Q14. A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square.
Sol. Given radius of cicle = 28 cm
We know that the circumference (C) of a circle having radius r is given by C = 2 π r
Therefore, Length of the wire = Circumference of the circle having radius r
$$ =2 \pi r=2 \times \frac {22}{7} \times 28 =176 \text { cm} $$Let’s assume that side of new formed square be x cm.
Then, Perimter of square = Circumference of the circle having radius r.
$$ \Rightarrow 4 x =176 \text { cm} $$ $$ \Rightarrow x=\frac{176}{4} = 44 \text { cm} $$ Hence, lenght of side of new formed square is 44 cm.
Q15. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
Sol. Given that total distance covered in 5000 revolutions = 11 km
$$ =11 \times 1000 \text { m} \quad \quad [\because 1 \text { km} = 1000 \text { m}] $$ $$ =11000 \text { m} $$ $$ \therefore \text {Distance covered in 1 revolution } =\frac{11000}{5000} $$ $$ =\frac{11}{5} \text { m} $$ $$ \text {Distance covered in 1 revolution } = \text {Circumference of the wheel } $$ $$ \Rightarrow \frac{11}{5}= \pi d \quad \quad [\text {Using : } C = \pi d \text { where d is diamenter of circle}] $$$$ \Rightarrow d=\frac{11 \times 7}{5 \times 22} \text { m} $$ $$ \Rightarrow d=0.7 \text { m} $$ Hence, the diameter of the wheel is 0.7 m.
Q16. A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling.
Sol. Given the diameter of the wheel =60 cm
Distance covered by the wheel in 1 revolution = Circumference of the wheel
\begin{array}{l} \text {Distance covered by the wheel in 1 revolution } =\pi d \quad \quad [\text {Using : } C = \pi d \text { where d is diamenter of circle}] \\ =\frac {22}{7} \times 60 \text { cm} \\\text {Distance covered by the wheel in 140 revolutions }=\frac {22}{7} \times 60 \times 140 =26400 \text { cm} \\ \text {As 140 revolutions are covered in 1 minutes.} \\ \therefore \text {Distance covered by wheel in 1 minute} = 26400 \text { cm} \\ \Rightarrow \text {Distance covered by wheel in 60 minutes} = 26400 \times 60 \text { cm} \\ \Rightarrow \text {Distance covered by wheel in 1 hr} = 26400 \times 60 \times \frac {1}{100000} \text { km} \quad \quad [\because 1 \text { cm} = \frac {1}{100000} \text { km}] \\ \Rightarrow \text {Distance covered by wheel in 1 hr} =15.84 \text { km} \\ \text {Hence, The cycling speed of the boy is 15.84 km/hr}\end{array}Q17. The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour?
Sol.
Given diameter of the wheel of the bus = 140 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
\begin{array}{l} \Rightarrow C =\frac{22 \times 140}{7} = 440 \text { cm} \\ \therefore \text {Bus cover 440 cm in one revolution.} \\\text {With a speed of 66 km per hour the bus will cover a distance of 66 km in 1 hr.} \\ \Rightarrow \quad \text {Distance covered in 1 hour or 60 mins} = 66 \text { km} \\ \Rightarrow \quad \text {Distance covered in 1 hour or 60 mins} = 66 \times 1000 \text { m} \quad \quad [\because 1 \text { km} = 1000 \text { m}] \\ \Rightarrow \quad \text {Distance covered in 1 min} = \frac {66 \times 1000}{60} \text { m} \\ \Rightarrow \quad \text {Distance covered in 1 min} = 1100 \text { m} \\ \Rightarrow \quad \text {Distance covered in 1 min} = 110000\text { cm} \\\text {Bus covers 440 cm in one revolution.} \\ \therefore \text {No. of revolution required to cover 110000 cm.} \\ = \frac {110000}{440} \\\end{array}= 250 revolutions
Hence, the bus need to take 250 revolutions per minute to keep the speed of 66 km/hr
Q18. A water sprinkler in a lawn sprays water as far as 7 m in all directions. Find the length of the outer edge of wet grass.
