 Study P4 Mathematics Maths - Area and Perimeter - Geniebook   # Area And Perimeter 1

1. Finding the maximum number of squares that can be fitted/cut from a rectangle
2. Finding unknown dimensions given area of a rectangle/square
3. 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}

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}

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}

$$15$$ 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}

$$12$$ 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}

$$8$$ 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}

$$18 \;cm$$

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:

\begin{align}​​ &= 72 \;cm^2 \div 9 \;cm\\[2ex] &= 8 \;cm \end{align}

$$8 \;cm$$

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 $$= 8 \;cm$$

$$8 \;cm$$

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 $$= 5 \;cm$$

$$5 \;cm$$

## 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}

$$13 \;cm$$

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}

$$12 \;cm$$

​\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}

$$15 \;cm$$

Question 4:

The area of a rectangular garden is $$168 \;m^2$$. Its breadth is 8 m. 1. Find the length of the garden.
2. Vincent jogged round the entire rectangular garden twice. Find the distance he jogged.

Solution:

1. Length of the rectangle\begin{align}​​\\[2ex] &= 168 \;m^2 \div 8 \;m \\[2ex] &= 21 \;m \end{align}
2. 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}

1. $$21 \;m$$

1. $$116 \;m$$

Question 5:

Elaine jogged $$36 \;m$$ round a square sand pit. 1. Find the length of one side of the sand pit.
2. Find the area of the square sand pit.

Solution:

1. Length of one side of the sand pit\begin{align}​​\\[2ex] &= 36 \;m \div 4\\[2ex] &= 9 \;m \end{align}
2. Area of the sand pit\begin{align}​​\\[2ex] &= 9 \;m \times 9 \;m\\[2ex] &= 81 \;m^2 \end{align}

1. $$9 \;m$$

1. $$81 \;m^2$$

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Factors And Multiples
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