IGCSE · Practice questions · Compound shapes and parts of shapes

Split, add, subtract.

Six original Cambridge-style questions on compound shapes and solids: perimeters, shaded areas, volumes and surface areas, and an Extended frustum problem.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Split into known shapes, then add or subtract.

Sketch the pieces and label every length you know.
Area and volume: add the pieces, subtract the holes.
Perimeter: outside edges only, never the joins.
Surface area: outside faces only; joined faces disappear.
Frustum (Extended): big cone minus small cone.
Accuracy: keep \(\pi\) and unrounded values until the end.

Calculation

[2 marks]

A rectangle 12 cm by 9 cm has a rectangle 5 cm by 4 cm cut from one corner. Find (a) the area (b) the perimeter of the shape that is left.

(a) \(12 \times 9 - 5 \times 4 = 108 - 20 = 88\text{ cm}^2\) ✓
(b) The cut corner does not change the perimeter: \(2(12 + 9) = 42\) cm ✓

Calculation

[2 marks]

A metal washer is a ring. The outer diameter is 20 mm and the hole has radius 6 mm. Calculate the area of metal.

Outer radius \(= 10\) mm. ✓
Area \(= \pi \times 10^2 - \pi \times 6^2 = 64\pi = 201.0... = 201\text{ mm}^2\) ✓

Calculation

[3 marks]

A shape is a rectangle 10 cm long and 4 cm wide with a semicircle on each of its two 4 cm ends.

Calculate (a) its area (b) its perimeter.

Two semicircles of radius 2 make one full circle. ✓
(a) \(10 \times 4 + \pi \times 2^2 = 40 + 4\pi = 52.56... = 52.6\text{ cm}^2\) ✓
(b) Two straight sides \(2 \times 10 = 20\), plus one full circumference \(4\pi\): \(20 + 4\pi = 32.6\) cm ✓

Calculation

[2 marks]

A solid is a cylinder of radius 5 cm and height 8 cm with a cone of radius 5 cm and vertical height 6 cm on top. Calculate its volume.

Cylinder \(= \pi \times 25 \times 8 = 200\pi\); cone \(= \tfrac{1}{3}\pi \times 25 \times 6 = 50\pi\). ✓
Total \(= 250\pi = 785.3... = 785\text{ cm}^3\) ✓

Analysis

[4 marks]

Calculate the total surface area of the solid in the previous question. The base of the cylinder is flat.

Cone slant height \(= \sqrt{5^2 + 6^2} = \sqrt{61} = 7.810...\); cone curved \(= \pi \times 5 \times 7.810... = 122.6...\) ✓
Cylinder curved \(= 2\pi \times 5 \times 8 = 80\pi = 251.3...\) ✓
Base \(= 25\pi = 78.53...\) (the top of the cylinder is covered by the cone). ✓
Total \(= 452.5... = 453\text{ cm}^2\) ✓

Analysis

[5 marks]

(Extended) A drinking glass is a frustum. Its top has radius 4 cm, its base has radius 3 cm and its height is 6 cm.

The frustum is a cone of height 24 cm and radius 4 cm with a cone of radius 3 cm removed from the bottom.

(a) Show that the volume of the glass is \(74\pi\text{ cm}^3\).

(b) How many glasses can be filled completely from 2 litres of juice?

Small cone height \(= 24 - 6 = 18\) cm (check: \(\tfrac{3}{4}\) of 24). ✓
(a) \(\tfrac{1}{3}\pi \times 16 \times 24 - \tfrac{1}{3}\pi \times 9 \times 18 = 128\pi - 54\pi = 74\pi\text{ cm}^3\) ✓
\(74\pi = 232.4...\text{ cm}^3\), and 2 litres \(= 2000\text{ cm}^3\). ✓
(b) \(2000 \div 232.4... = 8.60...\), so 8 glasses can be filled completely. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.