Composite Figures with Circular Parts

Find the perimeter and the area of figures made with semicircles, quarter circles and circles: arched windows, running tracks, shapes with pieces cut out, shaded regions in a square, and rings.

Before you start

A circle of radius $r$ has circumference $2\pi r$ (or $\pi d$ , where the diameter $d = 2r$ ) and area $\pi r^{2}$ . For example, a circle of radius 5 cm has circumference $10\pi \approx 31.4$ cm and area $25\pi \approx 78.5 \text{ cm}^{2}$ , correct to 3 significant figures.
A composite figure is made of simple shapes. Its area is found by adding the areas of its parts, or by subtracting a part from a larger shape. Its perimeter is the length of its boundary only: an edge where two parts join is inside the figure and is not on it.
The area of a rectangle is $l \times b$ , of a triangle $\frac{1}{2} \times$ base $\times$ height, of a parallelogram base $\times$ height, and of a trapezium $\frac{1}{2}(a + b)h$ .

Semicircles and Quarter Circles

A semicircle is half of a circle, cut off by a diameter. A quarter circle is a quarter of a circle, cut off by two radii at right angles to each other. The curved part of each is an arc.

An arc is a fraction of the whole circumference. The arc of a semicircle is $\frac{1}{2}$ of $2\pi r$ , and the arc of a quarter circle is $\frac{1}{4}$ of $2\pi r$ . In the same way, the area of a semicircle is $\frac{1}{2}$ of $\pi r^{2}$ and the area of a quarter circle is $\frac{1}{4}$ of $\pi r^{2}$ .

The perimeter of a semicircle or a quarter circle is not the arc alone. It is the arc and the straight edges: the diameter of the semicircle ( $2r$ ), or the two radii of the quarter circle ( $r + r = 2r$ ).

Parts of a circle

For a circle of radius $r$ :

Semicircle: arc $= \frac{1}{2} \times 2\pi r = \pi r$ ; perimeter $= \pi r + 2r$ ; area $= \frac{1}{2}\pi r^{2}$ .

Quarter circle: arc $= \frac{1}{4} \times 2\pi r = \frac{1}{2}\pi r$ ; perimeter $= \frac{1}{2}\pi r + 2r$ ; area $= \frac{1}{4}\pi r^{2}$ .

Answers with π

"Leave your answer in terms of $\pi$ " means keep $\pi$ as it is: $(7\pi + 14)$ cm is an exact answer.

"Use the $\pi$ button on your calculator" means work with the calculator's value of $\pi$ . "Take $\pi = 3.142$ " means use 3.142 in its place. Either way, give the answer correct to 3 significant figures unless the question asks for something else, and money to the nearest cent.

Keep the full value on the calculator until the last line and round once, at the end. Rounding a part of the working early can change the last figure of the answer.

Example 1: The arc, perimeter and area of a semicircle

The semicircle has a diameter of 14 cm.

Find (a) the length of its arc, (b) its perimeter, (c) its area. Give each answer in terms of $\pi$ , and then correct to 3 significant figures using the $\pi$ button on your calculator.

Solution

The radius is half the diameter: $r = 14 \div 2 = 7$ cm.

(a) Arc of a semicircle $= \pi r$

$= \pi \times 7$

$= 7\pi \text{ cm} \approx 22.0 \text{ cm}$

(b) Perimeter = arc + diameter (the straight edge is on the boundary)

$= 7\pi + 14$

$\approx 21.991 + 14 = 35.991$ , so the perimeter is $(7\pi + 14) \text{ cm} \approx 36.0 \text{ cm}$ .

$= \frac{1}{2} \times \pi \times 7^{2}$

$= 24.5\pi \text{ cm}^{2} \approx 77.0 \text{ cm}^{2}$

The arc is $7\pi \approx 22.0$ cm long, the perimeter is $(7\pi + 14) \approx 36.0$ cm and the area is $24.5\pi \approx 77.0 \text{ cm}^{2}$ .

Example 2: The perimeter and area of a quarter circle

The quarter circle has a radius of 8 cm.

Find (a) its perimeter, (b) its area. Leave each answer in terms of $\pi$ , and then give it correct to 3 significant figures using the $\pi$ button on your calculator.

Solution

(a) Arc of a quarter circle $= \frac{1}{2}\pi r = \frac{1}{2} \times \pi \times 8 = 4\pi$ cm

Perimeter = arc + two radii

$= 4\pi + 8 + 8$

$= (4\pi + 16) \text{ cm} \approx 28.6 \text{ cm}$

(b) Area of a quarter circle $= \frac{1}{4}\pi r^{2}$

$= \frac{1}{4} \times \pi \times 8^{2}$

$= 16\pi \text{ cm}^{2} \approx 50.3 \text{ cm}^{2}$

The perimeter is $(4\pi + 16) \approx 28.6$ cm and the area is $16\pi \approx 50.3 \text{ cm}^{2}$ .

