Home / IGCSE Maths 0580 / Mensuration / Compound shapes and parts of shapes
Cambridge IGCSE · Topic 5.5
IGCSE Maths 0580 / Topic 5 / IGCSE

Break it into pieces.

Exam shapes are rarely a single rectangle or circle. Split them into shapes you know, then add the pieces, or subtract the parts that have been cut away.

Full-screen labArea and perimeter lab → Topic 5All mensuration lessons →
The key idea

Area and volume: add the pieces, subtract the holes. Perimeter and surface area: include only the OUTSIDE edges and faces. A line or face where two pieces join is inside the shape, so it does not count.

8 cm 6 cm semicircle, r = 4 Area 48 + 8π = 73.1 cm² Perimeter 6 + 8 + 6 + 4π = 32.6 cm
Fig. 1: a window made of an 8 cm by 6 cm rectangle and a semicircle of diameter 8 cm (drawn to scale); the dashed join is not part of the perimeter
Section 01

Adding the pieces.

In the lab, build a compound shape and watch which edges the perimeter follows: only the outside ones. The area is the sum of the pieces, and a hole is subtracted.

Worked example An L-shape
A 10 cm by 8 cm rectangle has a 4 cm by 5 cm rectangle cut from one corner, leaving an L-shape. Find its area and perimeter.
  1. Area: whole rectangle minus the cut-out \(= 10 \times 8 - 4 \times 5 = 80 - 20 = 60\).
  2. Perimeter: the two inner edges (4 cm and 5 cm) replace the two missing pieces of the outer edges exactly.
  3. So the perimeter is the same as the rectangle's: \(2(10 + 8) = 36\).
Answer: Area \(60\text{ cm}^2\), perimeter 36 cm
Worked example Rectangle plus semicircle
Find the area and perimeter of the window in Fig. 1.
  1. Rectangle \(= 8 \times 6 = 48\).
  2. Semicircle, radius 4: \(\tfrac{1}{2}\pi \times 4^2 = 8\pi\).
  3. Area \(= 48 + 8\pi = 73.13...\)
  4. Perimeter: three sides of the rectangle \(6 + 8 + 6 = 20\), plus the curved edge \(\tfrac{1}{2} \times 2\pi \times 4 = 4\pi\).
  5. Perimeter \(= 20 + 4\pi = 32.56...\)
Answer: Area \(73.1\text{ cm}^2\), perimeter 32.6 cm
Examiner trap

Do not include the join. In Fig. 1 the 8 cm line under the semicircle is inside the window, so adding it (40.6 cm) is wrong.

Section 02

Parts of shapes and shaded regions.

For a shaded region, ask: what is the whole shape, and what has been taken away? Write the area as whole minus the unshaded part.

Worked example Circle in a square
A circle of radius 5 cm fits exactly inside a square of side 10 cm. Find the area of the square that is outside the circle.
  1. Square \(= 10^2 = 100\).
  2. Circle \(= \pi \times 5^2 = 25\pi = 78.53...\)
  3. Shaded \(= 100 - 25\pi = 21.46...\)
Answer: \(21.5\text{ cm}^2\)
Worked example A ring
A ring is the region between two circles with the same centre, of radii 7 cm and 4 cm. Find its area in terms of \(\pi\).
  1. Big circle \(= 49\pi\), small circle \(= 16\pi\).
  2. Ring \(= 49\pi - 16\pi = 33\pi\).
Answer: \(33\pi\text{ cm}^2\) (103.7 cm2)
Worked example A segment
A quarter circle has radius 6 cm. A straight line joins the ends of its arc. Find the area between the line and the arc.
  1. Quarter circle \(= \tfrac{1}{4}\pi \times 36 = 9\pi\).
  2. The line cuts off a right-angled triangle with legs 6 and 6: area \(= \tfrac{1}{2} \times 6 \times 6 = 18\).
  3. Segment \(= 9\pi - 18 = 10.27...\)
Answer: \(10.3\text{ cm}^2\)
Your turn
A circular pond of radius 3 m has a path 1 m wide all the way round it. Find the area of the path in terms of \(\pi\).
Examiner trap

For a ring, subtract the AREAS, not the radii: \(\pi(7 - 4)^2 = 9\pi\) is wrong. Square each radius first: \(\pi(7^2 - 4^2) = 33\pi\).

