CIE IGCSE Mathematics (0580) Compound shapes and parts of shapes Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Compound shapes and parts of shapes Study Notes
LEARNING OBJECTIVE
- Compound Shapes and Compound Solids
Key Concepts:
- Compound Shapes and Parts of Shapes
- Compound Solids and Parts of Solids
Compound Shapes and Parts of Shapes
Compound Shapes and Parts of Shapes
Compound shapes are made up of two or more basic shapes such as rectangles, triangles, semicircles, or trapeziums. To find the area or perimeter, break the compound shape into simpler parts.
Key Strategies:
- Split the shape into known basic shapes.
- Use known area formulas and add or subtract areas as needed.
- For perimeter, trace the outside of the figure and include only outer edges.
Example:
A compound shape consists of a rectangle 10 cm by 6 cm with a semicircle of radius 3 cm attached to one of its shorter sides. Find the total area of the shape.
▶️ Answer/Explanation
Area of rectangle = \( 10 \times 6 = 60 \ \text{cm}^2 \)
Area of semicircle = \( \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3)^2 = \frac{9}{2} \pi \approx 14.14 \ \text{cm}^2 \)
Total area = \( 60 + 14.14 = 74.14 \ \text{cm}^2 \)
Example:
A shape is formed by removing a semicircle of radius 2 cm from a rectangle of length 8 cm and width 4 cm. Find the area of the remaining shape.
▶️ Answer/Explanation
Area of rectangle = \( 8 \times 4 = 32 \ \text{cm}^2 \)
Area of semicircle = \( \frac{1}{2} \pi (2)^2 = \frac{1}{2} \pi \times 4 = 2\pi \approx 6.28 \ \text{cm}^2 \)
Remaining area = \( 32 – 6.28 = 25.72 \ \text{cm}^2 \)
Example:
A compound shape is made by joining a square of side 5 cm and a right-angled triangle with base 5 cm and height 3 cm. Find the perimeter of the shape, assuming the triangle is joined to one side of the square.
▶️ Answer/Explanation
Square sides = \( 5 \ \text{cm} \)
Triangle’s hypotenuse = \( \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83 \ \text{cm} \)
Perimeter = 3 sides of square + 2 sides of triangle (not including shared side):
\( 5 + 5 + 5 + 3 + 5.83 = 23.83 \ \text{cm} \)
Example :
Find the shaded area in the given figure. The shape consists of a parallelogram with a triangle removed from it.
▶️ Answer/Explanation
- Base of the parallelogram = \( 14 \, \text{cm} \)
- Height of the parallelogram = \( 6 \, \text{cm} \)
- The triangle removed has the same base \( 14 \, \text{cm} \), but a height of only \( 3 \, \text{cm} \)
Use area formulas.
- Area of parallelogram $\mathrm{= base × height}$
- Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \)
Parallelogram area = \( 14 \times 6 = 84 \, \text{cm}^2 \)
Triangle area = \( \frac{1}{2} \times 14 \times 3 = 21 \, \text{cm}^2 \)
Shaded area = \( 84 – 21 = \boxed{63 \, \text{cm}^2} \)
Compound Solids and Parts of Solids
Compound Solids and Parts of Solids
Compound solids are 3D shapes made by combining two or more basic solids (e.g. a cone on a cylinder, a cuboid on a prism). To solve problems involving surface area or volume:
- Break the shape into standard solids (cylinder, cone, cuboid, etc.).
- Use known formulas for volume and surface area of each part.
- For surface area, consider whether internal faces are hidden or exposed.
Key Surface Area Notes:
- Total surface area excludes the area of any faces that are joined together.
- Lateral/curved areas are often used when bases are not included.
Key Volume Notes:
- For compound solids, volume = sum of volumes of individual solids.
- For hollow parts, subtract the inner volume from the outer volume.
Example:
A solid consists of a cylinder of radius 4 cm and height 10 cm, with a cone of the same radius and height 6 cm on top. Find the total volume of the solid.
▶️ Answer/Explanation
Cylinder volume = \( \pi r^2 h = \pi \times 4^2 \times 10 = 160\pi \ \text{cm}^3 \approx 502.65 \ \text{cm}^3 \)
Cone volume = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 4^2 \times 6 = \frac{96\pi}{1} = 32\pi \ \text{cm}^3 \approx 100.53 \ \text{cm}^3 \)
Total volume = \( 160\pi + 32\pi = 192\pi \ \text{cm}^3 \approx 603.18 \ \text{cm}^3 \)
Example:
A metal pipe is a hollow cylinder with outer radius 5 cm, inner radius 3 cm, and height 20 cm. Find the volume of the metal used.
▶️ Answer/Explanation
Outer volume = \( \pi \times 5^2 \times 20 = 500\pi \ \text{cm}^3 \approx 1570.80 \ \text{cm}^3 \)
Inner volume = \( \pi \times 3^2 \times 20 = 180\pi \ \text{cm}^3 \approx 565.49 \ \text{cm}^3 \)
Metal volume = \( 500\pi – 180\pi = 320\pi \ \text{cm}^3 \approx 1005.31 \ \text{cm}^3 \)
Example:
A cuboid of dimensions $\mathrm{8~ cm × 5 ~cm × 4~ cm}$ is glued on top of a cube of side 5 cm. Calculate the total surface area of the new shape.
▶️ Answer/Explanation
Surface area of cube = \( 6 \times 5^2 = 150 \ \text{cm}^2 \)
Surface area of cuboid = \( 2(8 \times 5 + 5 \times 4 + 8 \times 4) = 2(40 + 20 + 32) = 2 \times 92 = 184 \ \text{cm}^2 \)
Since the cuboid is placed on top of the cube, one face (5 × 5) is hidden. So subtract one face:
Correct total surface area = \( 150 + 184 – 25 = 309 \ \text{cm}^2 \)
Example:
Find the volume of half of a sphere with radius 10 cm and its Total Surafce area.
▶️ Answer/Explanation
Volume of a full sphere = \( \frac{4}{3} \pi r^3 \)
= \( \frac{4}{3} \pi \times 10^3 = \frac{4}{3} \pi \times 1000= 1333.3\pi \ \text{cm}^3 \)
Half-sphere volume = \( \frac{1}{2} \times 1333.3\pi = 666.6\pi \ \text{cm}^3 \)
Total surface area of half of a sphere $= 3\pi r^2$
$= (3 \times 3.14 \times 10 \times 10)$ cm$^2$
$= (3 \times 314)$ cm$^2$
$= 942$ cm$^2$