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Sagot :
Let's address each part of the question step-by-step:
### Part (a)
1. Find the perimeter and area of a rectangle with length 10 cm and breadth 6 cm:
- Perimeter: The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (\text{length} + \text{breadth}) \][/tex]
Substituting the given values:
[tex]\[ P = 2 \times (10 + 6) = 2 \times 16 = 32 \text{ cm} \][/tex]
- Area: The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = \text{length} \times \text{breadth} \][/tex]
Substituting the given values:
[tex]\[ A = 10 \times 6 = 60 \text{ cm}^2 \][/tex]
### Part (b)
1. The perimeter of a square is 72 cm:
- Find its side length [tex]\( s \)[/tex]:
The perimeter [tex]\( P \)[/tex] of a square is given by:
[tex]\[ P = 4 \times s \][/tex]
Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{P}{4} = \frac{72}{4} = 18 \text{ cm} \][/tex]
- Find its area [tex]\( A \)[/tex]:
The area [tex]\( A \)[/tex] of a square is given by:
[tex]\[ A = s^2 \][/tex]
Substituting the side length:
[tex]\[ A = 18^2 = 324 \text{ cm}^2 \][/tex]
### Part (c)
1. The perimeter of a rectangle is 56 cm, and its length is 14 cm:
- Find its breadth [tex]\( b \)[/tex]:
The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (\text{length} + \text{breadth}) \][/tex]
Substituting the given values:
[tex]\[ 56 = 2 \times (14 + b) \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 14 + b = \frac{56}{2} = 28 \][/tex]
[tex]\[ b = 28 - 14 = 14 \text{ cm} \][/tex]
- Find its area [tex]\( A \)[/tex]:
The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = \text{length} \times \text{breadth} \][/tex]
Substituting the given values:
[tex]\[ A = 14 \times 14 = 196 \text{ cm}^2 \][/tex]
### Part (d)
1. The length of a rectangle is twice its breadth, and the area is 72 cm²:
- Let the breadth be [tex]\( b \)[/tex]. Then the length [tex]\( l \)[/tex] is [tex]\( 2b \)[/tex].
- Find the breadth [tex]\( b \)[/tex]:
The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = l \times b \][/tex]
Substituting [tex]\( l = 2b \)[/tex]:
[tex]\[ 72 = 2b \times b \][/tex]
[tex]\[ 72 = 2b^2 \][/tex]
[tex]\[ b^2 = 36 \][/tex]
[tex]\[ b = 6 \text{ cm} \][/tex]
- Find the length [tex]\( l \)[/tex]:
[tex]\[ l = 2b = 2 \times 6 = 12 \text{ cm} \][/tex]
- Find its perimeter [tex]\( P \)[/tex]:
[tex]\[ P = 2 \times (l + b) = 2 \times (12 + 6) = 2 \times 18 = 36 \text{ cm} \][/tex]
### Part (e)
1. The length of a rectangle is three times its breadth, and the area is 108 cm²:
- Let the breadth be [tex]\( b \)[/tex]. Then the length [tex]\( l \)[/tex] is [tex]\( 3b \)[/tex].
- Find the breadth [tex]\( b \)[/tex]:
The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = l \times b \][/tex]
Substituting [tex]\( l = 3b \)[/tex]:
[tex]\[ 108 = 3b \times b \][/tex]
[tex]\[ 108 = 3b^2 \][/tex]
[tex]\[ b^2 = 36 \][/tex]
[tex]\[ b = 6 \text{ cm} \][/tex]
- Find the length [tex]\( l \)[/tex]:
[tex]\[ l = 3b = 3 \times 6 = 18 \text{ cm} \][/tex]
- Find its perimeter [tex]\( P \)[/tex]:
[tex]\[ P = 2 \times (l + b) = 2 \times (18 + 6) = 2 \times 24 = 48 \text{ cm} \][/tex]
### Part (f)
1. A square room is 6 m long. If the area of the floor is equal to the area of the carpet, find the area of the carpet needed to cover its floor:
- Find the area [tex]\( A \)[/tex] of the square room (floor):
The side length [tex]\( s \)[/tex] of the square room is 6 m.