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

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A shaded region is a square with a circle cut out of it. How do you find its area?

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Find the area of a ring between circles of radii 5 cm and 3 cm, in terms of \(\pi\).

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A cylinder has a cone joined to its top. Which surfaces are in the total surface area?

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A toy is a cone of volume \(30\pi\text{ cm}^3\) on a hemisphere of volume \(18\pi\text{ cm}^3\). What is its volume?

Section 03

Compound solids.

In the lab, join the solids and watch the surface area: the faces that meet disappear inside the solid, so they are left out. Volumes just add.

Worked example Cone on a hemisphere
A toy is a cone of radius 3 cm and vertical height 10 cm on top of a hemisphere of radius 3 cm. Find its volume.
  1. Cone \(= \tfrac{1}{3}\pi \times 9 \times 10 = 30\pi\).
  2. Hemisphere \(= \tfrac{1}{2} \times \tfrac{4}{3}\pi \times 27 = 18\pi\).
  3. Total \(= 48\pi = 150.7...\)
Answer: \(151\text{ cm}^3\)
Worked example Surface area of a capsule shape
A solid is a cylinder of radius 4 cm and height 10 cm with a hemisphere of radius 4 cm on top. The bottom is flat. Find its total surface area.
  1. Cylinder curved surface \(= 2\pi \times 4 \times 10 = 80\pi\).
  2. Flat base \(= \pi \times 4^2 = 16\pi\). (There is no top circle: the hemisphere covers it.)
  3. Hemisphere curved surface \(= \tfrac{1}{2} \times 4\pi \times 16 = 32\pi\).
  4. Total \(= 80\pi + 16\pi + 32\pi = 128\pi = 402.1...\)
Answer: \(402\text{ cm}^2\)
Examiner trap

For surface area, leave out every face where two parts are joined. Counting the circle between the cylinder and the hemisphere adds an extra \(16\pi\).

Section 04

Parts of solids: the frustum.

(Extended) A frustum is a cone with its top cut off parallel to the base. Treat it as a big cone minus the small cone that was removed.

Worked example Volume of a frustum (Extended)
A cone of radius 6 cm and height 12 cm has a small cone of radius 2 cm and height 4 cm cut from the top. Find the volume of the frustum that is left.
  1. Big cone \(= \tfrac{1}{3}\pi \times 36 \times 12 = 144\pi\).
  2. Small cone \(= \tfrac{1}{3}\pi \times 4 \times 4 = \tfrac{16\pi}{3}\).
  3. Frustum \(= 144\pi - \tfrac{16\pi}{3} = \tfrac{416\pi}{3} = 435.6...\)
Answer: \(436\text{ cm}^3\)
Worked example Curved surface of the frustum (Extended)
Find the curved surface area of the same frustum.
  1. Slant heights: big cone \(\sqrt{6^2 + 12^2} = \sqrt{180} = 13.41...\); small cone \(\sqrt{2^2 + 4^2} = \sqrt{20} = 4.472...\)
  2. Curved surfaces: big \(\pi \times 6 \times 13.41... = 252.8...\); small \(\pi \times 2 \times 4.472... = 28.09...\)
  3. Frustum \(= 252.8... - 28.09... = 224.7...\)
Answer: \(225\text{ cm}^2\)
Examiner trap

You cannot use the frustum's own height in a single cone formula. Always work with the two complete cones and subtract.

Stage 2 · Exam

Exam-style questions

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A window is a rectangle 10 cm wide and 7 cm tall with a semicircle of diameter 10 cm on top. Find the perimeter of the window.

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A circle of radius 6 cm fits exactly inside a square of side 12 cm. Find the area inside the square but outside the circle.

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A solid toy is a cone of radius 3 cm and vertical height 4 cm on top of a hemisphere of radius 3 cm. Find its total surface area.

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(Extended) A cone of radius 9 cm and height 15 cm has a cone of radius 3 cm and height 5 cm removed from its top. Find the volume of the frustum.

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Go deeper · practice
Six original Cambridge-style questions
An L-shape, a ring, a running-track shape, a cone on a cylinder and the frustum of a glass.