A Rectangle with a Semicircle

An arched window, a door or the plan of a lawn is often a rectangle with a semicircle on one side. The diameter of the semicircle is the side of the rectangle where the two parts join.

For the area, add the two parts: the rectangle and the semicircle. For the perimeter, go round the outside. The side where they join is inside the figure, so it is not on the boundary: the perimeter is the arc and the other three sides of the rectangle.

Example 3: A lawn with a semicircular end

The plan of a lawn is a rectangle 20 m long and 10 m wide with a semicircle on one of its 10 m sides.

Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures. Find (a) the perimeter of the lawn, (b) the area of the lawn.

Solution

The diameter of the semicircle is 10 m, so its radius is $r = 5$ m.

(a) The boundary is the arc, the right side, the bottom and the left side. The dashed side is inside the lawn and is left out.

Arc $= \pi r = \pi \times 5 = 5\pi$ m

Perimeter $= 5\pi + 20 + 10 + 20$

$= (5\pi + 50) \text{ m} \approx 65.7 \text{ m}$

(b) Area = rectangle + semicircle

Rectangle $= 20 \times 10 = 200 \text{ m}^{2}$

Semicircle $= \frac{1}{2}\pi r^{2} = \frac{1}{2} \times \pi \times 5^{2} = 12.5\pi \text{ m}^{2}$

Area $= (200 + 12.5\pi) \text{ m}^{2} \approx 239 \text{ m}^{2}$

The perimeter of the lawn is $(5\pi + 50) \approx 65.7$ m and its area is $(200 + 12.5\pi) \approx 239 \text{ m}^{2}$ .

Running Tracks

A running track is a rectangle with a semicircle on each of its two ends. The two straight sides are the straights. The two semicircles have the same diameter, so together they make one whole circle: their two arcs add up to one circumference, $\pi d$ , and their two areas to the area of one circle, $\pi r^{2}$ .

The two ends of the rectangle, where the semicircles join it, are inside the figure and are not on the boundary.

A running track

For a track with straights of length $l$ and semicircular ends of diameter $d$ (radius $r$ ):

Perimeter $= 2l + \pi d$

Area $= l \times d + \pi r^{2}$

Example 4: The distance round a running track

The inner edge of a running track at a stadium has two straights of 100 m and a semicircle of diameter 60 m at each end.

Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures. (a) Find the distance once round the inner edge. (b) Siti runs 5 laps along the inner edge. How far does she run, in kilometres? (c) Find the area enclosed by the inner edge.

Solution

(a) The two semicircles make one circle of diameter 60 m.

Two arcs $= \pi d = \pi \times 60 = 60\pi$ m

Perimeter $= 100 + 100 + 60\pi$

$= (200 + 60\pi) \text{ m} \approx 388.496 \text{ m} \approx 388 \text{ m}$

(b) Distance $= 5 \times (200 + 60\pi) = (1000 + 300\pi)$ m

Since 1 km = 1000 m, divide the exact value by 1000: $(1000 + 300\pi) \div 1000 = (1 + 0.3\pi)$ km $\approx 1.942478$ km $\approx 1.94$ km.

Rectangle $= 100 \times 60 = 6000 \text{ m}^{2}$

Two semicircles = one circle $= \pi r^{2} = \pi \times 30^{2} = 900\pi \text{ m}^{2}$

Area $= (6000 + 900\pi) \text{ m}^{2} \approx 8827.4 \text{ m}^{2} \approx 8830 \text{ m}^{2}$

Once round is $(200 + 60\pi) \approx 388$ m, Siti runs about 1.94 km, and the area enclosed is $(6000 + 900\pi) \approx 8830 \text{ m}^{2}$ .

Cutting Out a Circle or a Semicircle

A circle cut from the middle of a shape leaves a hole. The area of what is left is the area of the whole shape minus the area of the circle.

A semicircle cut from the edge of a shape changes the boundary. The part of the side that was the diameter is gone, and the arc is on the boundary in its place. The arc ( $\pi r$ ) is longer than the diameter ( $2r$ ), so cutting the piece out makes the perimeter longer, while it makes the area smaller.

Cutting out and adding on

Area of what is left = area of the whole shape − area of the part cut out.

When a semicircle is cut from a side, or added on a side, the perimeter loses that part of the side (the diameter) and gains the arc.

Example 5: A circle cut from a rectangle

A rectangular piece of cardboard 24 cm by 18 cm has a circular hole of radius 4 cm cut from its centre.

Find the area of the cardboard that is left. Leave your answer in terms of $\pi$ , and then give it correct to 3 significant figures using the $\pi$ button on your calculator.

Solution

Rectangle $= 24 \times 18 = 432 \text{ cm}^{2}$

Circle $= \pi r^{2} = \pi \times 4^{2} = 16\pi \text{ cm}^{2}$

Area left = rectangle − circle

$= (432 - 16\pi) \text{ cm}^{2}$

$\approx 432 - 50.265 = 381.73$ , so the area left is $\approx 382 \text{ cm}^{2}$ .