Sol. The wet grass forms a circular area of radius 7 m.
$$ \therefore \text {The length of the outer edge of the wet grass} = \text {Circumference of the circular area having radius 7 m } $$ $$ =2 \times \frac{22}{7} \times 7 =44 \text{ m} $$Q19. A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 660 cm, then find the width of the parapet.
Sol.
\begin{array}{l} \text {Given that diameter of the well } =150 \text { cm} \\ \text {Radius of well} =\frac {150}{2}=75 \text {cm} \\ \text {Let’s assume that the width of the stone parapet be x cm.} \\ \therefore \text {Radius of outer circle including stone parapet } = (75 +x) \text { cm} \\\text {Given length of the outer edge of the parapet } =660 \text { cm} \\ \Rightarrow \text {Circumference of outer circle } = 660 \text { cm} \\ \Rightarrow \text 2 \pi \times (75+x)= 660 \text { cm} \\ \Rightarrow \text 75+x= \frac {660 \times 7}{2 \times 22} \text { cm} \\ \Rightarrow \text 75+x= 105 \text { cm} \\ \Rightarrow \text x= 105 -75 \text { cm} \\ \Rightarrow \text x= 30 \text { cm} \\ \text {Hence, the width of the parapet is 30 cm} \end{array}Q20. An Ox in a kolhu (an oil processing apparatus) is tethered to a rope 3 m long. How much distance does it cover in 14 rounds?
Sol. Radius of the circular path traced by the Ox in the Kolhu = 3 m.
Distance covered by the Ox in 1 round = Circumference of the circular path having radius 3 m
=2 π r
$$ =2 \times \frac{22}{7} \times 3 \text { m} $$ $$ \Rightarrow \text {Distance covered in 14 rounds }==2 \times \frac{22}{7} \times 3 \times 14 \text { m} $$ =264 m
Sol. Given radius of cicle = 28 cm
We know that the circumference (C) of a circle having radius r is given by C = 2 π r
Therefore, Length of the wire = Circumference of the circle having radius r
$$ =2 \pi r=2 \times \frac {22}{7} \times 28 =176 \text { cm} $$Let’s assume that side of new formed square be x cm.
Then, Perimter of square = Circumference of the circle having radius r.
$$ \Rightarrow 4 x =176 \text { cm} $$ $$ \Rightarrow x=\frac{176}{4} = 44 \text { cm} $$ Hence, lenght of side of new formed square is 44 cm.
Q15. A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
Sol. Given that total distance covered in 5000 revolutions = 11 km
$$ =11 \times 1000 \text { m} \quad \quad [\because 1 \text { km} = 1000 \text { m}] $$ $$ =11000 \text { m} $$ $$ \therefore \text {Distance covered in 1 revolution } =\frac{11000}{5000} $$ $$ =\frac{11}{5} \text { m} $$ $$ \text {Distance covered in 1 revolution } = \text {Circumference of the wheel } $$ $$ \Rightarrow \frac{11}{5}= \pi d \quad \quad [\text {Using : } C = \pi d \text { where d is diamenter of circle}] $$$$ \Rightarrow d=\frac{11 \times 7}{5 \times 22} \text { m} $$ $$ \Rightarrow d=0.7 \text { m} $$ Hence, the diameter of the wheel is 0.7 m.
Q16. A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling.
