Study MaterialsImportant QuestionsNCERT Solutions – Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

NCERT Solutions – Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3

Chapter 11 – Perimeter and Area Ex 11.3

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    Ex 11.3 Class 7 Maths Question 1.

    Find the circumference of the circles with the following radius. (Take π =\(\frac{22}{7}\))
    (a) 14 cm
    (b) 28 mm
    (c) 21 cm
    Solution:
    (a) Given: Radius (r) = 14 cm
    ∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 14
    = 88 cm
    (b) Given: Radius (r) = 28 mm
    ∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 28
    = 176 mm
    (c) Given: Radius (r) = 21 cm
    ∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 21
    = 132 cm

     

    Ex 11.3 Class 7 Maths Question 2.

    Find the area of the following circles, given that (Take π =\(\frac{22}{7}\))
    (a) radius = 14 mm
    (b) diameter = 49 m
    (c) radius = 5 cm
    Solution:
    (a) Here, r = 14 mm
    ∴ Area of the circle = πr2
    = π × 14 × 14 = \(\frac{22}{7}\) × 14 × 14
    (b) Here, diameter = 49 m 49
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 1

     

     

     

     

    (c) Here, radius = 5 cm
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 2

     

     

     

     

    Ex 11.3 Class 7 Maths Question 3.

    If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π =\(\frac{22}{7}\))
    Solution:
    Given: Circumference = 154 m
    ∴ 2πr = 154
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 3

     

    Ex 11.3 Class 7 Maths Question 4.

    A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds offence. Also find the cost of the rope, if it costs ₹ 4 per metre. (Take π =\(\frac{22}{7}\))
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 4
    Solution:
    Diameter of the circular garden = 21 m
    ∴ Radius = \(\frac{21}{2}\) m
    ∴ Circumference = 2πr = \(2 \times \frac{22}{7} \times \frac{21}{2}\)
    = 66 m
    Length of rope needed for 2 rounds
    = 2 × 66 m = 132 m
    Cost of the rope = ₹4 × 132 = ₹ 528

     

    Ex 11.3 Class 7 Maths Question 5.

    From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
    Solution:
    Radius of the circular sheet = 4 cm
    ∴ Area = πr2 = π × 4 × 4 = 16π cm2
    Radius of the circle to be removed = 3 cm
    ∴ Area of sheet removed = πr2 = 9π cm2
    Area of the remaining sheet
    = (16π – 9π) cm2 = 7π cm2
    = 7 × 3.14 cm2 = 21.98 cm2
    Hence, the required area = 21.98 cm2.

     

    Ex 11.3 Class 7 Maths Question 6.

    Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one metre of the lace costs ₹ 15. (Take π = 3.14)
    Solution:
    Diameter of the table cover = 1.5 m
    ∴ Radius = \(\frac{1.5}{2}\) = 0.75 m
    ∴ Length of the lace = 2πr = 2 × 3.14 × 0.75
    = 4.710 m
    Cost of the lace = ₹ 15 × 4.710 = ₹ 70.65

     

    Ex 11.3 Class 7 Maths Question 7.

    Find the perimeter of the given figure, which is a semicircle including its diameter.
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 5
    Solution:
    Given: Diameter = 10 cm
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 6
    Hence, the required perimeter
    = 25.7 cm. (approx.)

     

    Ex 11.3 Class 7 Maths Question 8.

    Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15 m2. (Take π = 3.14)
    Solution:
    Given:
    Diameter = 1.6 m
    ∴ Radius = \(\frac{1.6}{2}\) = 0.8 m
    Area of the table-top = πr2
    = 3.14 × 0.8 × 0.8 m2
    = 2.0096 m2
    ∴ Cost of polishing = ₹ 15 × 2.0096
    = ₹ 30.14 (approx.)

     

    Ex 11.3 Class 7 Maths Question 9.

    Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π =\(\frac{22}{7}\))
    Solution:
    Length of the wire to be bent into a circle = 44 cm
    2πr = 44
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 7
    Now, the length of the wire is bent into a square.
    Here perimeter of square
    = Circumference of line k
    Length of each side of the square
    \(=\frac{\text { Perimeter }}{4}=\frac{44}{4}=11 \mathrm{cm}\)
    Area of the square = (Side)2 = (11)2 = 121 cm2
    Since, 154 cm2 >121 cm2
    Thus, the circle encloses more area.