The area of the cardboard left is $(432 - 16\pi) \approx 382 \text{ cm}^{2}$ .

Example 6: A semicircle cut from the edge of a desk

The top of a desk is a rectangle 1.2 m long and 50 cm wide. A semicircle of diameter 30 cm is cut from the middle of one long side, so that a person can sit close.

Use the $\pi$ button on your calculator and give your answers in centimetres and square centimetres, correct to 3 significant figures. Find (a) the perimeter of the desk top, (b) its area.

Solution

Change the lengths to one unit first: 1.2 m = 120 cm. The radius of the semicircle is $30 \div 2 = 15$ cm.

(a) Go round the boundary. The long side with the cut-out has two straight parts, $120 - 30 = 90$ cm altogether, and the arc.

Arc $= \pi r = \pi \times 15 = 15\pi$ cm

Perimeter $= 120 + 50 + 50 + 90 + 15\pi$

$= (310 + 15\pi) \text{ cm} \approx 357.12 \text{ cm} \approx 357 \text{ cm}$

Check: the rectangle's perimeter is $2 \times (120 + 50) = 340$ cm; the cut-out loses 30 cm and gains $15\pi$ cm, and $340 - 30 + 15\pi = 310 + 15\pi$ .

(b) Area = rectangle − semicircle

Rectangle $= 120 \times 50 = 6000 \text{ cm}^{2}$

Semicircle $= \frac{1}{2}\pi r^{2} = \frac{1}{2} \times \pi \times 15^{2} = 112.5\pi \text{ cm}^{2}$

Area $= (6000 - 112.5\pi) \text{ cm}^{2} \approx 5646.6 \text{ cm}^{2} \approx 5650 \text{ cm}^{2}$

The perimeter of the desk top is $(310 + 15\pi) \approx 357$ cm and its area is $(6000 - 112.5\pi) \approx 5650 \text{ cm}^{2}$ .

Same perimeter, different area

If the same semicircle were added on the long side instead of cut out, the 30 cm of the side would again be inside the figure and the arc on the boundary: the perimeter would still be $(310 + 15\pi)$ cm. Only the area would change, to $(6000 + 112.5\pi) \text{ cm}^{2}$ .

Quarter Circles in a Square

A quarter circle fits in the corner of a square when its centre is a corner of the square and its radius is no more than a side. Its two radii then lie along the two sides that meet at that corner.

Shaded regions in a square are found by subtracting: the square minus the quarter circle, or the parts that are not shaded. Their perimeters are found by going round the boundary of the shaded region, piece by piece: some pieces are arcs and some are straight.

Example 7: A quarter circle cut from a corner

A square tile of side 10 cm has a quarter circle of radius 4 cm cut from one corner.

Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures. Find (a) the perimeter of the tile that is left, (b) its area.

Solution

(a) Go round the boundary: the bottom 10 cm, the left side 10 cm, the two cut sides, and the arc.

Each cut side $= 10 - 4 = 6$ cm

Arc of the quarter circle $= \frac{1}{2}\pi r = \frac{1}{2} \times \pi \times 4 = 2\pi$ cm

Perimeter $= 10 + 10 + 6 + 6 + 2\pi$

$= (32 + 2\pi) \text{ cm} \approx 38.3 \text{ cm}$

Check: the square's perimeter is $4 \times 10 = 40$ cm; the cut takes away $4 + 4 = 8$ cm of its sides and adds the arc, and $40 - 8 + 2\pi = 32 + 2\pi$ .

(b) Area = square − quarter circle

$= 10 \times 10 - \frac{1}{4} \times \pi \times 4^{2}$

$= (100 - 4\pi) \text{ cm}^{2} \approx 87.4 \text{ cm}^{2}$

The perimeter of the tile left is $(32 + 2\pi) \approx 38.3$ cm and its area is $(100 - 4\pi) \approx 87.4 \text{ cm}^{2}$ .

Example 8: The region between a square and a quarter circle

$ABCD$ is a square of side 10 cm. A quarter circle with centre $A$ is drawn from $B$ to $D$ . The region of the square outside the quarter circle is shaded.

Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures. Find (a) the area of the shaded region, (b) the perimeter of the shaded region.

Solution

The radius of the quarter circle is $AB = 10$ cm.

(a) Shaded area = square − quarter circle

$= 10 \times 10 - \frac{1}{4} \times \pi \times 10^{2}$

$= (100 - 25\pi) \text{ cm}^{2} \approx 21.5 \text{ cm}^{2}$

(b) Go round the shaded region: from $B$ up to $C$ (10 cm), across to $D$ (10 cm), and back along the arc to $B$ . The sides $AB$ and $AD$ are not on its boundary.

