Compound shapes and parts of shapes

1. Overview

This topic teaches you how to calculate areas, perimeters, surface areas, and volumes of compound shapes (2D shapes made of simpler shapes) and compound solids (3D objects made of simpler solids). You'll also learn to work with parts of shapes, like semicircles and sectors. These skills are essential for real-world applications and problem-solving.

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Key Definitions

• Compound Shape: A 2D shape made up of two or more basic shapes joined together.
• Compound Solid: A 3D object made up of two or more basic solids joined together.
• Perimeter: The total distance around the outside edge of a 2D shape.
• Surface Area: The total area of all the exterior faces of a 3D solid.
• Cross-section: The surface or shape that is or would be exposed by making a straight cut through a solid.

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Core Content

A. Compound 2D Shapes (Perimeter and Area)

To solve problems involving compound shapes, follow these steps:

1. Split the shape into basic shapes (rectangles, triangles, or circles).
2. Find missing lengths using the given dimensions.
3. Calculate the individual areas or perimeters.
4. Add or Subtract the parts to find the total.

Worked example 1 — Area and Perimeter of an L-Shape

Question: Calculate the perimeter and area of the following L-shaped figure. The base is 10cm, the total height on the left is 8cm. The top horizontal edge is 4cm and the right vertical edge is 3cm.

Step 1: Find missing lengths.

• Bottom horizontal = 10 cm. Top horizontal = 4 cm. Therefore, the middle horizontal is $10 - 4 = 6\text{ cm}$.
  • $10 \text{ cm} - 4 \text{ cm} = 6 \text{ cm}$ (Subtract the top length from the base)
• Left vertical = 8 cm. Right vertical = 3 cm. Therefore, the inner vertical is $8 - 3 = 5\text{ cm}$.
  • $8 \text{ cm} - 3 \text{ cm} = 5 \text{ cm}$ (Subtract the right height from the left height)

Step 2: Calculate Perimeter.

• Sum of all outer edges: $10 + 8 + 4 + 5 + 6 + 3 = 36\text{ cm}$.
  • $10 \text{ cm} + 8 \text{ cm} + 4 \text{ cm} + 5 \text{ cm} + 6 \text{ cm} + 3 \text{ cm} = 36 \text{ cm}$ (Add all the outside lengths)

Answer: The perimeter is $\boxed{36 \text{ cm}}$.

Step 3: Calculate Area.

• Split into two rectangles:
  • Rectangle A (left): $8\text{ cm} \times 4\text{ cm} = 32\text{ cm}^2$
    • $A = l \times w = 8 \text{ cm} \times 4 \text{ cm} = 32 \text{ cm}^2$ (Area of a rectangle)
  • Rectangle B (bottom right): $6\text{ cm} \times 3\text{ cm} = 18\text{ cm}^2$
    • $A = l \times w = 6 \text{ cm} \times 3 \text{ cm} = 18 \text{ cm}^2$ (Area of a rectangle)
• Total Area = $32 + 18 = 50\text{ cm}^2$.
  • $32 \text{ cm}^2 + 18 \text{ cm}^2 = 50 \text{ cm}^2$ (Add the areas of the two rectangles)

Answer: The area is $\boxed{50 \text{ cm}^2}$.

Worked example 2 — Area of a Compound Shape with a Triangle

Question: A shape is formed by a rectangle of length 12 cm and width 5 cm, with a triangle on top of the rectangle's width. The triangle has a height of 4 cm. Calculate the total area of the shape.

Step 1: Calculate the area of the rectangle.

• $A = l \times w = 12 \text{ cm} \times 5 \text{ cm} = 60 \text{ cm}^2$ (Area of a rectangle)

Answer: The area of the rectangle is $60 \text{ cm}^2$.

Step 2: Calculate the area of the triangle.

• $A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 5 \text{ cm} \times 4 \text{ cm} = 10 \text{ cm}^2$ (Area of a triangle)

Answer: The area of the triangle is $10 \text{ cm}^2$.

Step 3: Calculate the total area.

• $60 \text{ cm}^2 + 10 \text{ cm}^2 = 70 \text{ cm}^2$ (Add the areas of the rectangle and triangle)

Answer: The total area is $\boxed{70 \text{ cm}^2}$.

B. Parts of Shapes (Semicircles and Sectors)

Sometimes a shape includes a "part" of a circle.

