Area And Perimeter 1
In this article, the learning objectives are:
- Finding the maximum number of squares that can be fitted/cut from a rectangle
- Finding unknown dimensions given area of a rectangle/square
- Finding unknown dimensions given perimeter of a rectangle/square
Let’s recap P3 Area and Perimeter first!
Perimeter
The perimeter of a shape is the total distance around the shape.
Perimeter Of A Square
\(\small \begin{aligned} \textsf{Perimeter Of A Square } &\mathsf{= \text{Length + Length + Length + Length}}\\[2ex] &\mathsf{= 4 \times \text{Length}} \end{aligned}\)
Perimeter Of A Rectangle
\(\small{ \textsf{Perimeter Of A Rectangle} = \textsf{Length + Breadth + Length + Breadth} }\)
Area
The area is the space occupied by the figure.
Area Of A Square
\(\mathsf{\small{\text{Area Of A Square}= \text{Length} \times \text{Length}}}\)
Area Of A Rectangle
\(\mathsf{\small{\text{Area Of A Rectangle}= \text{Length} \times \text{Breadth}}}\)
Question 1:
Find the area and perimeter of the square below.
Solution:
Area of square
\(= 6 \text{ cm} \times 6 \text{ cm}\)
\(= 36 \text{ cm}^2 \)
Perimeter of square
\(= 4 \times 6 \text{ cm}\)
\(= 24 \text{ cm}\)
Answer:
Area: \(36 \text{ cm}^2\)
Perimeter: \(24 \text{ cm}\)
Question 2:
Find the area and perimeter of the rectangle below.
Solution:
Area of rectangle
\(= 6 \text{ cm} \times 4 \text{ cm}\)
\(= 24 \text{ cm}^2\)
Perimeter of rectangle
\(= 6 \text{ cm} + 4 \text{ cm} + 6 \text{ cm} + 4 \text{ cm}\)
\(= 20 \text{ cm}\)
Answer:
Area: \(24 \text{ cm}^2\)
Perimeter: \(20 \text{ cm}\)
-
Finding the maximum number of squares that can be fitted / cut from a rectangle
To find the maximum number of squares that can be fitted/cut from an area, we will first find out the number of squares that are able to fit along the length and the breadth of that area.
Question 1:
What is the maximum number of \(1 \text{ cm}\) squares that can be cut from the rectangle?
Solution:
Number of \(1 \text{ cm}\) squares along the length of the rectangle
\(= 5 \text{ cm} ÷ 1 \text{ cm}\)
\(= 5\)
Number of \(\small{1 \text{ cm}}\) squares along the breadth of the rectangle
\(= 3 \text{ cm} \div 1 \text{ cm}\)
\(= 3\)
Maximum number of squares that can be cut from the rectangle
\(= 5 \times 3\)
\(= 15\)
Answer:
\(15\) squares
Question 2:
What is the maximum number of \(2 \text{ cm}\) squares that can be cut from the rectangle?
Solution:
Number of \(2 \text{ cm}\) squares along the length of the rectangle
\(= 8 \text{ cm} \div 2 \text{ cm}\)
\(= 4\)
Number of \(\small{2 \text{ cm}}\) squares along the breadth of the rectangle
\(= 6 \text{ cm} \div 2 \text{ cm}\)
\(= 3\)
Maximum number of squares that can be cut from the rectangle
\(= 4 \times 3\)
\(= 12\)
Answer:
\(12\) squares
Question 3:
What is the greatest number of \(4 \text{ cm}\) squares that can be cut from the rectangle?
Solution:
Number of \(4 \text{ cm}\) squares along the length of the rectangle
\(= 16 \text{ cm} \div 4\text{ cm}\)
\(= 4\)
Number of \(4 \text{ cm}\) squares along the breadth of the rectangle
\(= 10 \text{ cm} \div 4 \text{ cm}\)
\(= 2 \text{ R } 2 \text{ cm}\)
We ignore the part which is represented by the remainder of \(2 \text{ cm}\) as no squares can be cut from it.
Greatest number of squares that can be cut from the rectangle
\(= 4 \times 2\)
\(= 8\)
Answer:
\(8\) squares
-
Finding unknown dimensions given area of a rectangle / square
\(\mathsf{ \small{\text{Area Of A Rectangle} = \text{Length} \times \text{Breadth}} }\)
Therefore,
\(\small \begin{align} \textsf{Length Of A Rectangle} &= \textsf{Area} \div \textsf{Breadth} \\[2ex] \textsf{Breadth Of A Rectangle} &= \textsf{Area} \div \textsf{Length} \end{align} \)
Question 1:
The area of a rectangle is \(126 \text{ cm}^2\). If its breadth is \(7 \text{ cm}\), what is the length of the rectangle?
Solution:
Length of the rectangle
\(= 126 \text{ cm}^2 ÷ 7 \text{ cm}\)
\(= 18 \text{ cm}\)
Answer:
\(18 \text{ cm}\)
Question 2:
The area of a rectangle is \(72 \text{ cm}^2\). Given that the length of the rectangle is \(9 \text{ cm}\), find the breadth of the rectangle.