Arc $= \frac{1}{2}\pi r = \frac{1}{2} \times \pi \times 10 = 5\pi$ cm

Perimeter $= 10 + 10 + 5\pi$

$= (20 + 5\pi) \text{ cm} \approx 35.7 \text{ cm}$

The shaded region has area $(100 - 25\pi) \approx 21.5 \text{ cm}^{2}$ and perimeter $(20 + 5\pi) \approx 35.7$ cm.

When two quarter circles are drawn in a square with their centres at opposite corners (each with the side of the square as its radius), they overlap in a shape like a leaf. Each quarter circle covers the leaf and one of the two corner regions. So the two quarter circles together cover the whole square once and the leaf a second time:

2 × quarter circle = square + leaf, so leaf = 2 × quarter circle − square.

Example 9: The leaf between two quarter circles

$ABCD$ is a square of side 10 cm. Two quarter circles of radius 10 cm are drawn, one with centre $A$ and one with centre $C$ , both from $B$ to $D$ . The leaf where they overlap is shaded.

Find (a) the area of the leaf, (b) its perimeter. Leave each answer in terms of $\pi$ , and then give it correct to 3 significant figures using the $\pi$ button on your calculator.

Solution

(a) Each quarter circle $= \frac{1}{4}\pi r^{2} = \frac{1}{4} \times \pi \times 10^{2} = 25\pi \text{ cm}^{2}$

Leaf = 2 × quarter circle − square

$= 2 \times 25\pi - 10 \times 10$

$= (50\pi - 100) \text{ cm}^{2} \approx 57.1 \text{ cm}^{2}$

Check another way: each corner region outside the leaf is the region of Example 8, $100 - 25\pi$ , so the leaf is $100 - 2(100 - 25\pi) = 50\pi - 100$ .

(b) The boundary of the leaf is two arcs only, one from each quarter circle; no side of the square is on it.

Each arc $= \frac{1}{2}\pi r = \frac{1}{2} \times \pi \times 10 = 5\pi$ cm

Perimeter $= 5\pi + 5\pi = 10\pi \text{ cm} \approx 31.4 \text{ cm}$

The leaf has area $(50\pi - 100) \approx 57.1 \text{ cm}^{2}$ and perimeter $10\pi \approx 31.4$ cm.

Rings: a Path Round a Circle

A ring is the region between two circles with the same centre: a path round a round pond, the stone edge of a fountain. If the outer circle has radius $R$ and the inner circle radius $r$ , the width of the ring is $R - r$ .

Area of a ring

Area of a ring = area of the outer circle − area of the inner circle $= \pi R^{2} - \pi r^{2}$ .

Example 10: A path round a pond

A round pond in a neighbourhood park has a radius of 5 m. A path 2 m wide is built round it.

(a) Find the area of the path, leaving your answer in terms of $\pi$ . (b) The path is paved at a cost of $35 per $\text{m}^{2}$ . Use the $\pi$ button on your calculator to find the cost, to the nearest cent.

Solution

(a) The inner radius is $r = 5$ m and the outer radius is $R = 5 + 2 = 7$ m.

Area of path $= \pi R^{2} - \pi r^{2}$

$= \pi \times 7^{2} - \pi \times 5^{2}$

$= 49\pi - 25\pi = 24\pi \text{ m}^{2}$

(b) Cost $= 24\pi \times 35 = 840\pi \approx 2638.937\ldots$

To the nearest cent, the cost is $2638.94.

The area of the path is $24\pi \text{ m}^{2}$ , and paving it costs $2638.94.

Working Backwards: Finding the Radius

When the arc, the perimeter or the area is given, write its formula with the radius $r$ unknown. This is an equation in $r$ : solve it by doing the same to both sides, and check by putting the answer back. Use the calculator's full value of $r$ in any later step.

Example 11: The radius from a perimeter and from an area

Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures.

(a) The perimeter of a semicircle is 36 cm. Find its radius.

(b) A lawn in the shape of a quarter circle has an area of 50 $\text{m}^{2}$ . Find its radius, and the length of its perimeter.

Solution

(a) Perimeter of a semicircle $= \pi r + 2r$ , so

$\pi r + 2r = 36$

Both terms on the left have $r$ : $\pi r + 2r = (\pi + 2)r$ .

$(\pi + 2)r = 36$

Divide both sides by $(\pi + 2)$ : $r = \frac{36}{\pi + 2} \approx 7.0017$

Check: $\pi \times 7.0017\ldots + 2 \times 7.0017\ldots \approx 21.997 + 14.003 = 36.000$ ✓

So $r \approx 7.00$ cm.