Semicircle Area: $A = \frac{1}{2} \pi r^2$ (Must be memorised)

Semicircle Perimeter (Arc only): $L = \pi r$ (Must be memorised)

Worked Example: The "Tombstone" Shape

Question: Calculate the area of the shape shown, which consists of a rectangle with a width of 6cm and height of 10cm, with a semicircle sitting on top of the 6cm width.

Step 1: Calculate the area of the rectangular base.

• $\text{Area} = \text{base} \times \text{height} = 6 \times 10 = 60\text{ cm}^2$.
  • $A = l \times w = 6 \text{ cm} \times 10 \text{ cm} = 60 \text{ cm}^2$ (Area of a rectangle)

Answer: The area of the rectangle is $60 \text{ cm}^2$.

Step 2: Calculate the area of the semicircular top.

• Radius ($r$) = $\text{width} \div 2 = 6 \div 2 = 3\text{ cm}$.
  • $r = \frac{d}{2} = \frac{6 \text{ cm}}{2} = 3 \text{ cm}$ (Radius is half the diameter)
• $\text{Area} = \frac{1}{2} \times \pi \times 3^2 = \frac{1}{2} \times \pi \times 9 \approx 14.14\text{ cm}^2$.
  • $A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \pi \times (3 \text{ cm})^2 = \frac{1}{2} \times \pi \times 9 \text{ cm}^2 \approx 14.14 \text{ cm}^2$ (Area of a semicircle)

Answer: The area of the semicircle is approximately $14.14 \text{ cm}^2$.

Step 3: Total Area.

• $60 + 14.14 = 74.14\text{ cm}^2$ (to 2 decimal places).
  • $60 \text{ cm}^2 + 14.14 \text{ cm}^2 = 74.14 \text{ cm}^2$ (Add the areas of the rectangle and semicircle)

Answer: The total area is $\boxed{74.14 \text{ cm}^2}$ (to 2 decimal places).

C. Compound Solids (Volume and Surface Area)

• Volume: Simply add the volumes of the individual parts.
• Surface Area: Be careful! When two solids are joined, the faces that touch are no longer on the "outside." You must subtract these hidden areas from the total.

Worked Example: Volume of a Compound Solid

Question: A solid is formed by a cylinder with a radius of 3cm and height of 5cm, with a hemisphere of the same radius sitting on top of the cylinder. Calculate the total volume of the solid.

Step 1: Calculate Volume of Cylinder.

• $V = \pi r^2 h = \pi \times 3^2 \times 5 = 45\pi \approx 141.37\text{ cm}^3$.
  • $V = \pi r^2 h = \pi \times (3 \text{ cm})^2 \times 5 \text{ cm} = \pi \times 9 \text{ cm}^2 \times 5 \text{ cm} = 45\pi \text{ cm}^3 \approx 141.37 \text{ cm}^3$ (Volume of a cylinder)

Answer: The volume of the cylinder is approximately $141.37 \text{ cm}^3$.

Step 2: Calculate Volume of Hemisphere.

• $V = \frac{1}{2} \times (\frac{4}{3} \pi r^3) = \frac{2}{3} \pi \times 3^3 = \frac{2}{3} \pi \times 27 = 18\pi \approx 56.55\text{ cm}^3$.
  • $V = \frac{1}{2} \times (\frac{4}{3} \pi r^3) = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi \times (3 \text{ cm})^3 = \frac{2}{3} \pi \times 27 \text{ cm}^3 = 18\pi \text{ cm}^3 \approx 56.55 \text{ cm}^3$ (Volume of a hemisphere)

Answer: The volume of the hemisphere is approximately $56.55 \text{ cm}^3$.

Step 3: Total Volume.

• $141.37 + 56.55 = 197.92\text{ cm}^3$.
  • $141.37 \text{ cm}^3 + 56.55 \text{ cm}^3 = 197.92 \text{ cm}^3$ (Add the volumes of the cylinder and hemisphere)

Answer: The total volume is $\boxed{197.92 \text{ cm}^3}$.

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Extended Content (Extended curriculum only)

While the core concepts of compound shapes and solids are the same for both Core and Extended curriculum students, Extended students will encounter more complex problems. These often involve:

• More complex shapes: Combinations of multiple shapes, requiring more steps to decompose and calculate.
• Frustums: A frustum is the solid formed when the top part of a cone or pyramid is cut off by a plane parallel to its base. Calculating the volume and surface area of frustums requires understanding similar shapes and proportions.
• Pythagoras' Theorem: Finding missing lengths within compound shapes or solids often requires using Pythagoras' Theorem ($a^2 + b^2 = c^2$). This is especially true when dealing with triangles within the shapes.

