Question 1: Find the circumference and area of a circle of radius 4.2 \ cm .

Answer:

Radius = 4.2 \ cm

Circumference of circle = 2 \pi r = 2 \times \frac{22}{7} \times 4.2 = 26.4 \ cm

Area of circle = \pi r^2 = \frac{22}{7} \times (4.2)^2 = 55.44 \ cm^2

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Question 2: Find the circumference of a circle whose area is 301.84 \ cm^2 .

Answer:

Area = 301.84 \ cm^2

Therefore \pi r^2 = 301.84

\Rightarrow r^2 = \frac{7}{22} \times 301.84 = 96.04

\Rightarrow r = 9.8 \ cm

Therefore circumference of the circle = 2 \pi r = 2 \times \frac{22}{7} \times 9.8 = 61.6 \ cm

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Question 3: Find the area of a circle whose circumference is 44 \ cm .

Answer:

Circumference = 44 \ cm

Therefore 2 \pi r = 44

\Rightarrow r = 2 \times \frac{7}{22} \times 44 = 28 \ cm

Therefore Area = \pi r^2 = \frac{22}{7} \times (28)^2 = 2764 \ cm^2

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Question 4: The circumference of a circle exceeds the diameter by 16.8 \ cm . Find the circumference of the circle.

Answer:

Given: Circumference - diameter = 16.8

\Rightarrow 2 \pi r - 2r = 16.8

\Rightarrow r = \frac{16.8}{2\pi -2} = \frac{16.8 \times 7}{2 \times 22 - 2 \times 7} = 3.92

Therefore Circumference = 2 \pi r = 2 \times \frac{22}{7} \times 3.92 = 24.64 \ cm

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Question 5: A horse is tied to a pole with 28 \ m long string. Find the area where the horse can graze.

Answer:

Area that the horse can graze = \pi r^2 = \frac{22}{7} \times (28)^2 = 2464 \ m^2

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Question 6: A steel wire when bent in the form of a square encloses an area of 121 \ cm^2 . If the same wire is bent in the form of a circle, find the area of the circle.

Answer:

Area of square = 121 \ cm^2

Let the side of the square = a

\Rightarrow a^2 = 121 \Rightarrow a = 11 \ cm

Therefore Perimeter = 4 \times 11 = 44 \ cm

Let r be the radius of the circle

Therefore 2 \pi r = 44

\Rightarrow r = \frac{7 \times 44}{2 \times 22} = 7 \ cm

Therefore Area of circle = \pi r^2 = \frac{22}{7} \times (7)^2 = 154 \ cm^2

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Question 7: The diameters of the front and rear wheels of a tractor are 80 \ cm and 2 \ m respectively. Find the number of revolutions that rear wheel will make to cover the distance which the front wheel covers in 1400 revolutions.

Answer:

Diameter of front wheel = 80 \ cm

Diameter of rear wheel = 2 \ m

Distance covered by front wheel = 1400 \times 2 \times \frac{22}{7} \times 0.40 = 3520 \ m

Let no of revolution that the rear wheel = n

Therefore n \times 2 \times \frac{22}{7} \times 1 = 3520

\Rightarrow n = \frac{7 \times 3520}{2 \times 22} = 560

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Question 8: A copper wire when bent in the form of a square encloses an area of 121 \ cm^2 . lf the same wire is bent into the form of a circle, find the area of the circle.

Answer:

Area of square = 121 \ cm^2

Let the side of the square = a

\Rightarrow a^2 = 121 \Rightarrow a = 11 \ cm

Therefore Perimeter = 4 \times 11 = 44 \ cm

Let r be the radius of the circle

Therefore 2 \pi r = 44

\Rightarrow r = \frac{7 \times 44}{2 \times 22} = 7 \ cm

Therefore Area of circle = \pi r^2 = \frac{22}{7} \times (7)^2 = 154 \ cm^2

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Question 9: The circumference of two circles are in the ratio 2: 3 . Find the ratio of their areas.