(b) Area of a quarter circle $= \frac{1}{4}\pi r^{2}$ , so

$\frac{1}{4}\pi r^{2} = 50$

Multiply both sides by 4: $\pi r^{2} = 200$

Divide both sides by $\pi$ : $r^{2} = \frac{200}{\pi} \approx 63.662$

A radius is positive, so $r = \sqrt{63.662\ldots} \approx 7.9788$

Check: $\frac{1}{4} \times \pi \times 7.9788\ldots^{2} \approx 50.000$ ✓

Perimeter $= \frac{1}{2}\pi r + 2r \approx \frac{1}{2} \times \pi \times 7.9788\ldots + 2 \times 7.9788\ldots \approx 12.533 + 15.958 = 28.491$

So $r \approx 7.98$ m and the perimeter $\approx 28.5$ m.

(a) The radius is about 7.00 cm. (b) The radius of the lawn is about 7.98 m and its perimeter about 28.5 m.

Common mistakes

The arc alone is not the perimeter. The perimeter of a semicircle of radius 7 cm is $7\pi + 14 \approx 36.0$ cm, not $7\pi \approx 22.0$ cm: the diameter is on the boundary too. A quarter circle has two radii on its boundary.
Counting an edge that is inside. When a semicircle sits on a side of a rectangle, that side is its diameter and is inside the figure: leave it out of the perimeter, but count both parts in the area.
Using the diameter as the radius. A semicircle of diameter 14 cm has $r = 7$ cm, so its area is $\frac{1}{2} \times \pi \times 7^{2}$ , not $\frac{1}{2} \times \pi \times 14^{2}$ .
Two circles at the ends of a track. The two semicircular ends make one circle, not two.
Squaring the width of a ring. The area of a ring is $\pi R^{2} - \pi r^{2}$ , not $\pi(R - r)^{2}$ . For $R = 7$ and $r = 5$ the first is $24\pi$ ; the second gives $4\pi$ , which is wrong.
Thinking a cut-out makes the perimeter shorter. A semicircle cut from a side takes away the diameter but adds the arc, which is longer.
Rounding too early. Keep $\pi$ , or the calculator's full value, until the last line, and round once.

Practice

Try each question before you open its solution. Each question says how to use π; give answers that are not exact correct to 3 significant figures, and money to the nearest cent.

Question 1

A quarter circle has a radius of 6 cm. Find (a) the length of its arc, (b) its perimeter, (c) its area. Leave your answers in terms of $\pi$ .

ShowHide solution

(a) Arc $= \frac{1}{2}\pi r = \frac{1}{2} \times \pi \times 6 = 3\pi$ cm

(b) Perimeter = arc + two radii $= 3\pi + 6 + 6 = (3\pi + 12)$ cm

(a) $3\pi$ cm (b) $(3\pi + 12)$ cm (c) $9\pi \text{ cm}^{2}$

Question 2

A semicircle has a diameter of 9 cm. Taking $\pi = 3.142$ , find (a) its perimeter, (b) its area, each correct to 3 significant figures.

ShowHide solution

The radius is $r = 9 \div 2 = 4.5$ cm.

(a) Perimeter $= \pi r + 2r = 3.142 \times 4.5 + 9 = 14.139 + 9 = 23.139$ cm $\approx 23.1$ cm

(b) Area $= \frac{1}{2}\pi r^{2} = \frac{1}{2} \times 3.142 \times 4.5^{2} = \frac{1}{2} \times 3.142 \times 20.25 = 31.81275 \text{ cm}^{2} \approx 31.8 \text{ cm}^{2}$

(a) The perimeter is about 23.1 cm. (b) The area is about $31.8 \text{ cm}^{2}$ .

Question 3

An arched window in a shophouse is a rectangle 70 cm wide and 1.2 m tall with a semicircle on top. Use the $\pi$ button on your calculator and give your answers in metres and square metres, correct to 3 significant figures. Find (a) the perimeter of the window, (b) its area.

ShowHide solution

Change to metres first: 70 cm = 0.7 m. The radius of the semicircle is $0.7 \div 2 = 0.35$ m.

(a) The boundary is the arc and three sides of the rectangle; the dashed side is inside.

Arc $= \pi r = 0.35\pi$ m

Perimeter $= 1.2 + 0.7 + 1.2 + 0.35\pi = (3.1 + 0.35\pi)$ m $\approx 4.1996$ m $\approx 4.20$ m

(b) Area = rectangle + semicircle

Rectangle $= 0.7 \times 1.2 = 0.84 \text{ m}^{2}$

Semicircle $= \frac{1}{2} \times \pi \times 0.35^{2} = 0.06125\pi \text{ m}^{2}$

Area $= (0.84 + 0.06125\pi) \text{ m}^{2} \approx 1.0324 \text{ m}^{2} \approx 1.03 \text{ m}^{2}$

(a) The perimeter of the window is about 4.20 m. (b) Its area is about $1.03 \text{ m}^{2}$ .

Question 4

In a Design and Technology lesson, Ravi cuts a circular hole of diameter 7 cm from the centre of a rectangular acrylic sheet 18 cm by 12 cm. Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures.

