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
| \(\begin{align} &= \text{Length + Length + Length + Length} \\[2ex] &= 4 \times \text{Length} \end{align}\) |  |
Perimeter Of A Rectangle
| \(= \text{Length + Breadth + Length + Breadth}\) |  |
Area
The area is the space occupied by the figure.
Area Of A Square
| \(\begin{align} = \text{Length} \times \text{Length} \end{align}\) | |
Area Of A Rectangle
| \(\begin{align} = \text{Length} \times \text{Breadth} \end{align}\) | |
Question 1:
Find the area and perimeter of the square below.
Solution:
Area of square
\(\begin{align} &= 6 \;cm \times 6 \;cm \\[2ex] &= 36 \;cm^2 \end{align}\)
Perimeter of square
\(\begin{align} &= 4 \times 6 \;cm \\[2ex] &= 24 \;cm \end{align}\)
Answer:
Area: \(36 \;cm^2\)
Perimeter: \(24 \;cm\)
Question 2:
Find the area and perimeter of the rectangle below.
Solution:
Area of rectangle
\(\begin{align} &= 6 \;cm \times 4 \;cm \\[2ex] &= 24 \;cm^2 \end{align}\)
Perimeter of rectangle
\(\begin{align} &= 6 \;cm + 4 \;cm + 6 \;cm + 4 \;cm \\[2ex] &= 20 \;cm \end{align}\)
Answer:
Area: \(24 \;cm^2\)
Perimeter: \(20 \;cm\)
1. 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 \(1 \;cm\) squares that can be cut from the rectangle?
Solution:
Number of \(1 \;cm\) squares along the length of the rectangle
\(\begin{align} &= 5 \;cm ÷ 1 \;cm \\[2ex] &= 5 \end{align} \)
Number of \(1 \;cm\) squares along the breadth of the rectangle
\(\begin{align} &= 3 \;cm ÷ 1 \;cm \\[2ex] &= 3 \end{align} \)
Maximum number of squares that can be cut from the rectangle
\(\begin{align} &= 5 \times 3 \\[2ex] &= 15 \end{align}\)
Answer:
\(15\) squares
Question 2:
What is the maximum number of \(2 \;cm\) squares that can be cut from the rectangle?
Solution:
Number of \(2 \;cm\) squares along the length of the rectangle
\(\begin{align} &= 8 \;cm \div 2 \;cm \\[2ex] &= 4 \end{align}\)
Number of \(2 \;cm\) squares along the breadth of the rectangle
\(\begin{align} &= 6 \;cm \div 2 \;cm \\[2ex] &= 3 \end{align}\)
Maximum number of squares that can be cut from the rectangle
\(\begin{align} &= 4 \times 3 \\[2ex] &= 12 \end{align}\)
Answer:
\(12\) squares
Question 3:
What is the greatest number of 4-cm squares that can be cut from the rectangle?
Solution:
Number of \(4 \;cm\) squares along the length of the rectangle
\(\begin{align} &= 16 \;cm \div 4\;cm \\[2ex] &= 4 \end{align}\)
Number of \(4 \;cm\) squares along the breadth of the rectangle
\(\begin{align} &= 10 \;cm \div 4\;cm \\[2ex] &= 2\;R \;2\;cm \end{align}\)
We ignore the part which is represented by the remainder of \(2\;cm\) as no squares can be cut from it.
Greatest number of squares that can be cut from the rectangle
\(\begin{align} &= 4 \times 2 \\[2ex] &= 8 \end{align}\)
Answer:
\(8\) squares
2. Finding unknown dimensions given area of a rectangle / square
\(\text{Area Of A Rectangle} = \text{Length} \times \text{Breadth}\)
Therefore,
\(\begin{align} \text{Length Of A Rectangle} &= \text{Area} \div \text{Breadth} \\[2ex] \text{Breadth Of A Rectangle} &= \text{Area} \div \text{Length} \end{align} \)
Question 1:
The area of a rectangle is \(126 \;cm^2\). If its breadth is \(7 \;cm\), what is the length of the rectangle?
Solution:
Length of the rectangle
\(\begin{align} &= 126 \;cm^2 ÷ 7 \;cm \\[2ex] &= 18 \;cm \end{align}\)
Answer:
\(18 \;cm\)
Question 2:
The area of a rectangle is 72 cm2. Given that the length of the rectangle is 9 cm, find the breadth of the rectangle.
