Year 9 Maths Measurement

Composite Shapes

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Theory

A composite shape is made of simpler shapes joined together. This Year 9 maths page revises the Year 8 work: finding its perimeter, including missing sides, and its area by adding pieces or subtracting a hole. It also covers shaded regions, semicircles and paths around a shape.

A composite shape is built from shapes you already know: rectangles, triangles, parallelograms, trapeziums and parts of circles.

The perimeter is the distance around the outside edge. Walk round the shape once and add every side. Lines drawn inside the shape do not count.

On a shape with right-angled corners, a missing side comes from the sides opposite it. The pieces going one way add up to the full length on the other side.

The area can be found two ways. Add: cut the shape into pieces and add their areas. Subtract: enclose it in a bigger shape and take away the missing part. Choose the way that needs fewer missing lengths.

A shaded region is usually "whole minus white": find the area of the whole shape and subtract the unshaded parts.

Add: cut the shape into pieces you know, find each area, and add them.

Subtract: complete the shape to a rectangle, then take away the missing piece (red). Both methods give the same area.

The shapes you build from:

\[A_{\text{rectangle}} = lb \qquad A_{\text{triangle}} = \dfrac{1}{2}bh \qquad A_{\text{parallelogram}} = bh\]

A_{triangle} = \frac{1}{2} b h

\[A_{\text{trapezium}} = \dfrac{h}{2}(a + b) \qquad A_{\text{circle}} = \pi r^2\]

A_{trapezium} = \frac{h}{2} (a + b)

A semicircle is half a circle:

\[A = \dfrac{1}{2}\pi r^2 \qquad \text{curved edge} = \dfrac{1}{2} \times 2\pi r = \pi r\]

A = \frac{1}{2} π r^{2}

A slanted side. If a side is slanted, draw a right-angled triangle under it and use Pythagoras. A trapezium with parallel sides \(9\) and \(13\) and height \(3\) has a slanted side of \(\sqrt{4^2 + 3^2} = 5\).

Working backwards. If the area is given, write it with the unknown and solve. A deck \(6\) m wide with a \(2\) m by \(1.5\) m hole cut out for a spa has \(45 \text{ m}^2\) of decking: \(6l - 3 = 45\), so \(l = 8\) m.

Equal marks. Sides with the same tick marks are equal in length. Use them to fill in unlabelled sides.

Corners cut off. If equal right-angled triangles are cut from the corners of a square, subtract all of them. A square joining the midpoints of a \(12\) cm square cuts off four triangles with legs of \(6\) cm, so it has area \(144 - 4 \times \dfrac{1}{2} \times 6 \times 6 = 72 \text{ cm}^2\): half the big square.

How to find the perimeter and area of a composite shape

Find any missing sides from the sides opposite them, or from matching tick marks.
Perimeter: walk round the outside once and add every side. A curved edge is part of a circumference.
Area: split the shape into pieces and add, or enclose it and subtract the missing part. For a shaded region, take the whole shape minus the white parts.
Use the right formula for each piece, then add or subtract.
Check the units: perimeter in m or cm, area in $\text{m}^2$ or $\text{cm}^2$.

Example 1 — Missing sides

A courtyard is a $14$ m by $10$ m rectangle with a notch $5$ m wide and $4$ m deep cut from the middle of one long side. Find its perimeter and area.

Solution

The two top pieces are $\dfrac{14 - 5}{2} = 4.5$ m each. Walk around the edge and add every side.

$P$	$=$	$14 + 10 + 4.5 + 4 + 5 + 4 + 4.5 + 10$
	$=$	$\mathbf{56}~\text{m}$

The notch removes $5$ m of edge but adds two $4$ m sides, so the perimeter is the rectangle's $48$ m plus $8$ m.

$A$	$=$	$14 \times 10 - 5 \times 4$
	$=$	$140 - 20 = \mathbf{120}~\text{m}^2$

A = 14 \times 10 - 5 \times 4 = 120

Example 2 — A semicircle on a rectangle

A rectangle $10$ cm wide and $6$ cm high has a semicircle on top, its diameter the full width. Find
(i) the area
(ii) the perimeter, each to two decimal places.

Solution

(i) The diameter is $10$ cm, so the radius is $5$ cm. Add the rectangle and half a circle.

$A$	$=$	$10 \times 6 + \dfrac{1}{2} \times \pi \times 5^2$
	$=$	$60 + 39.269\ldots$
	$\approx$	$\mathbf{99.27} \text{ cm}^2$

(ii) The top edge is now curved: half a circumference, $\dfrac{1}{2} \times 2\pi r = \pi r$. The other three sides are straight.

$P$	$=$	$10 + 6 + 6 + \pi \times 5$
	$=$	$22 + 15.707\ldots$
	$\approx$	$\mathbf{37.71} \text{ cm}$

P = 22 + 5 π \approx 37.71

Example 3 — A shape with a hole

A sign is a trapezium with parallel sides of $18$ cm and $30$ cm, $16$ cm apart. A triangular hole with a base of $10$ cm and a height of $6$ cm is cut out of it. Find the area of the sign.