The area [tex]\( A \)[/tex] of a square is given by:
[tex]\[ A = s^2 \][/tex]
Substituting the given values:
[tex]\[ A = 6^2 = 36 \text{ m}^2 \][/tex]
- Therefore, the area of the carpet needed to cover the floor is:
[tex]\[ 36 \text{ m}^2 \][/tex]
### Part (g)
1. The perimeter of a square garden is 120 m; find its area:
- Find its side length [tex]\( s \)[/tex]:
The perimeter [tex]\( P \)[/tex] of a square is given by:
[tex]\[ P = 4 \times s \][/tex]
Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{P}{4} = \frac{120}{4} = 30 \text{ m} \][/tex]
- Find its area [tex]\( A \)[/tex]:
The area [tex]\( A \)[/tex] of a square is given by:
[tex]\[ A = s^2 \][/tex]
Substituting the side length:
[tex]\[ A = 30^2 = 900 \text{ m}^2 \][/tex]
### Part (a)
1. Find the perimeter and area of a rectangle with length 10 cm and breadth 6 cm:
- Perimeter: The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (\text{length} + \text{breadth}) \][/tex]
Substituting the given values:
[tex]\[ P = 2 \times (10 + 6) = 2 \times 16 = 32 \text{ cm} \][/tex]
- Area: The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = \text{length} \times \text{breadth} \][/tex]
Substituting the given values:
[tex]\[ A = 10 \times 6 = 60 \text{ cm}^2 \][/tex]
### Part (b)
1. The perimeter of a square is 72 cm:
- Find its side length [tex]\( s \)[/tex]:
The perimeter [tex]\( P \)[/tex] of a square is given by:
[tex]\[ P = 4 \times s \][/tex]
Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{P}{4} = \frac{72}{4} = 18 \text{ cm} \][/tex]
- Find its area [tex]\( A \)[/tex]:
The area [tex]\( A \)[/tex] of a square is given by:
[tex]\[ A = s^2 \][/tex]
Substituting the side length:
[tex]\[ A = 18^2 = 324 \text{ cm}^2 \][/tex]
### Part (c)
1. The perimeter of a rectangle is 56 cm, and its length is 14 cm:
- Find its breadth [tex]\( b \)[/tex]:
The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (\text{length} + \text{breadth}) \][/tex]
Substituting the given values:
[tex]\[ 56 = 2 \times (14 + b) \][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[ 14 + b = \frac{56}{2} = 28 \][/tex]
[tex]\[ b = 28 - 14 = 14 \text{ cm} \][/tex]
- Find its area [tex]\( A \)[/tex]:
The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = \text{length} \times \text{breadth} \][/tex]
Substituting the given values:
[tex]\[ A = 14 \times 14 = 196 \text{ cm}^2 \][/tex]
### Part (d)
1. The length of a rectangle is twice its breadth, and the area is 72 cm²:
- Let the breadth be [tex]\( b \)[/tex]. Then the length [tex]\( l \)[/tex] is [tex]\( 2b \)[/tex].
- Find the breadth [tex]\( b \)[/tex]:
The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = l \times b \][/tex]
Substituting [tex]\( l = 2b \)[/tex]:
[tex]\[ 72 = 2b \times b \][/tex]
[tex]\[ 72 = 2b^2 \][/tex]
[tex]\[ b^2 = 36 \][/tex]
[tex]\[ b = 6 \text{ cm} \][/tex]
- Find the length [tex]\( l \)[/tex]:
[tex]\[ l = 2b = 2 \times 6 = 12 \text{ cm} \][/tex]
- Find its perimeter [tex]\( P \)[/tex]:
[tex]\[ P = 2 \times (l + b) = 2 \times (12 + 6) = 2 \times 18 = 36 \text{ cm} \][/tex]
### Part (e)
1. The length of a rectangle is three times its breadth, and the area is 108 cm²:
- Let the breadth be [tex]\( b \)[/tex]. Then the length [tex]\( l \)[/tex] is [tex]\( 3b \)[/tex].
- Find the breadth [tex]\( b \)[/tex]:
The area [tex]\( A \)[/tex] of a rectangle is given by:
[tex]\[ A = l \times b \][/tex]
Substituting [tex]\( l = 3b \)[/tex]:
[tex]\[ 108 = 3b \times b \][/tex]
[tex]\[ 108 = 3b^2 \][/tex]
[tex]\[ b^2 = 36 \][/tex]
[tex]\[ b = 6 \text{ cm} \][/tex]
- Find the length [tex]\( l \)[/tex]:
[tex]\[ l = 3b = 3 \times 6 = 18 \text{ cm} \][/tex]
- Find its perimeter [tex]\( P \)[/tex]:
[tex]\[ P = 2 \times (l + b) = 2 \times (18 + 6) = 2 \times 24 = 48 \text{ cm} \][/tex]
### Part (f)
1. A square room is 6 m long. If the area of the floor is equal to the area of the carpet, find the area of the carpet needed to cover its floor:
- Find the area [tex]\( A \)[/tex] of the square room (floor):
The side length [tex]\( s \)[/tex] of the square room is 6 m.
The area [tex]\( A \)[/tex] of a square is given by:
[tex]\[ A = s^2 \][/tex]
Substituting the given values:
[tex]\[ A = 6^2 = 36 \text{ m}^2 \][/tex]
- Therefore, the area of the carpet needed to cover the floor is:
[tex]\[ 36 \text{ m}^2 \][/tex]
### Part (g)
1. The perimeter of a square garden is 120 m; find its area:
- Find its side length [tex]\( s \)[/tex]:
The perimeter [tex]\( P \)[/tex] of a square is given by:
[tex]\[ P = 4 \times s \][/tex]
Solving for [tex]\( s \)[/tex]:
[tex]\[ s = \frac{P}{4} = \frac{120}{4} = 30 \text{ m} \][/tex]
- Find its area [tex]\( A \)[/tex]:
The area [tex]\( A \)[/tex] of a square is given by:
[tex]\[ A = s^2 \][/tex]
Substituting the side length:
[tex]\[ A = 30^2 = 900 \text{ m}^2 \][/tex]
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