Solution:
Breadth of the rectangle
\(\begin{align} &= 72 \;cm^2 \div 9 \;cm\\[2ex] &= 8 \;cm \end{align}\)
Answer:
\(8 \;cm\)
Question 3:
The area of a square is \(64 \;cm^2\). Find the length of one side of the square.
Solution:
Since \(8 \;cm \times 8 \;cm = 64 \;cm^2\),
Length of one side of each square \(= 8 \;cm\)
Answer:
\(8 \;cm \)
Question 4:
The figure below is made up of \(3\) identical squares. Given that the total area of the figure is \(75 \;cm^2\), find the length of one side of each square.
Solution:
Area of 1 square
\(\begin{align} &= 75 \;cm^2 \div 3\\[2ex] &= 25 \;cm^2 \end{align}\)
Since \(5 \;cm \times 5 \;cm = 25 \;cm^2\),
Length of one side of each square \(= 5 \;cm\)
Answer:
\(5 \;cm\)
3. Finding unknown dimensions given perimeter of a rectangle/square
\(\begin{align} \text{Perimeter Of Rectangle} &= \text{Length + Length + Breadth + Breadth}\\[2ex] \text{Length Of Rectangle} &= \text{(Perimeter - Breadth - Breadth) ÷ 2}\\[2ex] \text{Breadth Of Rectangle} &= \text{(Perimeter - Length - Length) ÷ 2} \end{align}\)
Question 1:
The perimeter of a rectangle is \(36 \;cm\). Given that its breadth is \(5 \;cm\), find its length.
Solution:
\(\begin{align} \text{Perimeter of a rectangle} &= \text{Length + Breadth + Length + Breadth} \\[3ex] \text{Total length of 2 lengths} &= 36 \;cm - 5 \;cm - 5 \;cm \\[2ex] &= 26 \;cm\\[3ex] \text{Length of rectangle} &= 26 \;cm ÷ 2 \\[2ex] &= 13 \;cm \end{align}\)
Answer:
\(13 \;cm\)
Question 2:
The perimeter of a rectangle is \(72 \;cm\). Given that its length is \(24 \;cm\), find its breadth.
Solution:
\(\begin{align} \text{Perimeter of a rectangle} &= \text{Length + Breadth + Length + Breadth} \\[3ex] \text{Total length of 2 breadths} &= 72 \;cm - 24 \;cm - 24 \;cm \\[2ex] &= 24 \;cm\\[3ex] \text{ Breadth of rectangle} &= 24 \;cm ÷ 2 \\[2ex] &= 12 \;cm \end{align}\)
Answer:
\(12 \;cm\)
\(\begin{align} \text{Perimeter of a square} &= 4 \times \text{Length} \\[2ex] \text{Length of one side of a square} &= \text{Perimeter} \div 4 \end{align}\)
Question 3:
The perimeter of a square is \(60 \;cm\). Find the length of one side of the square.
Solution:
Perimeter of a square \(= 4 \;×\) Length
Length of one side of the square\(\begin{align}\\[2ex] &= 60 \;cm \div 4\\[2ex] &= 15 \;cm \end{align}\)
Answer:
\(15 \;cm\)
Question 4:
The area of a rectangular garden is \(168 \;m^2\). 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:
- Length of the rectangle\(\begin{align}\\[2ex] &= 168 \;m^2 \div 8 \;m \\[2ex] &= 21 \;m \end{align}\)
- Perimeter of garden\(\begin{align}\\[2ex] &= 21 \;m + 8 \;m + 21 \;m + 8 \;m \\[2ex] &= 58 \;m \end{align}\)
Distance he jogged \(\begin{align}\\[2ex] &= 58 \;m \times 2\\[2ex] &= 116 \;m \end{align}\)
Answer:
- \(21 \;m\)
Answer:
- \(116 \;m\)
Question 5:
Elaine jogged \(36 \;m\) round a square sand pit.
- Find the length of one side of the sand pit.
- Find the area of the square sand pit.
Solution:
- Length of one side of the sand pit\(\begin{align}\\[2ex] &= 36 \;m \div 4\\[2ex] &= 9 \;m \end{align}\)
- Area of the sand pit\(\begin{align}\\[2ex] &= 9 \;m \times 9 \;m\\[2ex] &= 81 \;m^2 \end{align}\)
Answer:
- \(9 \;m\)
Answer:
- \(81 \;m^2\)
| Continue Learning | |
|---|---|
| Multiplication | Whole Numbers |
| Multiplication And Division | Decimals |
| Model Drawing Strategy | Division |
| Fractions | Factors And Multiples |
| Area And Perimeter 1 | Line Graphs |
| Time | |
Primary
Secondary