Solution

The sign is the whole trapezium minus the triangular hole.

$\text{trapezium}$	$=$	$\dfrac{1}{2} \times (18 + 30) \times 16 = 384 \text{ cm}^2$
$\text{hole}$	$=$	$\dfrac{1}{2} \times 10 \times 6 = 30 \text{ cm}^2$
$A$	$=$	$384 - 30 = \mathbf{354} \text{ cm}^2$

A = 384 - 30 = 354

The $16$ cm height is the perpendicular distance between the parallel sides, not the sloping side.

Example 4 — A path around a pond

A pond $6$ m by $3$ m has a path $1$ m wide all the way round. Paving costs $30 per square metre. Find the area of the path and the cost.

Solution

The path adds $1$ m on both sides, so the outer rectangle is $8$ m by $5$ m. The path is the outer rectangle minus the pond.

$\text{outer}$	$=$	$8 \times 5 = 40 \text{ m}^2$
$\text{pond}$	$=$	$6 \times 3 = 18 \text{ m}^2$
$\text{path}$	$=$	$40 - 18 = \mathbf{22} \text{ m}^2$

$\text{cost}$

$=$

$22 \times \$30 = \mathbf{\$660}$

(8 \times 5 - 6 \times 3) \times 30 = 660

The path costs $660.

Common pitfalls

A cut line is not part of the perimeter. It only helps with the area. The perimeter follows the outside edge.

Find the missing sides first. A shape can have two unmarked sides. Both come from the marked sides opposite them.

A notch does not shrink the perimeter. Cutting a rectangular notch from an edge removes area but adds the two side edges of the notch.

A path goes on both sides. A path $2$ m wide around a $10$ m pool makes the outer length $14$ m, not $12$ m.

Use the radius in the circle formula. If the diameter is $10$, the radius is $5$. A semicircle is half of $\pi r^2$.

Frequently asked questions

What is a composite shape?

A composite shape is made of two or more simple shapes joined together, such as a rectangle with a triangle or a semicircle on top.

How do you find the area of a composite shape?

Either split it into simple shapes, find each area and add them, or enclose it in a bigger shape and subtract the part that is missing. Both give the same answer.

How do you find the shaded area?

Find the area of the whole shape and subtract the area of the white or unshaded parts. Sometimes it is quicker to add the shaded pieces directly.

How do you find a missing side on a composite shape?

On a shape with right-angled corners, the sides going one way add up to the full side opposite them. If the bottom is $14$ m and one top piece is $5$ m, the other top pieces add to $9$ m.

How do you find the perimeter of a shape with a semicircle?

Add the straight sides, then add the curved edge. The curved edge of a semicircle is half the circumference: $\pi r$.

How do you find the area of a path around a rectangle?

Find the area of the outer rectangle, including the path on both sides, then subtract the area of the inner rectangle.

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Area

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Surface Area of Prisms and Pyramids

\(P\)	\(=\)	\(14 + 10 + 4.5 + 4 + 5 + 4 + 4.5 + 10\)
\(\)	\(=\)	\(\mathbf{56}~\text{m}\)

\(A\)	\(=\)	\(14 \times 10 - 5 \times 4\)
\(\)	\(=\)	\(140 - 20 = \mathbf{120}~\text{m}^2\)

\(A\)	\(=\)	\(10 \times 6 + \dfrac{1}{2} \times \pi \times 5^2\)
\(\)	\(=\)	\(60 + 39.269\ldots\)
\(\)	\(\approx\)	\(\mathbf{99.27} \text{ cm}^2\)

\(P\)	\(=\)	\(10 + 6 + 6 + \pi \times 5\)
\(\)	\(=\)	\(22 + 15.707\ldots\)
\(\)	\(\approx\)	\(\mathbf{37.71} \text{ cm}\)

\(\text{trapezium}\)	\(=\)	\(\dfrac{1}{2} \times (18 + 30) \times 16 = 384 \text{ cm}^2\)
\(\text{hole}\)	\(=\)	\(\dfrac{1}{2} \times 10 \times 6 = 30 \text{ cm}^2\)
\(A\)	\(=\)	\(384 - 30 = \mathbf{354} \text{ cm}^2\)

\(\text{outer}\)	\(=\)	\(8 \times 5 = 40 \text{ m}^2\)
\(\text{pond}\)	\(=\)	\(6 \times 3 = 18 \text{ m}^2\)
\(\text{path}\)	\(=\)	\(40 - 18 = \mathbf{22} \text{ m}^2\)

Composite Shapes

📖 Theory

How to find the perimeter and area of a composite shape

Common pitfalls

Frequently asked questions

Theory