Find (a) the area of the sheet that is left, (b) the total length of all its edges, the four sides and the edge of the hole, which Ravi files smooth.

ShowHide solution

The radius of the hole is $7 \div 2 = 3.5$ cm.

(a) Area left = rectangle − circle

Rectangle $= 18 \times 12 = 216 \text{ cm}^{2}$

Circle $= \pi r^{2} = \pi \times 3.5^{2} = 12.25\pi \text{ cm}^{2}$

Area left $= (216 - 12.25\pi) \text{ cm}^{2} \approx 177.515 \text{ cm}^{2} \approx 178 \text{ cm}^{2}$

(b) The sheet that is left has two edges: the outer edge, which is the four sides of the rectangle, and the edge of the hole, which is a whole circle.

Outer edge $= 2 \times (18 + 12) = 60$ cm

Edge of the hole = circumference $= \pi d = \pi \times 7 = 7\pi$ cm

Total length $= (60 + 7\pi)$ cm $\approx 81.991$ cm $\approx 82.0$ cm

To 3 significant figures, 81.991 is 82.0: the 0 is the third significant figure, so it is written.

(a) The area of the sheet left is $(216 - 12.25\pi) \approx 178 \text{ cm}^{2}$ . (b) The total length of its edges is $(60 + 7\pi) \approx 82.0$ cm.

Question 5

The field inside a running track at a stadium is a rectangle 85 m by 70 m with a semicircle on each 70 m end. Use the $\pi$ button on your calculator.

(a) Mei Ling runs 4 times round the edge of the field. How far does she run, in kilometres, correct to 3 significant figures?

(b) The field is turfed at $6 per $\text{m}^{2}$ . Find the cost, to the nearest cent.

ShowHide solution

(a) The two ends make one circle of diameter 70 m.

Once round $= 85 + 85 + \pi \times 70 = (170 + 70\pi)$ m $\approx 389.91$ m

Four times round $= 4 \times (170 + 70\pi) = (680 + 280\pi)$ m

Since 1 km = 1000 m, divide the exact value by 1000: $(680 + 280\pi) \div 1000 = (0.68 + 0.28\pi)$ km $\approx 1.559646$ km $\approx 1.56$ km

(b) The radius of each end is $70 \div 2 = 35$ m.

Area $= 85 \times 70 + \pi \times 35^{2} = (5950 + 1225\pi) \text{ m}^{2} \approx 9798.451 \text{ m}^{2}$

Cost $= 6 \times (5950 + 1225\pi) \approx 58\,790.706$

To the nearest cent, the cost is $58 790.71.

(a) Mei Ling runs about 1.56 km. (b) Turfing the field costs $58 790.71.

Question 6

A piece of a puzzle is made from a rectangle 16 cm by 10 cm. A semicircle of diameter 10 cm is cut from its left end and the same semicircle is added on its right end. Use the $\pi$ button on your calculator.

Find (a) the area of the piece, (b) its perimeter, correct to 3 significant figures.

ShowHide solution

(a) The semicircle added on the right has the same area as the semicircle cut from the left, so the area is the area of the rectangle: $16 \times 10 = 160 \text{ cm}^{2}$ .

(b) Neither end of the rectangle is on the boundary. The boundary is the top 16 cm, the bottom 16 cm and two arcs of radius 5 cm.

Each arc $= \pi r = 5\pi$ cm

Perimeter $= 16 + 16 + 5\pi + 5\pi = (32 + 10\pi)$ cm $\approx 63.4$ cm

(a) The area is $160 \text{ cm}^{2}$ . (b) The perimeter is $(32 + 10\pi) \approx 63.4$ cm.

Question 7

A square metal plate of side 12 cm has a quarter circle of radius 3 cm cut from each of its four corners. Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures.

Find (a) the perimeter of the plate that is left, (b) its area.

ShowHide solution

(a) Each side of the square keeps a straight part of $12 - 3 - 3 = 6$ cm, and each corner gives an arc.

Each arc $= \frac{1}{2}\pi r = \frac{1}{2} \times \pi \times 3 = 1.5\pi$ cm

Perimeter $= 4 \times 6 + 4 \times 1.5\pi = (24 + 6\pi)$ cm $\approx 42.8$ cm

(b) The four quarter circles together make one circle of radius 3 cm.

Area $= 12 \times 12 - \pi \times 3^{2} = (144 - 9\pi) \text{ cm}^{2} \approx 116 \text{ cm}^{2}$

(a) The perimeter is $(24 + 6\pi) \approx 42.8$ cm. (b) The area is $(144 - 9\pi) \approx 116 \text{ cm}^{2}$ .

Question 8

A semicircle is drawn inside a square of side 12 cm, with one side of the square as its diameter. The part of the square outside the semicircle is shaded. Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures.

Find (a) the area of the shaded region, (b) the perimeter of the shaded region.

ShowHide solution

The radius of the semicircle is $12 \div 2 = 6$ cm.