Solution:
Breadth of the rectangle
\(= 72 \text{ cm}^2 \div 9 \text{ cm}\)
\(= 8 \text{ cm}\)
Answer:
\(8 \text{ cm}\)
Question 3:
The area of a square is \(64 \text{ cm}^2\). Find the length of one side of the square.
Solution:
Since,
\(8 \text{ cm} \times 8 \text{ cm} = 64 \text{ cm}^2\),
Length of one side of each square \(= 8 \text{ cm}\)
Answer:
\(8 \text{ cm}\)
Question 4:
The figure below is made up of \(3\) identical squares. Given that the total area of the figure is \(75 \text{ cm}^2\), find the length of one side of each square.
Solution:
Area of 1 square
\(= 75 \text{ cm}^2 \div 3\)
\(= 25 \text{ cm}^2\)
Since,
\(5 \text{ cm} \times 5 \text{ cm} = 25 \text{ cm}^2\),
Length of one side of each square \(= 5 \text{ cm}\)
Answer:
\(5 \text{ cm}\)
-
Finding unknown dimensions given perimeter of a rectangle/square
\(\small\begin{align} \mathsf{\text{Perimeter Of Rectangle }} &\mathsf{= \text{Length + Length + Breadth + Breadth}}\\[2ex] \mathsf{\text{Length Of Rectangle }} &\mathsf{= \text{(Perimeter - Breadth - Breadth)} \div 2}\\[2ex] \mathsf{\text{Breadth Of Rectangle }} &\mathsf{= \text{(Perimeter - Length - Length)} \div 2} \end{align}\)
Question 1:
The perimeter of a rectangle is \(36 \text{ cm}\). Given that its breadth is \(5 \text{ cm}\), find its length.
Solution:
Perimeter of a rectangle \(= \textsf{Length + Breadth + Length + Breadth}\)
Total length of 2 lengths
\(= 36 \text{ cm} - 5 \text{ cm} - 5 \text{ cm}\)
\(= 26 \text{ cm}\)
Length of rectangle
\(= 26 \text{ cm} \div 2\)
\(= 13 \text{ cm}\)
Answer:
\(13 \text{ cm}\)
Question 2:
The perimeter of a rectangle is \(72 \text{ cm}\). Given that its length is \(24 \text{ cm}\), find its breadth.
Solution:
\(= \textsf{Length + Breadth + Length + Breadth}\)
Perimeter of a rectangle
\(= 72 \text{ cm} - 24 \text{ cm} - 24 \text{ cm}\)
\(= 24 \text{ cm}\)
\(= 24 \text{ cm} \div 2\)
\(= 12 \text{ cm}\)
Answer:
\(12 \text{ cm}\)
\(\small\begin{align} \textsf{Perimeter of a square} &= \mathsf{4 \times Length} \\[2ex] \textsf{Length of one side of a square} &= \mathsf{Perimeter \div 4} \end{align}\)
Question 3:
The perimeter of a square is \(60 \text{ cm}\). Find the length of one side of the square.
Solution:
Perimeter of a square \(= 4 \times \small{\textsf{ Length }}\)
Length of one side of the square
\(= 60 \text{ cm} \div 4\)
\(= 15 \text{ cm}\)
Answer:
\(15 \text{ cm}\)
Question 4:
The area of a rectangular garden is \(168 \text{ m}^2\). Its breadth is 8 m.
- Find the length of the garden.
- Vincent jogged round the entire rectangular garden twice. Find the distance he jogged.
Solution:
A. Length of the rectangle\(\begin{align} &= 168 \text{ m}^2 \div 8 \text{ m} \\[2ex] &= 21 \text{ m} \end{align}\)
B. Perimeter of garden\(\begin{align}&= 21 \text{ m} + 8 \text{ m} + 21 \text{ m} + 8 \text{ m} \\[2ex] &= 58 \text{ m} \end{align}\)
Distance he jogged \(\begin{align} &= 58 \text{ m} \times 2\\[2ex] &= 116 \text{ m} \end{align}\)
Answer:
A. \(21 \text{ m}\)
B. \(116 \text{ m}\)
Question 5:
Elaine jogged \(36 \text{ m}\) round a square sand pit.
- Find the length of one side of the sand pit.
- Find the area of the square sand pit.
Solution:
A. Length of one side of the sand pit\(\begin{align} &= 36 \text{ m} \div 4\\[2ex] &= 9 \text{ m} \end{align}\)
B. Area of the sand pit\(\begin{align} &= 9 \text{ m} \times 9 \text{ m}\\[2ex] &= 81 \text{ m}^2 \end{align}\)
Answer:
\(9 \text{ m}\)
B. \(81 \text{ m}^2\)
Continue Learning | |
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Multiplication | Whole Numbers |
Multiplication And Division | Decimals |
Model Drawing Strategy | Division |
Fractions | Factors And Multiples |
Area And Perimeter 1 | Line Graphs |
Conversion Of Time |