Arcs and Sectors

A whole circle has 360° at its centre.

Work out the area of the whole circle in terms of \pi first.

The area of a whole circle is \pi r^2. What fraction of the circle is the sector?

The arc is a fraction of the circumference of the whole circle, 2\pi r.

Find the area of the whole circle, then the fraction of it that the sector takes up.

Look at the edges of the shaded region. Is it bounded by two radii and an arc, or by a straight line and an arc?

Part (a)

1. The shaded region lies between the straight line AB and the arc.
2. A region between a chord and an arc is a segment. (A sector would be between two radii and an arc.)

Part (b)

1. AB joins two points on the circle but does not pass through the centre O.
2. So AB is a chord. (It is not a diameter, because it misses the centre.)

The curved edge is half of the circumference of a whole circle with diameter 12 cm.

1. Circumference of the whole circle =\pi\times\text{diameter}=\pi\times12=12\pi cm.
2. The curved edge is half of this: 12\pi\div2=6\pi.
3. Length of the curved edge =6\pi cm.

The formula for the area of a circle uses the radius, which is half of the diameter.

1. Radius =18\div2=9 cm.
2. Area of the whole circle =\pi\times9^2=254.46...\text{ cm}^2.
3. The sector is \frac{120}{360} of the circle: \frac{120}{360}\times254.46...=84.823...
4. Correct to 1 decimal place: 84.8\text{ cm}^2.

The arc is \frac{135}{360} of the circumference of the whole circle.

1. Circumference of the whole circle =2\times\pi\times7=43.982... cm.
2. Arc length =\frac{135}{360}\times43.982...=16.493... cm.
3. Correct to 3 significant figures: 16.5 cm.

The angles at the centre of all the slices add up to 360°.

Part (a)

1. The 8 angles share the 360° at the centre equally.
2. 360\div8=45°.

Part (b)

1. Each slice uses 24° of the 360°.
2. 360\div24=15 slices.

The perimeter is the arc length plus the two straight edges, which are both radii.

1. Arc length =\frac{100}{360}\times2\times\pi\times6=10.471... cm.
2. The two straight edges are radii: 6+6=12 cm.
3. Perimeter =10.471...+12=22.471... cm.
4. Correct to 1 decimal place: 22.5 cm.

Work out each area in terms of \pi. A semicircle is a sector with angle 180°.

1. Area of A =\frac{80}{360}\times\pi\times6^2=\frac{2}{9}\times36\pi=8\pi\text{ cm}^2.
2. Area of B =\frac{180}{360}\times\pi\times4^2=\frac{1}{2}\times16\pi=8\pi\text{ cm}^2.
3. Both areas are 8\pi\text{ cm}^2, so they have the same area. The bigger angle of B is balanced by the bigger radius of A.

First work out the angle of the shaded sector. The angles at the centre add up to 360°.

1. Angle of the shaded sector =360-72=288°.
2. Area of the whole circle =\pi\times5^2=78.539...\text{ cm}^2.
3. Shaded area =\frac{288}{360}\times78.539...=62.831...\text{ cm}^2.
4. Correct to 1 decimal place: 62.8\text{ cm}^2.

Work out the area of the whole circle in terms of \pi. What fraction of it is 6\pi?

1. Area of the whole circle =\pi\times6^2=36\pi\text{ cm}^2.
2. The sector is \frac{6\pi}{36\pi}=\frac{1}{6} of the circle.
3. So the angle is \frac{1}{6} of 360°: x=360\div6=60.

Find the length of the arc, then decide how many whole 2 m lengths are needed.

1. Arc length =\frac{120}{360}\times2\times\pi\times3=6.283... m.
2. 6.283...\div2=3.14..., so 3 lengths (6 m) are not enough. 4 lengths are needed.
3. Cost =4\times6.80= £27.20

Work out the area of the large sector with radius OB, then take away the small sector with radius OA.

1. OB =6+4=10 m.
2. Large sector =\frac{60}{360}\times\pi\times10^2=52.359...\text{ m}^2.
3. Small sector =\frac{60}{360}\times\pi\times6^2=18.849...\text{ m}^2.
4. Area of path =52.359...-18.849...=33.510...\text{ m}^2.
5. Correct to 1 decimal place: 33.5\text{ m}^2.

Go round the outside of the shape: two sides of the rectangle, the curved arc, and the long bottom edge DE. The dashed line BC is inside the shape.

1. The radius of the quarter circle is CB =6 cm, so CE =6 cm and DE =10+6=16 cm.
2. Arc BE =\frac{1}{4}\times2\times\pi\times6=9.424... cm.
3. Perimeter = DA + AB + arc BE + ED =6+10+9.424...+16=41.424... cm.
4. Correct to 1 decimal place: 41.4 cm.

The arc is \frac{150}{360} of the circumference 2\pi r. Set this equal to 10\pi.

1. Arc length =\frac{150}{360}\times2\pi r=\frac{5}{12}\times2\pi r=\frac{5\pi r}{6}.
2. So \frac{5\pi r}{6}=10\pi, which gives 5r=60.
3. r=12.

Work out the area of the whole quarter circle, then take away the area of triangle OAB.

1. Area of the quarter circle =\frac{1}{4}\times\pi\times40^2=1256.63...\text{ cm}^2.
2. Triangle OAB has a right angle at O: area =\frac{1}{2}\times40\times40=800\text{ cm}^2.
3. Blue area =1256.63...-800=456.63...\text{ cm}^2.
4. To the nearest square centimetre: 457\text{ cm}^2.

The radius of the sector is CB, which is the same length as AD. Find the total area, then round the number of bags up.

1. Area of the rectangle =10\times6=60\text{ m}^2.
2. The radius of the sector is CB =6 m. Area of the sector =\frac{120}{360}\times\pi\times6^2=37.699...\text{ m}^2.
3. Total area =60+37.699...=97.699...\text{ m}^2.
4. 97.699...\div12=8.14..., so 8 bags are not enough. 9 bags are needed.
5. Cost =9\times6.49= £58.41