(a) Shaded area = square − semicircle $= 12 \times 12 - \frac{1}{2} \times \pi \times 6^{2} = (144 - 18\pi) \text{ cm}^{2} \approx 87.5 \text{ cm}^{2}$

(b) The shaded region touches the bottom side only at its two ends, so the bottom side is not on its boundary. Its boundary is the other three sides and the arc.

Arc $= \pi r = 6\pi$ cm

Perimeter $= 12 + 12 + 12 + 6\pi = (36 + 6\pi)$ cm $\approx 54.8$ cm

(a) The shaded area is $(144 - 18\pi) \approx 87.5 \text{ cm}^{2}$ . (b) Its perimeter is $(36 + 6\pi) \approx 54.8$ cm.

Question 9

$PQRS$ is a square of side 6 cm. Quarter circles of radius 6 cm are drawn with centres $P$ and $R$ , both from $Q$ to $S$ . Find the area and the perimeter of the shaded leaf, leaving your answers in terms of $\pi$ .

ShowHide solution

Each quarter circle $= \frac{1}{4} \times \pi \times 6^{2} = 9\pi \text{ cm}^{2}$

Leaf = 2 × quarter circle − square $= 2 \times 9\pi - 6 \times 6 = (18\pi - 36) \text{ cm}^{2}$

The boundary of the leaf is two arcs, each $\frac{1}{2} \times \pi \times 6 = 3\pi$ cm.

Perimeter $= 3\pi + 3\pi = 6\pi$ cm

The leaf has area $(18\pi - 36) \text{ cm}^{2}$ and perimeter $6\pi$ cm.

Question 10

A logo is made from a semicircle of diameter 12 cm with two semicircles of diameter 6 cm cut out of it, side by side on the same diameter. The region that is left is shaded.

Find (a) the area of the shaded region, (b) its perimeter. Leave your answers in terms of $\pi$ , and then give them correct to 3 significant figures using the $\pi$ button on your calculator.

ShowHide solution

The large semicircle has radius 6 cm; each small semicircle has diameter $12 \div 2 = 6$ cm and radius 3 cm.

(a) Shaded area = large semicircle − two small semicircles

$= \frac{1}{2} \times \pi \times 6^{2} - 2 \times \frac{1}{2} \times \pi \times 3^{2}$

$= 18\pi - 9\pi = 9\pi \text{ cm}^{2} \approx 28.3 \text{ cm}^{2}$

(b) The dashed diameter is not on the boundary: the shaded region is above the two small arcs. Its boundary is three arcs.

Perimeter $= \pi \times 6 + \pi \times 3 + \pi \times 3 = 12\pi$ cm $\approx 37.7$ cm

(a) The shaded area is $9\pi \approx 28.3 \text{ cm}^{2}$ . (b) Its perimeter is $12\pi \approx 37.7$ cm.

Question 11

A round fountain at a community club has a diameter of 6 m. Use the $\pi$ button on your calculator.

(a) A brass strip is fixed round the edge of the fountain at $18 per metre. Find the cost, to the nearest cent.

(b) A path 1.5 m wide is laid round the fountain. Find the area of the path, correct to 3 significant figures.

ShowHide solution

(a) Length of strip = circumference $= \pi d = \pi \times 6 = 6\pi$ m

Cost $= 6\pi \times 18 = 108\pi \approx 339.292$

To the nearest cent, the cost is $339.29.

(b) The inner radius is $r = 6 \div 2 = 3$ m; the outer radius is $R = 3 + 1.5 = 4.5$ m.

Area of path $= \pi R^{2} - \pi r^{2} = \pi \times 4.5^{2} - \pi \times 3^{2} = 20.25\pi - 9\pi = 11.25\pi \text{ m}^{2} \approx 35.3 \text{ m}^{2}$

(a) The brass strip costs $339.29. (b) The area of the path is $11.25\pi \approx 35.3 \text{ m}^{2}$ .

Question 12

A flower bed at Gardens by the Bay is a semicircle. Its curved edge is 22 m long. Use the $\pi$ button on your calculator.

(a) Find the radius of the flower bed, correct to 3 significant figures.

(b) A low fence is put round the whole flower bed, along its curved edge and its straight edge, at $12.50 per metre. Find the cost, to the nearest cent.

ShowHide solution

(a) Arc of a semicircle $= \pi r$ , so $\pi r = 22$ .

Divide both sides by $\pi$ : $r = \frac{22}{\pi} \approx 7.00282$

Check: $\pi \times 7.00282 \approx 22.000$ ✓, so $r \approx 7.00$ m.

(b) Straight edge = diameter $= 2r = \frac{44}{\pi} \approx 14.0056$ m

Length of fence $= 22 + 14.0056\ldots \approx 36.0056$ m

Cost $= 12.50 \times 36.0056\ldots \approx 450.070$

To the nearest cent, the cost is $450.07.