Worked example 3 — Volume of a Frustum

Question: A frustum is made by cutting off the top of a cone. The original cone had a height of 12 cm and a base radius of 5 cm. The smaller cone that was removed had a height of 4 cm. Calculate the volume of the frustum.

Step 1: Calculate the volume of the original cone.

• $V_{original} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times (5 \text{ cm})^2 \times 12 \text{ cm} = 100\pi \text{ cm}^3 \approx 314.16 \text{ cm}^3$ (Volume of a cone)

Answer: The volume of the original cone is approximately $314.16 \text{ cm}^3$.

Step 2: Find the radius of the smaller cone.

• Since the smaller cone is similar to the larger cone, the ratio of their radii is equal to the ratio of their heights.
  • $\frac{r_{small}}{r_{large}} = \frac{h_{small}}{h_{large}}$
  • $\frac{r_{small}}{5 \text{ cm}} = \frac{4 \text{ cm}}{12 \text{ cm}}$
  • $r_{small} = \frac{4 \text{ cm}}{12 \text{ cm}} \times 5 \text{ cm} = \frac{5}{3} \text{ cm} \approx 1.67 \text{ cm}$ (Multiply both sides by 5 cm)

Answer: The radius of the smaller cone is approximately $1.67 \text{ cm}$.

Step 3: Calculate the volume of the smaller cone.

• $V_{small} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times (\frac{5}{3} \text{ cm})^2 \times 4 \text{ cm} = \frac{100}{27} \pi \text{ cm}^3 \approx 11.64 \text{ cm}^3$ (Volume of a cone)

Answer: The volume of the smaller cone is approximately $11.64 \text{ cm}^3$.

Step 4: Calculate the volume of the frustum.

• $V_{frustum} = V_{original} - V_{small} = 314.16 \text{ cm}^3 - 11.64 \text{ cm}^3 = 302.52 \text{ cm}^3$ (Subtract the volume of the smaller cone from the volume of the original cone)

Answer: The volume of the frustum is $\boxed{302.52 \text{ cm}^3}$.

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Key Equations

Shape Part	Formula	Notes
Semicircle Area	$A = \frac{1}{2} \pi r^2$	$r$ = radius
Semicircle Arc	$L = \pi r$	Length of the curved part only
Sector Area	$A = \frac{\theta}{360} \times \pi r^2$	$\theta$ = angle of the sector
Compound Volume	$V_{total} = V_1 + V_2$	Sum of all parts
Compound Surface Area	$SA_{total} = \sum(\text{visible faces})$	Exclude internal touching faces

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Common Mistakes to Avoid

• ❌ Wrong: Including internal lines when calculating perimeter.
  • ✓ Right: Only sum the lines that form the outer boundary of the shape.
• ❌ Wrong: Using the diameter instead of the radius in circle formulas.
  • ✓ Right: Always check if the given line goes across the whole circle ($d$) or halfway ($r$). If given the diameter, remember to divide by 2 to find the radius: $r = \frac{d}{2}$.
• ❌ Wrong: Forgetting to divide the volume of a sphere by 2 when dealing with a hemisphere.
  • ✓ Right: Write "Hemisphere = $\frac{1}{2}$ Sphere" at the start of your working.
• ❌ Wrong: Assuming that all faces of joined solids contribute to the total surface area.
  • ✓ Right: Identify and subtract the area of any faces that are joined together and therefore hidden from the exterior.

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Exam Tips

• Show Your Split: When calculating the area of a compound shape, draw a dotted line on the diagram to show how you divided it. This helps the examiner award method marks even if your final answer is wrong.
• Calculator Tip: Use the $\pi$ button on your calculator rather than $3.14$ for better accuracy, unless the question tells you otherwise.
• Command Words:
  • "Calculate": Perform the math and give a numerical answer.
  • "Show that": You are given the answer; you must show every step of the working to reach it.
• Units: Always check your units. If dimensions are in $cm$, area is $cm^2$ and volume is $cm^3$. If lengths are mixed (e.g., $m$ and $cm$), convert them all to the same unit before calculating.
• Formulae: Formulas for the area of a circle and volume of a prism must be memorised. Complicated formulas (like the volume of a sphere or cone) are usually provided on the IGCSE formula sheet, but check your specific exam series to be sure.