     

    Ex 11.3 Class 7 Maths Question 10.

    From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure below). Find the area of the remaining sheet. (Take π = \(\frac{22}{7}\))
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 8
    Solution:
    Radius of the circular sheet = 14 cm
    ∴ Area = πr2 = \(\frac{22}{7}\) × 14 × 14 cm2
    = 616 cm2
    Area of 2 small circles = 2 × πr2
    = 2 × \(\frac{22}{7}\) × 3.5 × 3.5 cm2
    = 77.0 cm2
    Area of the rectangle = l × b
    = 3 × 1 cm2 = 3 cm2
    Area of the remaining sheet after removing the 2 circles and 1 rectangle
    = 616 cm2 – (77 + 3) cm2
    = 616 cm2 – 80 cm2 = 536 cm2

     

    Ex 11.3 Class 7 Maths Question 11.

    A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
    Solution:
    Side os the square sheet = 6 m
    ∴ Area of the sheet = (Side)2 = (6)2 = 36 cm2
    Radius of the circle = 2 cm
    ∴ Area of the circle to be cut out = πr2
    = \(=\frac{22}{7} \times 2 \times 2=\frac{88}{7} \mathrm{cm}^{2}\)
    Area of the left over sheet

    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 9

     

    Ex 11.3 Class 7 Maths Question 12.

    The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)
    Solution:
    Circumference of the circle = 31.4 cm
    2πr = 31.4
    ∴ \(r=\frac{31.4}{2 \times 3.14}\) = 5cm
    Area of the circle = 7πr2 = 3.14 × 5 × 5 = 78.5 cm2
    Hence, the required radius = 5 cm and area = 78.5 cm2.

     

    Ex 11.3 Class 7 Maths Question 13.

    A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Take π = 3.14)
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 10
    Solution:
    Diameter of the flower bed = 66 m .
    ∴ Radius = \(\frac{66}{2}\) = 33 m
    Let r1 = 33 m
    Width of the path = 4 m
    Radius of the flower bed included path
    = 33 m + 4 m = 37m
    Let r2 = 37m
    Area of the circular path = \(\pi\left(r_{2}^{2}-r_{1}^{2}\right)\)
    = 3.14 (372– 332)
    = 3.14 × (37 + 33) (37 – 33) [Y a2 – b2 = (a + b)(a-b)] = 3.14 × 70 × 4 = 879.20 m2
    Hence, the required area = 879.20 m2

     

    Ex 11.3 Class 7 Maths Question 14.

    A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler can water the entire garden?
    [Take π = 3.14] Solution:
    Area of the flower garden = 314 m2
    Radius of the circular portion covered by the sprinkler = 12 m
    ∴ Area = 7πr2 = 3.14 × 12 × 12
    = 3.14 × 144 m2 = 452.16 m2
    Since 452.16 m2 > 314 m2
    Yes, the sprinkler will water the entire garden.

     

    Ex 11.3 Class 7 Maths Question 15.

    Find the circumference of the inner and the outer circles, shown in the given figure. (Take π = 3.14)
    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 11
    Solution:
    Radius of the outer circle = 19 m
    ∴ Circumference of the outer circle = 2πr
    = 2 × 3.14 × 19 = 3.14 × 38 m
    = 119.32 m
    Radius of the inner circle
    = 19m – 10m = 9m
    ∴ Circumference = 2πr = 2 × 3.14 × 9
    = 56.52 m
    Here the required circumferences are 56.52 m and 119.32 m.

     

    Ex 11.3 Class 7 Maths Question 16.

    How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = \(\frac{22}{7}\))
    Solution:
    Radius of the wheel = 28 cm
    ∴ Circumference = 2πr = 2 × \(\frac{22}{7}\) × 28 = 176 cm
    Number of rotations made by the wheel in going 352 m or 35200 cm
    \(=\frac{35200}{176}=200\)
    Hence, the required number of rotation = 200.

     

    Ex 11.3 Class 7 Maths Question 17.

    The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
    Solution:
    Length of minute hand = 15 cm
    ∴ Radius = 15 cm
    Circumference = 2πr
    = 2 × 3.14 × 15 cm = 94.2 cm
    Since the minute hand covers the distance in 1 hour equal to the circumference of the circle. Here the required distance covered by the minute hand = 94.2 cm.

     

    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Q1

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    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Q3

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    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Q6

    NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Q7

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