POP

C5.5 Compound Shapes and Parts of Shapes

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01 — Explanation

How do I break a compound shape into pieces I can calculate?

A compound (composite) shape or solid is made by joining or removing simple shapes — rectangles, triangles, semicircles, cylinders, cones. Split it into pieces you already have a formula for, calculate each separately, then add (joined shapes) or subtract (a hole or removed piece). Sketching each piece separately, with its own dimensions labelled, before calculating anything is the most reliable way to avoid mixing up which lengths belong to which shape.

02 — Explanation

How do I find the area of a compound 2D shape?

Worked example. A shape is a rectangle $16\text{ cm}\times10\text{ cm}$ with a semicircle of diameter $10\text{ cm}$ attached to one short end. Find the total area to 3 s.f.

A_{\text{rect}} = 16\times10 = 160\text{ cm}^2 \qquad r = 10\div2 = 5\text{ cm} \qquad A_{\text{semicircle}} = \frac{1}{2}\pi(5)^2 = 39.27\ldots\text{ cm}^2

\text{Total} = 160 + 39.27\ldots \approx 199\text{ cm}^2 \ (3\text{ s.f.})

Common mistake

including the internal line where two shapes join when calculating perimeter. Perimeter only measures the outer boundary of the finished compound shape — the straight edge where the rectangle and semicircle meet is inside the combined shape, not part of its outline, so it must be left out of any perimeter calculation entirely.

03 — Explanation

How do I find the area of a shape with a piece removed?

Worked example. A square of side $12\text{ cm}$ has a circle of radius $3\text{ cm}$ cut out of its centre. Find the remaining area to 3 s.f.

A_{\text{square}} = 12^2 = 144\text{ cm}^2 \qquad A_{\text{circle}} = \pi(3)^2 = 28.27\ldots\text{ cm}^2

\text{Remaining area} = 144 - 28.27\ldots \approx 116\text{ cm}^2 \ (3\text{ s.f.})

Exam tip

with a "hole" question, always check whether you're subtracting a radius-based shape (circle) from a shape whose given measurement is a full side or diagonal — mixing up a diameter with a radius partway through a subtraction is an easy way to lose accuracy on an otherwise correct method.

04 — Explanation

How do I find the volume of part of a solid, like a hemisphere or a compound solid?

A hemisphere is exactly half a sphere, so halve the volume formula — but the radius itself stays the same.

Worked example. Find the volume of a solid hemisphere with radius $6\text{ cm}$ , in terms of $\pi$ .

V = \frac{1}{2}\times\frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(6)^3 = \frac{2}{3}\pi(216) = 144\pi\text{ cm}^3

Common mistake

halving the radius instead of the volume. "Half a sphere" means the volume (and curved surface area) is halved — the radius is unchanged, since the hemisphere is still cut from a sphere of the original size.

Worked example. A solid is a cylinder of radius $4\text{ cm}$ and height $10\text{ cm}$ , topped with a hemisphere of the same radius. Find the total volume, in terms of $\pi$ .

V_{\text{cylinder}} = \pi r^2h = \pi(4)^2(10) = 160\pi\text{ cm}^3 \qquad V_{\text{hemisphere}} = \frac{2}{3}\pi(4)^3 = \frac{128}{3}\pi\text{ cm}^3

V_{\text{total}} = 160\pi + \frac{128}{3}\pi = \frac{480\pi+128\pi}{3} = \frac{608}{3}\pi\text{ cm}^3