Find the area of a semicircle of radius 5 cm. Use π = 3.14. [2 marks]

Find the volume of a cone of radius 6 cm and height 7 cm. Use π = 3.14. [3 marks]

ScotlandApplications of MathematicsSyllabus dot point

How do you find the area of a composite shape including part of a circle, and the volume of a composite solid built from prisms, cylinders, cones, spheres or pyramids?

Solving a problem involving the area of a composite shape including part of a circle, and the volume of a composite solid made from standard solids such as cuboids, cylinders, cones, spheres and pyramids.

A focused answer to the SQA National 5 Applications of Mathematics geometry content on composite shapes, covering finding the area of a composite shape including part of a circle, and the volume of a composite solid built from cuboids, cylinders, cones, spheres and pyramids.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-15

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Composite area including part of a circle
Composite volume
Examples in context
Try this

What this dot point is asking

The SQA wants you to solve problems involving the area of a composite shape, including part of a circle, and the volume of a composite solid built from standard solids such as cuboids, cylinders, cones, spheres and pyramids, by splitting the figure into familiar parts.

Composite area including part of a circle

A composite shape is split into rectangles, triangles and parts of circles. Find each area, then add them, or subtract a piece that has been removed. The first step is always to look at the figure and decide which simple shapes it is built from, marking the dimensions of each before calculating, because a missing or wrongly read length is the most common reason a composite-area answer goes wrong.

Find a composite area

A shape is a rectangle $10$ cm by $4$ cm with a quarter circle of radius $4$ cm on one corner.

Step 1: Identify the component shapes

Look at the figure and decide which standard shapes it consists of. Here the shape is a rectangle with a quarter circle attached. Label each part with its dimensions before calculating, so that no measurement is used for the wrong component.

Step 2: Calculate the rectangle area

The rectangle area formula is length times width:

10 \times 4 = 40 \text{ cm}^2

Step 3: Calculate the quarter circle area

A quarter circle is one quarter of a full circle. Use the radius (not the diameter) in the area formula:

\tfrac{1}{4} \times 3.14 \times 4^2 = \tfrac{1}{4} \times 50.24 = 12.56 \text{ cm}^2

Step 4: Add the parts

Because the quarter circle is attached to the rectangle (not cut out), the total area is the sum of the two parts:

\text{Total area} = 40 + 12.56 = 52.56 \text{ cm}^2

Final answer: Total area $= 52.56$ cm $^2$ .

Composite volume

A composite solid is split into standard solids. Calculate each volume with its formula, then add them, taking care to use the radius and the right height for each part. Sometimes a volume is subtracted instead of added, for example a pipe is a large cylinder with a smaller cylinder removed from the middle, so its volume is the outer cylinder minus the inner one. Read the figure to decide whether each part is added or taken away.

Find a composite volume

A solid is a cuboid $6 \times 4 \times 10$ cm topped by a square-based pyramid with the same $6 \times 4$ base and a height of $9$ cm.

Step 1: Calculate the cuboid volume

The cuboid volume formula is length times width times height. All three dimensions are given directly:

V_{\text{cuboid}} = 6 \times 4 \times 10 = 240 \text{ cm}^3

Step 2: Calculate the pyramid volume

A pyramid's volume includes the factor of $\tfrac{1}{3}$ , which is the most common source of errors in composite volume questions. The base area here is $6 \times 4 = 24$ cm $^2$ :

V_{\text{pyramid}} = \tfrac{1}{3} \times 24 \times 9 = 72 \text{ cm}^3

Step 3: Add the two volumes

Because the pyramid sits on top of the cuboid (it is not cut out), the volumes are added:

\text{Total volume} = 240 + 72 = 312 \text{ cm}^3

Final answer: Total volume $= 312$ cm $^3$ .

Examples in context

These problems model real objects: a sports pitch with rounded ends, a window with a curved top, a grain silo (cylinder plus cone), or a fuel tank (cylinder with hemispherical ends). Each rests on splitting the figure into standard parts, applying the right area or volume formula, and combining them, the skills here. The radius trap and the one-third factor on cones and pyramids are where marks are most often lost.

A hemisphere is half a sphere, so its volume is $\tfrac{1}{2} \times \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi r^3$ ; this appears often in composite solids such as a dome on top of a cylinder. When a question asks for capacity in litres, find the volume in cm $^3$ first and then convert, remembering that $1000$ cm $^3$ equals $1$ litre.

Try this

Q1. Find the area of a semicircle of radius $5$ cm. Use $\pi = 3.14$ . [2 marks]

Cue. $\tfrac{1}{2} \times 3.14 \times 5^2 = 39.25$ cm $^2$ .

Q2. Find the volume of a cylinder of radius $3$ cm and height $10$ cm. Use $\pi = 3.14$ . [2 marks]

Cue. $3.14 \times 3^2 \times 10 = 282.6$ cm $^3$ .

Q3. Find the volume of a cone of radius $6$ cm and height $7$ cm. Use $\pi = 3.14$ . [3 marks]

Cue. $\tfrac{1}{3} \times 3.14 \times 6^2 \times 7 = 263.76$ cm $^3$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 Apps style4 marksA shape is a rectangle

8

cm by

6

cm with a semicircle of diameter

6

cm on one short end. Calculate its total area, to

1

decimal place. Use

\pi = 3.14

Show worked answer →

Split the shape into a rectangle and a semicircle. Rectangle area: $8 \times 6 = 48$ cm $^2$ (1 mark). The semicircle has diameter $6$ cm, so radius $3$ cm; a full circle is $\pi r^2 = 3.14 \times 3^2 = 28.26$ cm $^2$ , so the semicircle is half: $28.26 \div 2 = 14.13$ cm $^2$ (2 marks). Add the parts: $48 + 14.13 = 62.13 \approx 62.1$ cm $^2$ (1 mark). Markers reward the rectangle area, the semicircle area using the correct radius, and the rounded total.

SQA N5 Apps style4 marksA solid is a cylinder of radius

5

cm and height

12

cm with a cone of the same radius and height

6

cm on top. Calculate the total volume, to the nearest cm

^3

. Use

\pi = 3.14

Show worked answer →

Find each volume and add them. Cylinder: $\pi r^2 h = 3.14 \times 5^2 \times 12 = 3.14 \times 25 \times 12 = 942$ cm $^3$ (2 marks). Cone: $\tfrac{1}{3}\pi r^2 h = \tfrac{1}{3} \times 3.14 \times 25 \times 6 = 157$ cm $^3$ (1 mark). Total volume: $942 + 157 = 1099$ cm $^3$ (1 mark). Markers reward the cylinder volume, the cone volume with the one-third factor, and the total. Forgetting the one-third on the cone is the usual error.

Related dot points

Sources & how we know this

SQA National 5 Applications of Mathematics Course Specification — SQA (2023)
SQA National 5 Applications of Mathematics - Course overview and resources — SQA (2023)