(a) The radius is about 7.00 m. (b) The fence costs $450.07.

Question 13

A running track has two straights, each $x$ m long, and a semicircle of diameter 64 m at each end. The distance once round the track is 400 m. Find $x$ , using the $\pi$ button on your calculator and giving your answer correct to 3 significant figures.

ShowHide solution

The two ends make one circle of diameter 64 m, so the two arcs are $64\pi$ m.

Distance round $= 2x + 64\pi$ , so

$2x + 64\pi = 400$

Subtract $64\pi$ from both sides: $2x = 400 - 64\pi$

Divide both sides by 2: $x = 200 - 32\pi \approx 99.469$

Check: $2 \times 99.469 + 64\pi \approx 198.938 + 201.062 = 400$ ✓

$x = 200 - 32\pi \approx 99.5$ , so each straight is about 99.5 m long.

Question 14

A flower bed in a corner of the school field is a quarter circle. Mr Tan needs exactly 30 m of edging to go all the way round it, along its two straight edges and its curved edge. Use the $\pi$ button on your calculator and give your answers correct to 3 significant figures.

(a) Find the radius of the flower bed.

(b) Find the area of the flower bed.

ShowHide solution

(a) The perimeter of a quarter circle is its arc and its two radii: $\frac{1}{2}\pi r + 2r$ .

$\frac{1}{2}\pi r + 2r = 30$

Both terms on the left have $r$ : $\frac{1}{2}\pi r + 2r = (\frac{1}{2}\pi + 2)r$ .

$(\frac{1}{2}\pi + 2)r = 30$

Divide both sides by $(\frac{1}{2}\pi + 2)$ : $r = \frac{30}{\frac{1}{2}\pi + 2} \approx 8.4015$

Check: $\frac{1}{2} \times \pi \times 8.4015\ldots + 2 \times 8.4015\ldots \approx 13.197 + 16.803 = 30.000$ ✓, so $r \approx 8.40$ m.

(b) Use the full value of $r$ , not the rounded one.

Area $= \frac{1}{4}\pi r^{2} \approx \frac{1}{4} \times \pi \times 8.4015\ldots^{2} \approx 55.437 \text{ m}^{2} \approx 55.4 \text{ m}^{2}$

(a) The radius is about 8.40 m. (b) The area of the flower bed is about $55.4 \text{ m}^{2}$ .

Question 15

A lawn in front of a community centre in Sengkang is a semicircle with an area of 150 $\text{m}^{2}$ . Its straight edge runs along the wall of the building. Use the $\pi$ button on your calculator.

(a) Find the radius of the lawn, correct to 3 significant figures.

(b) A fence is put along the curved edge only, at $16.80 per metre. Find the cost of the fence, to the nearest cent.

ShowHide solution

(a) Area of a semicircle $= \frac{1}{2}\pi r^{2}$ , so

$\frac{1}{2}\pi r^{2} = 150$

Multiply both sides by 2: $\pi r^{2} = 300$

Divide both sides by $\pi$ : $r^{2} = \frac{300}{\pi} \approx 95.493$

A radius is positive, so $r = \sqrt{95.493\ldots} \approx 9.7721$

Check: $\frac{1}{2} \times \pi \times 9.7721\ldots^{2} \approx 150.000$ ✓, so $r \approx 9.77$ m.

(b) The straight edge is along the wall, so only the arc is fenced.

Arc $= \pi r \approx \pi \times 9.7721\ldots \approx 30.6998$ m

Cost $= 16.80 \times 30.6998\ldots \approx 515.757$

To the nearest cent, the cost is $515.76. (With the rounded radius 9.77 m the cost would come out as $515.65: keep the full value until the end.)

(a) The radius of the lawn is about 9.77 m. (b) The fence costs $515.76.

Key points

A circle of radius $r$ : circumference $2\pi r = \pi d$ ; area $\pi r^{2}$ .
A semicircle: arc $\pi r$ ; perimeter $\pi r + 2r$ ; area $\frac{1}{2}\pi r^{2}$ .
A quarter circle: arc $\frac{1}{2}\pi r$ ; perimeter $\frac{1}{2}\pi r + 2r$ ; area $\frac{1}{4}\pi r^{2}$ .
The perimeter of a composite figure is its boundary only: a diameter or a radius where two parts join is not on it.
Two semicircles of the same diameter make one circle: a track with straights $l$ and ends of diameter $d$ has perimeter $2l + \pi d$ .
A semicircle cut from a side, or added on it, takes that part of the side off the boundary and puts its arc $\pi r$ on.
The leaf in a square: area = 2 × quarter circle − square.
A ring: area $= \pi R^{2} - \pi r^{2}$ .
To find a radius, write the formula as an equation in $r$ , solve it, and check by putting it back.
Leave an answer in terms of $\pi$ when asked; otherwise use the $\pi$ the question gives and round once, at the end.