Grade 8 Problems and Questions on Circles with Answers

Detailed solutions and full explanations to grade 8 problems and questions on circles are presented.

  1. 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?

    circles problem 1 .

    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 = Pi (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 = Pi (2.5)2

    The ratio of A to B is given by

    A / B = Pi (10)2 / Pi (2.5)2

    Simplify

    = (10)2 / (2.5)2

    = (10 / 2.5)2 = 42 = 16



  2. 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.

    .

    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

    Pi (2)2 = 4 Pi

    The total area of the garden is equal to

    16 + 4 Pi = 28.56 square meters. (with Pi = 3.14)



  3. 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

    Pi 122 = 144 Pi = 452 square meters

  4. 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).

    circles problem 4.

    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

    Pi 112 - Pi 102 = 121 Pi - 100 Pi = 21 Pi square meters



  5. 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

  6. 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 r2 = 5

    r2 = 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

  7. 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 r2

    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 r2

    Change in area

    Change = Area after increase - Area before increase = 1.44 Pi r2 - Pi r2

    = Pi r2 (1.44 - 1) = 0.44 Pi r2

    Percent change in area

    (Change / Area before change) 100% = (0.44 Pi r2/ Pi r2) 100%

    = 0.44 100% = 44%

  8. 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

  9. 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?

    circles problem 9.



    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 = Pi r2

    Solve for r2

    r2 = 100 / Pi

    We now use Pythagora's theorem in triangle ABC to find the length of AC

    AC2 = AB2 + AC2

    Both lengths of AB and BC are equal to r. Hence

    AC2 = r2 + r2

    = 2 r2 = 2 (100 / Pi) = 200 / Pi

    AC = √ (2 100 / Pi) = √ (100) √ (2 / Pi)

    = 10 √ (2 / Pi) cm

  10. 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

    (Pi 2 Ra) / (Pi 2 Rb) = 3/1 = Ra / Rb

    which gives

    Ra = 3 Rb

    We now express the areas Aa and Ab of the two cricles

    Aa = Pi Ra2 = Pi (3 Rb)2

    Ab = Pi Rb2

    The ratio of the areas is given by

    Aa / Ab = [ Pi (3 Rb)2 ] / [ Pi Rb2 ]

    = Pi 9 Rb2 / Pi Rb2

    = 9

    The ratio is 9:1


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Updated: 9 April 2009 (A Dendane)

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