
b) 141.12
c) 179.52
d) 225.792
8. The volume of a cylinder with a radius of 7 inches and height of 10 inches is _______ cubic inches.
a) 308
b) 440
c) 616
d) 840
9. The volume of a cylinder with a radius of 6.5 cm and height of 15 cm is _______ cubic centimeters.
a) 496.125
b) 769.5
c) 858.75
d) 1,002.75
10. The volume of a cylinder with a radius of 9 meters and height of 12 meters is _______ cubic meters.
a) 3,456
b) 7,344
c) 9,936
d) 15,552
11. The volume of a cylinder with a radius of 3.8 inches and height of 7 inches is _______ cubic inches.
a) 248.6
b) 318.5
c) 417.9
d) 560.8
12. The volume of a cylinder with a radius of 5.5 cm and height of 10 cm is _______ cubic centimeters.
a) 953.75
b) 1,325
c) 1,815
d) 2,475
13. The volume of a cylinder with a radius of 7.2 meters and height of 9 meters is _______ cubic meters.
a) 1,365.12
b) 1,884.48
c) 2,352.96
d) 3,211.52
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WEEK NINE & TEN
WEIGHT
The units used for measuring weight are kilogram (kg) and gram (g).
1000 gram = 1kg
1000kg = 1 ton
EXAMPLE 1
Change to kilogram
2000g to kilogram
1000g = 1kg
2000g = m
Cross multiply
1000 × m = 2000 x 1
1000m = 2000
m = 2000 = 2kg
1000
EXAMPLE 2
Change 5kg to grams
Solution
To change from kg to grams, multiply by 1000
5kg = 5 x 1000
= 5000 grams
EXERCISE
Change the following to kilograms:
- 2000g
- 5000g
- 3500g
- 1205g
- 750g
Change to grams
- 6.5 kg
- 0.5kg
- 4.298kg
- 8.765kg
- 13.020kg
- 0.07kg
[mediator_tech]
Example 1:
Convert 2500 grams to kilograms.
Solution:
2500 grams = 2500/1000 kilograms = 2.5 kilograms
Therefore, 2500 grams is equal to 2.5 kilograms.
Example 2:
Convert 3.7 kilograms to grams.
Solution:
3.7 kilograms = 3.7 * 1000 grams = 3700 grams
Therefore, 3.7 kilograms is equal to 3700 grams.
Example 3:
Convert 4.5 kilograms to tons.
Solution:
4.5 kilograms = 4.5/1000 tons = 0.0045 tons
Therefore, 4.5 kilograms is equal to 0.0045 tons.
Example 4:
Convert 7500 grams to tons.
Solution:
7500 grams = 7500/1000 kilograms = 7.5 kilograms
7.5 kilograms = 7.5/1000 tons = 0.0075 tons
Therefore, 7500 grams is equal to 0.0075 tons.
Example 5:
Convert 0.9 kilograms to grams.
Solution:
0.9 kilograms = 0.9 * 1000 grams = 900 grams
Therefore, 0.9 kilograms is equal to 900 grams.
Example 6:
Convert 1200 grams to kilograms.
Solution:
1200 grams = 1200/1000 kilograms = 1.2 kilograms
Therefore, 1200 grams is equal to 1.2 kilograms.
Example 7:
Convert 6.3 kilograms to tons.
Solution:
6.3 kilograms = 6.3/1000 tons = 0.0063 tons
Therefore, 6.3 kilograms is equal to 0.0063 tons.
Example 8:
Convert 0.005 tons to kilograms.
Solution:
0.005 tons = 0.005 * 1000 kilograms = 5 kilograms
Therefore, 0.005 tons is equal to 5 kilograms.
These examples demonstrate the conversion between kilograms, grams, and tons. It is important to understand the conversion factors (1 kilogram = 1000 grams and 1 ton = 1000 kilograms) in order to convert weights accurately between these units.
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1. 500 milligrams = _______ centigrams.
a) 5
b) 50
c) 5000
d) 50000
2. 8 centigrams = _______ milligrams.
a) 0.8
b) 8
c) 80
d) 800
3. 3 decigrams = _______ centigrams.
a) 0.03
b) 0.3
c) 3
d) 30
4. 2 grams = _______ milligrams.
a) 0.002
b) 0.02
c) 20
d) 2000
5. 6 kilograms = _______ grams.
a) 0.6
b) 6
c) 60
d) 6000
6. 2500 milligrams = _______ grams.
a) 2.5
b) 25
c) 250
d) 2500
7. 12 decigrams = _______ grams.
a) 0.12
b) 0.012
c) 1.2
d) 12
8. 5 centigrams = _______ milligrams.
a) 5
b) 50
c) 500
d) 5000
9. 3 kilograms = _______ grams.
a) 0.03
b) 3
c) 30
d) 3000
10. 5000 milligrams = _______ grams.
a) 0.005
b) 0.05
c) 5
d) 50
11. 4 centigrams = _______ decigrams.
a) 0.04
b) 0.4
c) 4
d) 40
12. 7 kilograms = _______ grams.
a) 0.007
b) 0.07
c) 7
d) 7000
13. 1800 grams = _______ kilograms.
a) 0.18
b) 1.8
c) 18
d) 180
14. 250 decigrams = _______ grams.
a) 0.025
b) 0.25
c) 2.5
d) 25
15. 0.05 kilograms = _______ grams.
a) 0.005
b) 0.05
c) 0.5
d) 5
Choose the correct options for each question based on the given conversion units and relationships between them.
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3RD TERM EXAM QUESTIONS PRIMARY 5 MATHEMATICS
Primary 5 Third Term Examination Mathematics
