GCSE Maths

Area and Perimeter

Edexcel
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Introduction Area and Perimeter of Different Shapes Real life Examples of Area and Perimeter Solved Examples

Introduction

Arrow pointing to the shaded area of a purple rectangle representing the concept of area
  • The amount of space inside of any 2-Dimensional shape is Area.
  • Calculation of area for every shape is done in different way.
  • The Area is measured in square units.
Arrow pointing to the border of a rectangle representing the perimeter
  • The sum of sides of any shape it is made up of is known as Perimeter (the distance around a shape)
  • Calculation of Perimeter for almost every shape is common.
  • The Perimeter is measured in linear units.

Area and Perimeter of Different Shapes

SQUARE:

  • Area: $a \times a = (\text{side})^2$
  • Perimeter: $a + a + a + a = 4a$ ($4 \times \text{sides}$)
Diagram of a square with both sides labeled 'a' to illustrate.

RECTANGLE:

  • Area: $l \times b$
  • Perimeter: $2 \times (l + b)$
Blue rectangle labeled with length L and breadth B.

CIRCLE:

  • Area: $\pi \times \text{radius}^2$
  • Perimeter: $2 \times \pi \times \text{radius}$
Green circle with radius 'r' marked to demonstrate formulas.

TRIANGLE:

  • Area: $\frac{1}{2} \times \text{base} \times \text{height}$
  • Perimeter: sum of all three sides
Pink triangle labeled with base 'b' and height 'h'.

TRAPEZIUM:

  • Area: $\frac{1}{2} \times (a + b) \times h$
  • Perimeter: $a + b + c + d$
Teal trapezium with sides a, b, c, d and height.

Real life Examples of Area and Perimeter

Illustration showing football field, pizza, house, and a thinking child to explain area and perimeter concepts

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Solved Examples

Solved Example
Problem: Find the area and perimeter of circles with radius equal to: (i) $r = 5\text{ cm}$ (ii) $r = 7\text{ cm}$ (Use $\pi \approx 3.14$ or $\frac{22}{7}$)
Two coloured circles with radii labeled 5cm and 7cm to compare area and perimeter
SOLUTION

(i) The Area and Perimeter of a circle with radius 5cm:

$$ \begin{aligned} \text{Area} &= \pi \times \text{radius}^2 \\ &= \pi \times (5)^2 \\ &= (3.14 \times 25) \\ &= 78.5\text{ cm}^2 \end{aligned} $$
$$ \begin{aligned} \text{Perimeter} &= 2 \times \pi \times \text{radius} \\ &= (2 \times 3.14 \times 5) \\ &= 31.4\text{ cm} \end{aligned} $$

(ii) The Area and Perimeter of a circle with radius 7cm:

$$ \begin{aligned} \text{Area} &= \pi \times \text{radius}^2 \\ &= \frac{22}{7} \times (7)^2 \\ &= \left(\frac{22}{7} \times 49\right) \\ &= 154\text{ cm}^2 \end{aligned} $$
$$ \begin{aligned} \text{Perimeter} &= 2 \times \pi \times \text{radius} \\ &= \left(2 \times \frac{22}{7} \times 7\right) \\ &= 44\text{ cm} \end{aligned} $$

Final Answer: (i) Area $\approx 78.5\text{ cm}^2$, Perimeter $\approx 31.4\text{ cm}$ | (ii) Area $\approx 154\text{ cm}^2$, Perimeter $\approx 44\text{ cm}$

Solved Example
Problem: If a rectangular field is of following dimensions find its area and length of required fence for the field: Length = 20cm, Breadth = 15cm
Farmers working in a crop field with fences on the side to represent area and perimeter visually
SOLUTION

To find the length of required fence for the field we will find out the perimeter:

$$ \begin{aligned} \text{Area} &= l \times b \\ &= (20 \times 15) \\ &= 300\text{ cm}^2 \end{aligned} $$
$$ \begin{aligned} \text{Perimeter} &= 2 \times (l + b) \\ &= 2 \times (20 + 15) \\ &= (2 \times 35) \\ &= 70\text{ cm} \end{aligned} $$