Answer:

Let the radius of first circle = r_1

Let the radius of second circle = r_2

Therefore \frac{2\pi r_1}{2 \pi r_2} = \frac{2}{3} 

\Rightarrow \frac{r_1}{r_2} = \frac{2}{3} 

Therefore Ratios of their areas = \frac{\pi {r_1}^2}{\pi {r_2}^2} = \Big( \frac{r_1}{r_2} \Big)^2 = \frac{4}{3} 

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Question 10: The side of a square is 10 \ cm . Find the area of circumscribed and inscribed circles.

Answer:2019-01-02_22-45-04

When square is inscribed in the circle

Therefore diameter = 10\sqrt{2}  

Therefore radius of circle = 5\sqrt{2}

Therefore are of circle = \pi (5\sqrt{2})^2 = 50\pi = 157.143 \ cm^2

When the circle is inscribed in the square

Diameter = 10 \ cm

Therefore radius = 5 \ cm

Therefore Area of circle = \pi (5)^2 = 25 \pi = 78.57 \ cm^2

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Question 11: The sum of the radii of two circles is 140 \ cm and the difference of their circumferences is 88 \ cm . Find the diameters of the circles.

Answer:

Let the two radii be r_1 and r_2

Therefore r_1 + r_2 = 140 … … … … … i)

2\pi r_1 - 2\pi r_2 = 88 … … … … … ii)

From i) and ii) r_2 = 140 - r_1

Therefore 2\pi (r_1-r_2) = 88

\Rightarrow 2 \pi (r_1 - (140 - r_1)) = 88

2\pi (2r_1 - 140) = 88

2r_1 = \Big( \frac{88}{2} \times \frac{22}{7} \Big) + 140

\Rightarrow 2r_1 = 154

\Rightarrow r_1 = 77 \ cm

Therefore r_2 = 140 - r_1 = 140 - 77 = 63 \ cm

Diameter of circles are 154 \ cm and 126 \ cm

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Question 12: Find the area of the circle in which a square of area 64 \ cm^2 is inscribed. ( Take \pi = 3.14)

Answer:

Area of square = 64 \ cm^2

\Rightarrow Side of square= 8 \ cm

\Rightarrow Diameter of circle = 8\sqrt{2} \ cm

\Rightarrow Radius of the circle = 4\sqrt{2} \ cm

Hence Area of circle = \pi (4\sqrt{2})^2 = 100.48 \ cm^2

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Question 13: A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be Rs. \ 2640 at the rate of Rs. \ 12 per meter. Then, the field is to be thoroughly ploughed at the cost of Rs. \ 0.50 \ per \ m . What is the amount required to plough the field? ( Take \pi = \frac{22}{7} )

Answer:

Cost of fencing = 2640 \ Rs.

Circumference of the field = \frac{2640}{12} = 220 \ m

Therefore 2 \pi r = 220

\Rightarrow r = \frac{220 \times 7}{2 \times 22} = 35 \ m

Area of the field = \pi r^2 = \pi (35)^2 = 3850 \ m^2

Therefore cost of ploughing = 3850 \times 0.5 = 1925 \ Rs.

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Question 14: If square is inscribed in a circle, find the ratio of the areas of the circle and the square.

Answer:2019-01-03_20-05-12

Let the side of the square = a

Therefore diameter of circle = a\sqrt{2}

Therefore radius of the circle = \frac{a}{\sqrt{2}}

Therefore ratios of their areas = \frac{\pi (\frac{a}{\sqrt{2}})^2}{a^2} = \frac{\pi}{2}

Hence the ratios of their area is \pi : 2

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Question 15: A park is in the form of a rectangle 120 \ m \times 100 \ m . At the center of the park there is a circular lawn. The area of park excluding lawn is 8700 \ m^2 . Find the radius of the circular lawn. ( Take \pi = \frac{22}{7} )

Answer:2019-01-03_20-18-53.jpg

Let the radius of the circle = r

Area of the park = 100 \times 120 = 12000 \ m^2

Are of circle = \pi r^2

Therefore 12000 = \pi r^2 = 8700

\Rightarrow \pi r^2 = 12000 - 8700

\Rightarrow \pi r^2 = 3300

\Rightarrow r^2 = 1050

Therefore r = 32.40 \ m

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Question 16: The radii of two circles are 8 \ cm and 6 \ cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.

