Grade 8 Problems and Questions on Circles with Detailed Solutions
Detailed solutions and full explanations to grade 8 problems and questions on circles are presented.

The three circles C1, C2 and C3 have their centers O1, O2 and O3 on the line L and are all tangent at the same point. If the diameter of the largest circle is 20 units, what is the ratio of the area of the largest circle to the area of the smallest circle?
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Solution
The diameter of circle C1 is equal to 20 units and therefore its radius is equal to 10 units. The area A of the largest circle C1 is equal to
A = π (10)^{2}
The diameter of circle C2 is equal to the radius of circle C1 which is 10 unis. The diameter of circle C3 is equal to the radius of circle C2 which is 5 units. The radius of circle C3 is equal to 2.5. We now calculate the area B of the smallest circle C3.
B = π (2.5)^{2}
The ratio of A to B is given by
A / B = π (10)^{2} / π (2.5)^{2}
Simplify
= (10)^{2} / (2.5)^{2}
= (10 / 2.5)^{2} = 4^{2} = 16

Mrs Parkinson's garden is made up of 4 squares and 2 semicircles as shown below. Each small square has an area of 4 square meters. Find the total area of the garden.
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Solution
The garden is made up of 4 squares and 2 semi circles. The total area of the 4 squares is equal to
4 × 4 = 16 square meters
Since the area of a small square is equal to 4 square meters, the side of a small square is equal to 2 meters. Also the radius of a semicircle is equal to length of the side of the square which 2 meters. The area of the surface inside the two semicircles is equal to the area inside one whole circle and is equal to
π (2)^{2} = 4 π
The total area of the garden is equal to
16 + 4 π = 28.56 square meters. (with π = 3.14)

A water sprinkler can spray water at a maximum distance of 12 m in all directions. What area of the garden can this sprinkler irrigate? round your answer to the nearest square meter.
Solution
One full rotation of the sprinkler would irrigate an area enclosed by a circle of radius 12 m. Hence the area of the garden that the sprinkler can irrigate is given by
π 12^{2} = 144 π = 452 square meters

A circular garden with a diameter of 10 meters is surrounded by a walkway of width 1 meter. Find the area of the walkway (shaded part).
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Solution
The walkway is enclosed between a smaller circle of radius 10 meters and a larger circle of radius 11 meters and therefore the area of the walkway is equal to the area enclosed by the larger circle minus the area enclosed by the smaller circle and is equal to
π × 11^{2}  π × 10^{2} = 121 π  100 π = 21 π square meters

A circular pizza costs $19.99. What is the cost of 1 square centimeter if the diameter of the pizza is 36 cm?
Solution
The $19.99 is the total cost of the whole pizza whose area is
Pi × (36/2)^{2} = 1017 square cm
The cost of 1 square cm is equal to
19.99 / 1017 = $0.02 = 2 cents per square cm

How much fencing is needed for the Robinsons circular flower garden that has an area of 5 square meters?(round your answer to the nearest meter.)
Solution
The fencing will be put around the circular garden and therefore the length of the fencing is equal to its circumference. The radius r of the garden is found using the area
Pi × r^{2} = 5
r^{2} = 5 / Pi
r = √ (5 / Pi) = 1.26 meters
The circumference of the garden is equal to
2r × Pi= 2.52 Pi = 8 meters (rounded to the nearest meter)
The length of the fencing needed is equal to 8 meters

The radius of circular disk is increased by 20%. What is the percent increase in the area of the disk?
Solution
If r is the radius of the disk, its area (before increase) is equal to
Pi r^{2}
If r is increased by 20% it becomes
r + 20% r = r + (20/100) r = 4 + 0.2 r = 1.2 r
and the area (after increase) of the disk becomes
Pi (1.2 r)^{2} = 1.44 Pi r^{2}
Change in area
Change = Area after increase  Area before increase = 1.44 Pi r^{2}  Pi r^{2}
= Pi r^{2} (1.44  1) = 0.44 Pi r^{2}
Percent change in area
(Change / Area before change) × 100% = (0.44 Pi r^{2}/ Pi r^{2}) × 100%
= 0.44 × 100% = 44%

A circular table has a dimeter of 100 inches. A circular tablecloth hangs over the table 15 inches around the table. What is the area of the tablecloth?
Solution
If the table has a diameter of 100 inches and the table cloth hangs 15 inches around the table, then the diameter of the tablecloth is equal to
100 + 15 + 15 = 130 inches
The area of the tablecloth is equal to
Pi (130 / 2) = 4,225 Pi = 13,267 square inches

ABCD is a square with one vertex at the center of the circle and two vertices on the circle. What is the length of AC if the area of the circle is 100 square cm?
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Solution
Note that the length of the side of the square AB has equal length as the radius r of the circle. Since the area is given, we can write
100 = π r^{2}
Solve for r^{2}
r^{2} = 100 / π
We now use Pythagora's theorem in triangle ABC to find the length of AC
AC^{2} = AB^{2} + AC^{2}
Both lengths of AB and BC are equal to r. Hence
AC^{2} = r^{2} + r^{2}
= 2 r^{2} = 2 (100 / π) = 200 / π
AC = √ (2 × 100 / π) = √ (100) × √ (2 / π)
= 10 √ (2 / π) cm

The ratio of the perimeter of circle A to the perimeter of circle B is 3:1. What is the ratio of the area of circle A to the area of circle B?
Solution
Let Ra the radius of circle A and Rb the radius of circle B. The ratio of perimeter of circle A to perimeter of circle B gives
(π 2 Ra) / (π 2 Rb) = 3/1 = Ra / Rb
which gives
Ra = 3 Rb
We now express the areas Aa and Ab of the two cricles
Aa = π Ra^{2} = π (3 Rb)^{2}
Ab = π Rb^{2}
The ratio of the areas is given by
Aa / Ab = [ π (3 Rb)^{2} ] / [ π Rb^{2} ]
= π 9 Rb^{2} / π Rb^{2}
= 9
The ratio is 9:1

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