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Area Related to Circles Test - 12

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Area Related to Circles Test - 12
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Weekly Quiz Competition
  • Question 1
    1 / -0

    The area of a circle is 38.5 sq. cm. Its circumference is


    Solution

    Given area of circle = πr²

    r²= 38.5/π

    r = 3.5

    Circumference of circle = 2πr

    = 2 × π × 3.5

    = 22 

  • Question 2
    1 / -0

    The area of a circle is 2464 sq. cm, then its diameter is given by

    Solution

    Area of circle= 2464

    πr²= 2464

    r²=(2464×7)/22

    r²=784

    r= 28

    Diameter= 2 × 28 = 56 cm 

  • Question 3
    1 / -0

    The length of the wire is 66cm. The number of circles of circumference 13.2cm can be made from the wire is


    Solution

    let the no.of circles with 13.2 cm circumference be ' x'

    x × circumference of each circle=66cm

    x(13.2cm)=66cm

    x=66/13.2

    =5 circles 

  • Question 4
    1 / -0

    The perimeter of a semicircular protractor whose radius is 7cm is

    Solution

    Perimeter of circular protractor = πr + 2r = (22/7 × 7) + (2 × 7) = 36 cm 

  • Question 5
    1 / -0

    The area of a sector of a circle with sector angle θ is given by


    Solution

    The area of a sector of a circle with sector angle θ is given by πr2θ/360°, where r = radius of the circle 

  • Question 6
    1 / -0

    The area of a square inscribed in a circle of diameter ‘d’ is


    Solution

    Area of square = (BD)²/2

    = d²/2 
  • Question 7
    1 / -0

    The circumference of a circle whose diameter is 4.2cm is

    Solution

    Given diameter= 4.2

    Circumference = πd

    = π × 4.2

    = 13.2 cm

  • Question 8
    1 / -0

    If the sum of the areas of two circles with radii r1 and r2 is equal to the area of the circle of radius R, then


    Solution

    According to the question

    πR²=πr1²+πr2² = π(r1²+r2²)

    R² = r1²+r2² 

  • Question 9
    1 / -0

    The radius of the circle whose area is equal to the sum of the areas of the two circles of radii 24cm and 7cm is

    Solution

    Let required radius be R.

    Then according to the questions.

    πR²=πr1²+πr2² 

    R²= 24²+7²

    =576+49=625

    R=25 cm 

  • Question 10
    1 / -0

    If ‘r’ is the radius of a circle, then it's circumference is given by

    Solution

    If the radius of a circle is given, the circumference or perimeter can be calculated using the formula below:-

     Circumference =2πr 
  • Question 11
    1 / -0

    The perimeter of a protractor is 

    Solution

    Let the radius of the protractor be  r 

     ∴ Perimeter of protractor = Perimeter of semicircle + Diameter of a semicircle

    ⇒ Perimeter of protractor = πr + 2r 

  • Question 12
    1 / -0

    The distance covered by a circular wheel of diameter  ‘d’ in 100 revolutions is

    Solution

    Number of revolution = total distancecircumference of wheel 

    100 = total distance/πd

    Total distance = 100πd 

  • Question 13
    1 / -0

    The radius of the circle whose circumference is equal to the sum of the circumferences of the two circles of radii 24cm and 7cm is

    Solution

    Let required radius be R.

    Then according to the question,

    2πR = 2πr1+2πr2

    = 2π(r1 + r2)

    ⇒R = r1 + r2

    ⇒R = 24 + 7 = 31cm 

  • Question 14
    1 / -0

    The circumference of a circle exceeds its diameter by 120cm, then its radius is 

    Solution

    Let radius be r, then according to the question,

    2πr = 2r + 120

    2r(π-1) = 120

    r = 28 cm 

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