Circles word problems :
Here we are going to see some practical problems based on the concept circles.
Area of circle = πr²
Circumference of circle = 2πr
Example 1 :
The diameter of a cart wheel is 2.1 m. Find the distance travelled when it completes 100 revolutions.
In order to find distance covered in one revolution, we have to find the circumference of the circle.
radius of wheel = 2.1 / 2 = 1.05 m
Distance travelled when it completes 100 revolutions = 1000 x circumference of the wheel
= 100 x 2 x (22/7) x (1.05)
= 100 x 2 x 22 x 0.15
= 660 m²
Example 2 :
The diameter of a circular park is 98 m. Find the cost of fencing it at $4 per metre.
Diameter of the circular park = 98 m
radius = 98/2 = 49 m
Cost of fencing = $ 4 per meter
length covered by fencing = circumference of the circular park
Circumference of park = 2πr
= 2 (22/7) x 49
= 2 x 22 x 7
= 308 m
Required cost for fencing = 308 x 4 = $1232
Example 3 :
A wheel makes 20 revolutions to cover a distance of 66 m. Find the diameter of the wheel.
A wheel makes 20 revolutions to cover a distance of 66 m
20 x one revolution = 66 m
20 x 2πr = 66
20 x 2 x (22/7) x r = 66
r = 66 x (7/22) x (1/20) x (1/2)
r = 21/40
r = 0.525
Diameter = 2r = 2(0.525) = 1.05 m
Example 4 :
The radius of a cycle wheel is 35 cm. How many revolutions does it make to cover a distance of 81.40 m?
Radius of cycle wheel = 35 cm = 0.35 m
n x Revolution of cycle wheel = 81.40
n x 2πr = 81.40
n x 2 x (22/7) x 0.35 = 81.40
n = 81.40/(2 x 22 x 0.05)
n = 81.40/2.2
n = 37
Hence the cycle wheel has to revolve 374 times to cover the distance.
Example 5 :
The radius of a circular park is 63 m. Find the cost of fencing it at $12 per metre.
Radius of the circular park = 63 m
Length of fencing = Circumference of circular park
= 2πr = 2 x (22/7) x 63
= 2 x 22 x 9
= 396 m
Cost of fencing = $ 12 per meter
Total cost for fencing = 396 x 12
Example 6 :
A goat is tethered by a rope 3.5 m long. Find the maximum area that the goat can graze.
Radius of the circle = Length of the rope
radius r = 3.5 m = 7/2 m
Maximum area grazed by the goat = πr²sq. units.
= (22/7) x (7/2) x (7/2)
= 77/2 = 38.5 sq. m
Maximum area grazed by the goat is 38.5 sq. m.
Example 7 :
The circumference of a circular park is 176 m. Find the area of the park.
Circumference = 176 m (given)
2πr = 176
2 x (22/7) x r = 176
44 r /7 = 176
r = (176 x 7)/44
r = 28 m
Area of the park = πr²
= (22/7) x 28 x 28
= 22 x 4 x 28
= 2464 sq. m.
Example 8 :
A silver wire when bent in the form of a square encloses an area of 121 sq. cm. If the same wire is bent in the form of a circle. Find the area of the circle.
Let a be the side of the square
Area of the square = 121 sq. cm. (given)
a² = 121, a² = 11 cm (11 x 11 = 121)
Perimeter of the square = 4a units
= 4 x 11 cm
= 44 cm
Length of the wire = Perimeter of the square
= 44 cm
The wire is bent in the form of a circle
The circumference of the circle = Length of the wire
Circumference of a circle = 44 cm
2πr = 44
2 x (22/7) x r = 44
r = 44 x (7/22) x (1/2)
r = 7 cm
Area of the circle = πr²
= (22/7) x 7 x 7
Area of the circle = 154 cm²
After having gone through the stuff given above, we hope that the students would have understood "Circles word problems".
Apart from the stuff given above, if you want to know more about "Circles word problems", please click here
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
APTITUDE TESTS ONLINE
ACT MATH ONLINE TEST
TRANSFORMATIONS OF FUNCTIONS
ORDER OF OPERATIONS
MATH FOR KIDS
HCF and LCM word problems