Answer:

Let the radius of the circle = r

Therefore \pi r^2 = \pi (8)^2 + \pi (6)^2

\Rightarrow r^2 = 64 + 36

\Rightarrow r = 10 \ cm

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Question 17: The radii of two circles are 19 \ cm and 9 \ cm respectively. Find the radius and area of the circle which has its circumference equal to the sum of the circumferences of the two circles.

Answer:

Let the radius of the circle = r

Therefore 2\pi r = 2\pi (19) + 2\pi (9)

\Rightarrow r = 19 + 9 = 28 \ cm

Therefore Area = \pi (28)^2 = 2464 \ cm^2

Circumference = 2 \pi (28) = 176 \ cm

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Question 18: A car travels 1-kilometer distance in which each wheel makes 450 complete revolutions. Find the radius of its wheels.

Answer:

Let the radius of the wheel = r

Therefore 2 \pi r \times 450 = 1000

\Rightarrow r = \frac{1000 \times 7}{2 \times 450 \times 22} = 0.3535 \ m \ or \  35.35 \ cm

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Question 19: The area enclosed between the concentric circles is 770 \ cm^2 . If the radius of the outer circle is 21 \ cm , find the radius of the inner circle.

Answer:2019-01-03_21-24-39.jpg

Let the radius of the inner circle = r

Therefore \pi (21)^2 - \pi (r)^2 = 770

\Rightarrow \pi r^2 = \frac{22}{7} \times 21^2 - 770

\Rightarrow \pi r^2 = 616 \ cm^2

Therefore r^2 = \frac{7}{22} \times 616 = 196

\Rightarrow r = 14 \ cm

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Question 20: The wheel of a car is making 5 revolutions per second. If the diameter of the wheel is 84 \ cm , find its speed in km/hr . Give your answer, correct to nearest km.

Answer:

No. of revolutions per second = 5

Radius of the wheel = 42 \ cm

Therefore circumference of the wheel = 2 \pi r = 2 \times \frac{22}{7} \times 42 = 264 \ cm

Hence the distance covered in 1 second = 5 \times 264 = 1320 \ cm \ or \  13.20 \ m

Hence distance covered in one hour = 13.20 \times 3600 = 47520 \ m \ or \ 47.52 \ km

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Question 21: A sheet is 11 \ cm long and 2 \ cm wide. Circular pieces of 0.5 \ cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.

Answer:

Number of disks by length = 22

No of disks by breadth = 4

Therefore total number of disks that can be cut = 4 \times 22 = 88

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Question 22: A copper wire when bent in the form of an equilateral triangle has area 121\sqrt{3} \ cm^2 . If the same wire is bent into the form of a circle, find the area enclosed.

Answer:

Let the side of the equilateral triangle = a

Therefore h = \sqrt{a^2 - (\frac{a}{2})^2} = \frac{\sqrt{3}}{2} a

Therefore 121 \sqrt{3} = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a

\Rightarrow a^2 = 2^2 \times 11^2

\Rightarrow a = 22 \ cm

Therefore perimeter = 66 \ cm

Let the radius of the circle = r

Therefore 2\pi r = 66

\Rightarrow r = \frac{7 \times 66}{2 \time 22} = 10.5 \ cm

Area of circle = \pi (10.5)^2 = 346.5 \ cm^2

Click here for Second half of Exercise 16

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