In the realm of geometry, understanding how to calculate the perimeters and areas of compound shapes and parts of shapes is crucial. This topic delves into the methods for determining the lengths around and the spaces within various combined geometric figures, a skill essential for tackling both academic challenges and real-world problems. Our focus will be on breaking down these compound shapes into simpler, manageable parts to make calculations straightforward.

Introduction to Compound Shapes

Compound shapes, also known as composite shapes, are figures that consist of two or more simple geometric shapes combined. The strategy for calculating the perimeter and area of compound shapes involves dissecting them into simpler shapes whose properties we already know, such as rectangles, triangles, circles, and trapezoids.

Image courtesy of TickTockMaths

Calculating Perimeters of Compound Shapes

Example 1: Rectangle Combined with Triangle

Consider a compound shape made up of a rectangle with dimensions 6 cm (length) and 4 cm (width), and a right-angled triangle with a base of 3 cm and height of 4 cm, sharing one of its height sides with the rectangle.

Solution:

• Perimeter calculation steps:
  • Rectangle perimeter without shared side: $2 \times (6 \, \text{cm} + 4 \, \text{cm}) - 4 \, \text{cm}$
  • Triangle perimeter without the hypotenuse: $3 \, \text{cm} + 4 \, \text{cm}$
  • Total Perimeter: Sum of the two calculations above.

• Rectangle Perimeter without Shared Side:

• Triangle Perimeter without Hypotenuse:

• Total Perimeter:

Calculating Areas of Compound Shapes

Example 2: Rectangle and Semi-circle

A shape consisting of a rectangle (10 cm by 4 cm) and a semi-circle with a diameter equal to the width of the rectangle (4 cm).

Solution:

• Rectangle area: $\text{Area} = 10 \, \text{cm} \times 4 \, \text{cm} = 40 \, \text{cm}^2$
• Semi-circle area: $\text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2 \, \text{cm})^2 \approx 6.283 \, \text{cm}^2$
• Total area: $\text{Total Area} = 40 \, \text{cm}^2 + 6.283 \, \text{cm}^2 \approx 46.283 \, \text{cm}^2$

Example 3: Sector of a Circle

A sector with a central angle of 60° in a circle of radius 5 cm.

Solution:

• Circle area: $\text{Area} = \pi (5 \, \text{cm})^2 = 25\pi \, \text{cm}^2$
• Sector area: $\text{Sector Area} = \frac{60}{360} \times 25\pi \, \text{cm}^2 \approx 13.09 \, \text{cm}^2$

Practice Problems

Problem 1

Question: A compound shape consists of a rectangle with dimensions 8 cm by 5 cm and a right-angled triangle with a base of 3 cm and height of 5 cm. Calculate both the perimeter and area of the compound shape.

Solution:

1. Calculate the Area of the Rectangle:

$\text{Area}_{\text{rectangle}} = \text{length} \times \text{width} = 8 \, \text{cm} \times 5 \, \text{cm} = 40 \, \text{cm}^2$

2. Calculate the Area of the Triangle:

$\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \, \text{cm} \times 5 \, \text{cm} = 7.5 \, \text{cm}^2$

3. Total Area of the Compound Shape:

$\text{Total Area} = \text{Area}{\text{ rectangle}} + \text{Area}{\text{ triangle}} = 40 \, \text{cm}^2 + 7.5 \, \text{cm}^2 = 47.5 \, \text{cm}^2$

4. Calculate the Perimeter:

• The perimeter includes the two lengths and one width of the rectangle, and two legs and the hypotenuse of the triangle. Calculate the hypotenuse of the triangle using Pythagoras' theorem: $\sqrt{3^2 + 5^2}$.
• Total Perimeter involves adding the unique sides, considering shared sides appropriately.

Problem 2

Question: A circular flower bed has a diameter of 10 meters. A path 1 meter wide runs around the outside of the flower bed. Calculate the total area covered by the flower bed and the path.

Solution:

1. Calculate the Radius of the Flower Bed:

$\text{Radius}_{\text{flower bed}} = \frac{\text{Diameter}}{2} = \frac{10 \, \text{m}}{2} = 5 \, \text{m}$

2. Calculate the Radius of the Outer Circle (Flower Bed + Pathway):

$\text{Radius}{\text{ outer circle}} = \text{Radius}{\text{ flower bed}} + \text{ Width of Pathway } = 5 \, \text{m} + 1 \, \text{m} = 6 \, \text{m}$

3. Calculate the Area of the Flower Bed:

$\text{ Area }{\text{ flower bed }} = \pi \times (\text{Radius }{\text{ flower bed}})^2 = \pi \times (5 \, \text{m})^2$

4. Calculate the Total Area (Flower Bed and Pathway):

$\text{ Area }{\text{ total }} = \pi \times (\text{Radius } {\text{ outer circle}})^2 = \pi \times (6 \, \text{m})^2$

5. Calculate the Area of the Pathway:

$\text{Area }{\text{ pathway}} = \text{Area }{\text{ total }} - \text{ Area }_{\text{flower bed}}$