Final Answer: Area = $300\text{ cm}^2$ and Perimeter = $70\text{ cm}$

Solved Example
Problem: Find the Area and Perimeter of the following 2 Dimensional shapes
Labeled diagram showing a trapezium (parallel sides 3cm and 5cm, vertical height 4cm), square (side 4cm), and a triangle (base 4cm, height 4cm)
SOLUTION

(a) The Area and Perimeter of Trapezium (assuming parallel sides are 3 cm and 5 cm, and vertical height is 4 cm):

$$ \begin{aligned} \text{Area} &= \frac{1}{2} \times (\text{sum of parallel sides}) \times h \\ &= \frac{1}{2} \times (3 + 5) \times 4 \\ &= \left(\frac{1}{2} \times 8 \times 4\right) \\ &= 16\text{ cm}^2 \end{aligned} $$

Perimeter of Trapezium (assuming an isosceles trapezium based on the height):

$$ \begin{aligned} \text{Perimeter} &= \text{Sum of all sides} \\ &= (3 + 5 + 4.12 + 4.12) \\ &\approx 16.24\text{ cm} \end{aligned} $$

(b) Area and Perimeter of a square having a side equal to 4cm:

$$ \begin{aligned} \text{Area} &= \text{side} \times \text{side} \\ &= (4 \times 4) \\ &= 16\text{ cm}^2 \end{aligned} $$
$$ \begin{aligned} \text{Perimeter} &= 4 \times \text{side} \\ &= (4 \times 4) \\ &= 16\text{ cm} \end{aligned} $$

(c) Area and Perimeter of a Triangle with base equal to 4cm and height equal to 4cm (assuming it is an isosceles triangle):

$$ \begin{aligned} \text{Area} &= \frac{1}{2} \times \text{base} \times \text{height} \\ &= \left(\frac{1}{2} \times 4 \times 4\right) \\ &= 8\text{ cm}^2 \end{aligned} $$
$$ \begin{aligned} \text{Perimeter} &= \text{Sum of all sides} \\ &= (4 + 4.47 + 4.47) \\ &\approx 12.94\text{ cm} \end{aligned} $$

Final Answer: (a) Area = $16\text{ cm}^2$ & Perimeter $\approx 16.24\text{ cm}$ | (b) Area = $16\text{ cm}^2$ & Perimeter = $16\text{ cm}$ | (c) Area = $8\text{ cm}^2$ & Perimeter $\approx 12.94\text{ cm}$

Solved Example
Problem: Work out the area of the following shapes
Diagram of two compound shapes: (a) A square of side 6cm with two semicircles attached to opposite sides, (b) A compound shape made of two rectangles (5x2 cm and 7x2 cm) and a square (3x3 cm)
SOLUTION

(a) The shape is a combination of a central square and two semi-circles on either side. If the square has a side of 5 cm, the attached semi-circles have a diameter of 5 cm and a radius ($r$) of 2.5 cm:

$$ \begin{aligned} \text{Area of Square} &= 5 \times 5 \\ &= 25\text{ cm}^2 \\[10pt] \text{Area of 2 semi-circles} &= \pi \times r^2 \\ &\approx 3.14 \times (2.5)^2 \\ &= (3.14 \times 6.25) \\ &= 19.625\text{ cm}^2 \\[10pt] \text{Total Area} &= (25 + 19.625) \\ &= 44.625\text{ cm}^2 \end{aligned} $$

(b) The shape is a combination of two rectangles and one square:

$$ \begin{aligned} \text{Area of Rectangle 1} &= (5 \times 2) \\ &= 10\text{ cm}^2 \\[10pt] \text{Area of Rectangle 2} &= (7 \times 2) \\ &= 14\text{ cm}^2 \\[10pt] \text{Area of Square} &= (3 \times 3) \\ &= 9\text{ cm}^2 \\[10pt] \text{Total Area} &= (10 + 14 + 9) \\ &= 33\text{ cm}^2 \end{aligned} $$

Final Answer: (a) Area $\approx 44.625\text{ cm}^2$ and (b) Area = $33\text{ cm}^2$

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