GCSE MATHEMATICS Intermediate Tier, topic sheet. CIRCLES

GCSE MATHEMATICS Intermediate Tier, topic sheet.
CIRCLES
CIRCUMFERENCE = × diameter
AREA = × radius 2
1. A circular disc has diameter 36 mm.
a) Calculate the circumference of the disc.
b) Calculate the area of the disc.
2. A gardener is making a circular lawn of radius 8 m.
a) Calculate the area of the lawn.
b) The gardener wishes to put an edging around the circumference of the lawn.
Calculate the length of edging needed.
3. A circle is cut from a square sheet of card as shown. The card left after the circle is cut out is
wasted.
20 cm
What percentage of each sheet of card is wasted?
4. Find the area of the following shape which shows two semi-circles drawn upon the sides of a
rectangle.
6 cm
10 cm
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5. Find the area of the following shape which shows 3 semi-circles.
6 cm 6 cm
6. Find the perimeters of the following shapes.
a) 6 cm
b)
10 cm
6 cm 6 cm
7. The drive wheel on a lift has a diameter of 2.4 metres.
a) How far will the lift rise when the drive wheel revolves once?
Drive
wheel
b) How many complete times does the wheel revolve
when it raises the lift 30 metres?
Lift
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SOLUTIONS / ANSWERS.
1. Circumference = × diameter = × 36 = 113.097… mm
Area = × radius 2 = × 18 2 = 1017.876 mm 2 .
2. a) Area = × radius 2 = × 8 2 = 201.0619… m 2 .
b) Circumference = × diameter = × 16 = 50.265… m.
3. First calculate the area of wasted card.
Wasted card
= area of square area of circle
= 20 × 20 × 10 2 {radius of circle = 10 cm !}
= 400 314.1592654
= 85.84073464 cm 2 .
For the percentage wasted, we compare the actual wasted with the original square card.
% wasted =
85.84073464
400
× 100% = 21.46 %.
area of square card = 20 × 20 = 400 cm 2 .
4. Area = area of large half-circle + area of small half-circle + area of rectangle
6 cm
=
× 5 2
2
+
× 3 2
2
+ 6 × 10
10 cm
= 113.4070751 cm 2 .
5. Area = area of large half-circle + 2 × area of small half-circle
=
× 6 2
2
× 3 2
+ 2 ×
2
= 84.82300165 cm 2 .
6 cm 6 cm
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6. a) Perimeter = total distance around the outside edges
6 cm
= length of large half-circle + length of small half-circle + 6 + 10
10 cm
=
×10
2
+
× 6
2
+ 16
= 41.13274123 cm.
b) Perimeter = total distance around the outside edges
= length of large half circle + 2 × length of small
half-circle
×12 × 6
= + 2 ×
2
2
= 37.69911184 cm.
6 cm 6 cm
7. Tricky one!
a) If the drive wheel revolves once then the lift will raise the distance of the
circumference. This is because the cable is in direct contact with the circumference
of the circle.
Distance = circumference = × diameter = × 2.4 = 7.539822369 m.
b) 1 revolution of the drive wheel will raise the lift 7.539822369 m.
So, the number of revolutions =
30
7.539822369 = 3.978…
This gives 3 complete revolutions are required.
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