Sol. Given the diameter of the wheel =60 cm
Distance covered by the wheel in 1 revolution = Circumference of the wheel
\begin{array}{l} \text {Distance covered by the wheel in 1 revolution } =\pi d \quad \quad [\text {Using : } C = \pi d \text { where d is diamenter of circle}] \\ =\frac {22}{7} \times 60 \text { cm} \\\text {Distance covered by the wheel in 140 revolutions }=\frac {22}{7} \times 60 \times 140 =26400 \text { cm} \\ \text {As 140 revolutions are covered in 1 minutes.} \\ \therefore \text {Distance covered by wheel in 1 minute} = 26400 \text { cm} \\ \Rightarrow \text {Distance covered by wheel in 60 minutes} = 26400 \times 60 \text { cm} \\ \Rightarrow \text {Distance covered by wheel in 1 hr} = 26400 \times 60 \times \frac {1}{100000} \text { km} \quad \quad [\because 1 \text { cm} = \frac {1}{100000} \text { km}] \\ \Rightarrow \text {Distance covered by wheel in 1 hr} =15.84 \text { km} \\ \text {Hence, The cycling speed of the boy is 15.84 km/hr}\end{array}Q17. The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour?
Sol.
Given diameter of the wheel of the bus = 140 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
\begin{array}{l} \Rightarrow C =\frac{22 \times 140}{7} = 440 \text { cm} \\ \therefore \text {Bus cover 440 cm in one revolution.} \\\text {With a speed of 66 km per hour the bus will cover a distance of 66 km in 1 hr.} \\ \Rightarrow \quad \text {Distance covered in 1 hour or 60 mins} = 66 \text { km} \\ \Rightarrow \quad \text {Distance covered in 1 hour or 60 mins} = 66 \times 1000 \text { m} \quad \quad [\because 1 \text { km} = 1000 \text { m}] \\ \Rightarrow \quad \text {Distance covered in 1 min} = \frac {66 \times 1000}{60} \text { m} \\ \Rightarrow \quad \text {Distance covered in 1 min} = 1100 \text { m} \\ \Rightarrow \quad \text {Distance covered in 1 min} = 110000\text { cm} \\\text {Bus covers 440 cm in one revolution.} \\ \therefore \text {No. of revolution required to cover 110000 cm.} \\ = \frac {110000}{440} \\\end{array}= 250 revolutions
Hence, the bus need to take 250 revolutions per minute to keep the speed of 66 km/hr
Q18. A water sprinkler in a lawn sprays water as far as 7 m in all directions. Find the length of the outer edge of wet grass.
Sol. The wet grass forms a circular area of radius 7 m.
$$ \therefore \text {The length of the outer edge of the wet grass} = \text {Circumference of the circular area having radius 7 m } $$ $$ =2 \times \frac{22}{7} \times 7 =44 \text{ m} $$Q19. A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 660 cm, then find the width of the parapet.
Sol.
\begin{array}{l} \text {Given that diameter of the well } =150 \text { cm} \\ \text {Radius of well} =\frac {150}{2}=75 \text {cm} \\ \text {Let’s assume that the width of the stone parapet be x cm.} \\ \therefore \text {Radius of outer circle including stone parapet } = (75 +x) \text { cm} \\\text {Given length of the outer edge of the parapet } =660 \text { cm} \\ \Rightarrow \text {Circumference of outer circle } = 660 \text { cm} \\ \Rightarrow \text 2 \pi \times (75+x)= 660 \text { cm} \\ \Rightarrow \text 75+x= \frac {660 \times 7}{2 \times 22} \text { cm} \\ \Rightarrow \text 75+x= 105 \text { cm} \\ \Rightarrow \text x= 105 -75 \text { cm} \\ \Rightarrow \text x= 30 \text { cm} \\ \text {Hence, the width of the parapet is 30 cm} \end{array}Q20. An Ox in a kolhu (an oil processing apparatus) is tethered to a rope 3 m long. How much distance does it cover in 14 rounds?
Sol. Radius of the circular path traced by the Ox in the Kolhu = 3 m.
Distance covered by the Ox in 1 round = Circumference of the circular path having radius 3 m
=2 π r
$$ =2 \times \frac{22}{7} \times 3 \text { m} $$ $$ \Rightarrow \text {Distance covered in 14 rounds }==2 \times \frac{22}{7} \times 3 \times 14 \text { m} $$ =264 m