Circles - Circumference and Area

The diameter is twice the radius.

The radius goes only from the centre to the edge.

Use circumference equals pi times diameter.

Use twice pi times the radius.

Square the radius before multiplying by pi.

Find the radius before using the area formula.

1. r=10\div2=5 cm.
2. A=\pi\times5^2=25\pi\text{ cm}^2.

Use the pi key, multiply by the diameter, then round.

1. C=\pi\times7.2=22.6194\ldots m.
2. To 1 decimal place, the circumference is 22.6 m.

Square 2.5, multiply by pi and round at the end.

1. A=\pi\times2.5^2=6.25\pi=19.6349\ldots\text{ cm}^2.
2. The area is 19.6\text{ cm}^2 to 1 decimal place.

Area measures the surface inside the circle.

1. Area is \pi r^2, so use \pi\times6^2.
2. The result is 36\pi\text{ cm}^2. 2\pi\times6 would find the circumference.

Find the whole circle’s area, then take half.

1. Full area: \pi\times6^2=36\pi\text{ cm}^2.
2. Semicircle area: 36\pi\div2=18\pi\text{ cm}^2.

Half the full circumference gives the curved part. Include the diameter too.

1. Curved part: \pi\times10\div2=5\pi cm.
2. Perimeter: 5\pi+10=25.7079\ldots cm.
3. The perimeter is 25.7 cm to 1 decimal place.

Find one quarter of the circle’s area, then multiply by the price. Keep pi unrounded.

1. Quarter-circle area: \pi\times4^2\div4=4\pi\text{ m}^2.
2. Cost: 8\times4\pi=32\pi=100.5309\ldots pounds.
3. The cost to the nearest penny is £100.53.

One full turn travels one circumference.

1. One turn: \pi\times0.6=0.6\pi m.
2. Twenty turns: 20\times0.6\pi=12\pi=37.6991\ldots m.
3. The distance is 37.7 m to 1 decimal place.

Divide by pi to find the diameter, then halve it.

1. C=2\pi r, so r=47.1\div(2\pi)=7.4961\ldots cm.
2. The radius is 7.5 cm to 1 decimal place.

Use the circumference to find the diameter and radius before finding area.

1. d=26\pi\div\pi=26 cm, so r=26\div2=13 cm.
2. A=\pi\times13^2=169\pi\text{ cm}^2.

Divide the area by pi, then find the positive square root. Use that radius for circumference.

Part (a)

1. \pi r^2=81\pi, so r^2=81.
2. r=\sqrt{81}=9 cm.

Part (b)

1. C=2\pi\times9=18\pi cm.

Halve both diameters, then subtract the smaller circle’s area from the larger circle’s area.

1. The radii are 12\div2=6 cm and 4\div2=2 cm.
2. Outer area: 36\pi\text{ cm}^2. Hole area: 4\pi\text{ cm}^2.
3. Material area: 36\pi-4\pi=32\pi\text{ cm}^2.

The semicircles make one whole circle. Their straight diameters are internal joins, not part of the outside perimeter.

Part (a)

1. Rectangle: 10\times6=60\text{ m}^2. The circle radius is 6\div2=3 m.
2. Total: 60+9\pi=88.2743\ldots\text{ m}^2, which rounds to 88.3\text{ m}^2.

Part (b)

1. The straight edges total 10+10=20 m.
2. The two arcs total one circumference, 6\pi m.
3. Perimeter: 20+6\pi=38.8495\ldots m, which rounds to 38.8 m.

Subtract the pond area from the square area, then divide by the coverage and round up if needed.

1. Square area: 8^2=64\text{ m}^2. Pond radius: 4\div2=2 m.
2. Gravel area: 64-4\pi=51.4336\ldots\text{ m}^2.
3. Bags: (64-4\pi)\div5=10.2867\ldots.
4. Ten bags cover only 50\text{ m}^2, so 11 bags are needed.

The shaded region is the quarter circle with triangle OAB removed. Both radii have the same length.

1. The quarter circle has area \frac14\times\pi\times10^2=25\pi\text{ cm}^2.
2. OA=OB=10 cm and the angle at O is 90^\circ, so triangle OAB has area \frac12\times10\times10=50\text{ cm}^2.
3. The shaded area is 25\pi-50=28.5398\ldots\text{ cm}^2.
4. To 1 decimal place, the shaded area is 28.5\text{ cm}^2.