Circles and Semicircles

A quarter circle is \frac{1}{4} of the whole circle.

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

The area of a whole circle is \pi r^2. A semicircle is half of the circle.

The curved edge is \frac{1}{4} of the circumference of the whole circle, 2\pi r.

Work out the area of the whole circle, then halve it.

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. One piece is \frac{1}{4} of the circle: 254.46...\div4=63.617...\text{ cm}^2.
4. Correct to 1 decimal place: 63.6\text{ cm}^2.

The edge of the rug is the curved edge and the straight edge. The straight edge is the diameter.

1. Curved edge =\pi\times1.2\div2=1.884... m.
2. Add the straight edge, the diameter: 1.884...+1.2=3.084... m.
3. Correct to 3 significant figures: 3.08 m.

The distance once round the track is the circumference, \pi d.

Part (a)

1. Circumference of the track =\pi\times60=188.49... m.
2. The 4 flags cut it into 4 equal arcs, each a quarter of the circle: 188.49...\div4=47.12... m.
3. Correct to 1 decimal place: 47.1 m.

Part (b)

1. The mean distance is the distance once round the track divided by the number of gaps between flags.
2. The track is the same circle, and there are still 4 flags, so still 4 gaps.
3. So the mean distance has not changed, even though some gaps are now longer and some shorter.

Go all the way round: the curved edge and the two straight edges. Both straight edges are radii.

1. Curved edge =\frac{1}{4}\times2\times\pi\times60=94.247... cm.
2. The two straight edges are radii: 60+60=120 cm.
3. Length of edging =94.247...+120=214.247... cm.
4. Correct to the nearest centimetre: 214 cm.

Which length goes into \pi r^2?

1. The area of a circle is \pi r^2, where r is the radius.
2. The radius is half of the diameter: 10\div2=5 cm, but Sam used 10.
3. The area of the semicircle is \pi\times5^2\div2=39.3\text{ cm}^2 to 1 decimal place. Dividing by 2 was right.

Work out the area of the square, then take away the area of the quarter circle. The radius of the quarter circle is AB.

1. Area of the square =8\times8=64\text{ cm}^2.
2. The quarter circle has radius AB =8 cm: area =\frac{1}{4}\times\pi\times8^2=50.265...\text{ cm}^2.
3. Shaded area =64-50.265...=13.734...\text{ cm}^2.
4. Correct to 1 decimal place: 13.7\text{ cm}^2.

Add only the edges round the outside. The dashed line is inside the sign, so it is not part of the ribbon.

1. Curved edge of the semicircle =\pi\times80\div2=125.66... cm.
2. Straight edges round the outside: 100+80+100=280 cm.
3. Length of ribbon =125.66...+280=405.66... cm.
4. Correct to the nearest centimetre: 406 cm.

Find the length of the curved edge, then work out how many whole 1.5 m lengths are needed.

1. Curved edge =\frac{1}{4}\times2\times\pi\times5=7.853... m.
2. 7.853...\div1.5=5.23..., so 5 lengths are not enough. 6 lengths are needed.
3. Cost =6\times4.60= £27.60

Work out the area of the rectangle, then take away the two semicircles. The radius of each semicircle is half of 12 cm.

1. Area of the rectangle =30\times10=300\text{ cm}^2.
2. Radius of each semicircle =12\div2=6 cm. Area of one semicircle =\pi\times6^2\div2=56.548...\text{ cm}^2.
3. Two semicircles =2\times56.548...=113.097...\text{ cm}^2.
4. Shaded area =300-113.097...=186.902...\text{ cm}^2.
5. Correct to 3 significant figures: 187\text{ cm}^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.

Use Pythagoras' theorem in triangle PQR to find RQ first. PR is the longest side, because it is opposite the right angle.

1. PR is the longest side of the right-angled triangle, so \text{RQ}^2=11^2-7^2=121-49=72.
2. So RQ =\sqrt{72}=8.485... cm. This is the radius of the quarter circle.
3. Area of the quarter circle =\frac{1}{4}\times\pi\times8.485...^2=\frac{1}{4}\times\pi\times72=56.548...\text{ cm}^2.
4. Correct to 3 significant figures: 56.5\text{ cm}^2.

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.

Split the lawn into the triangle, the rectangle and the semicircle. The radius of the semicircle is half of BC. Round the number of bags up.

1. Area of the triangle =\frac{1}{2}\times6\times8=24\text{ m}^2.
2. Area of the rectangle =14\times8=112\text{ m}^2.
3. Radius of the semicircle =8\div2=4 m. Area =\pi\times4^2\div2=25.132...\text{ m}^2.
4. Total area =24+112+25.132...=161.132...\text{ m}^2.
5. 161.132...\div20=8.05..., so 8 bags are not enough. 9 bags are needed.
6. Cost =9\times12.40= £111.60