Q7. A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. It is re-bent to form a circular ring. What is the diameter of the ring?
Sol.
Given length of side of equilateral triangle }=6.6 cm
Therefore, length of the wire = Perimeter of equilateral triangle
=3 × 6.6 =19.8 cm
Let’s assume that diameter of new formed circle be d cm.
Then, Circumference of circle (C) } = Perimeter of the rectangle =19.8 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
$$ \Rightarrow d = \frac {C}{\pi} $$ $$ \Rightarrow d =\frac{19.8 \times 7}{22} = 6.3 \text { cm} $$Q8. The diameter of a wheel of a car is 63 cm. Find the distance travelled by the car during the period, the wheel makes 1000 revolutions.
Sol. Given diameter of the wheel of the car =63 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
$$ \Rightarrow C =\frac{22 \times 63}{7} = 198 \text { cm} $$ $$ \therefore \text {Car cover 198 cm in one revolution.} $$ $$ \Rightarrow \text {The distance covered by car in 1000 revolutions } = 1000 \times 198 $$ = 198000 cm
$$ = 198000 \times \frac {1}{100} \text { m} \quad \quad [\because 1 \text { cm} = \frac {1}{100} \text { m}] $$ = 1980 m
Q9. The diameter of a wheel of a car is 98 cm. How many revolutions will it make to travel 6160 metres.
Sol. Given diameter of the wheel of the car = 98 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
$$ \Rightarrow C =\frac{22 \times 98}{7} = 308 \text { cm} $$ $$ \therefore \text {Car cover 308 cm in one revolution.} $$Given that total distance covered by car = 6160 m
$$ =6160 \times 100 \text { cm} \quad \quad [\because 1 \text { m} = 100 \text { cm}] $$ = 616000 cm
$$ \therefore \text {Total no. of revolutions } = \frac {\text {Total distance travelled by car}}{\text {Distance covered in 1 revolution}} $$ $$ = \frac {616000}{308} $$ = 2000
Q10. The moon is about 384400 km from the earth and its path around the earth is nearly circular. Find the circumference of the path described by the moon in lunar month.
Sol. The radius of the path of the moon around the earth =384400 km
$$ \therefore \text {The circumference of the path of the moon (C)}=2 \pi r $$ $$ =2 \times \frac{22}{7} \times 384400 \text { km} $$ =2416228.57 km
Q11. How long will John take to make a round of a circular field of radius 21 m cycling at the speed of 8 km/hr
Sol. Given radius (r) of the circular field =21 m
We know that the circumference (C) of a circle having radius r is given by C = 2 π r
$$ \therefore \text {Circumference of the circular field }=2 \times \frac{22}{7} \times 21 =132 \text { m} $$ Given that John cyclying speed is 8 km/hr
$$ \Rightarrow \text { Distance travel by John in 1 hr } = 8 \text { km} $$ $$ =8 \times 1000 \text { km} \quad \quad [\because 1 \text { mm} = 1000 \text { m}] $$ =8000 m
$$ \Rightarrow \text {Time taken by John to travel 8000 m } = 1 \text { hr} $$ $$ \Rightarrow \text {Time taken by John to travel 1 m } = \frac {1}{8000} \text { hr} $$ $$ \Rightarrow \text {Time taken by John to travel 138 m } = 138 \times \frac {1}{8000} \text { hr} $$ $$ = 138 \times \frac {1}{8000} \times 3600 \text { sec} \quad \quad [\because 1 \text { hr} = 3600 \text { sec}] $$ = 59.4 sec
Q12. The hour and minute hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days.
Sol. We know that the circumference (C) of a circle having radius r is given by C = 2 π r
$$ \text {The radius } r_{1}\text { of the path of hour hand } = \text {Length of the hour hand } =4 \text { cm} $$ $$ \text {The radius } r_{2}\text { of the path of minute hand }= \text {Length of the minute hand }=6 \text { cm} $$ $$ \text {Distance covered by the hour hand in 1 revolution } = \text {Circumference } C_{1}\text { of the path of hour hand } $$ $$ =2 \pi r_{1} =2 \times \frac{22}{7} \times 4=\frac{176}{7} \text { cm} $$Hour hand makes 2 revolutions in one day.
$$ \therefore \text {Distance covered by the hour hand in 2 days }=\frac{176}{7} \times 2 \times 2=100.57 \text { cm} $$$$ \text {Now, Distance covered by the minute hand in 1 revolution } $$ $$ = \text {Circumference } C_{2}\text { of the path of minute hand } $$ $$ =2 \pi r_{2} =2 \times \frac{22}{7} \times 6=\frac{264}{7} \text { cm} $$Minute hand makes 24 revolutions in one day.
$$ \therefore \text {Distance covered by the minute hand in 2 days } $$ $$=\frac{176}{7} \times 2 \times 24=1810.28 \text { cm} $$Sum of the distances travelled by the hour and minute hands in 2 days = 100.57 + 1810.28 =1910.85 cm
Q13. A rhombus has the same perimeter as the circumference of a circle. If the side of the rhombus is 2.2 m, find the radius of the circle.
Sol. Given that the side of a rhombus =2.2 m
We know that perimeter of rhombus = 4 × (Side)
=4 × 2.2 = 8.8 m
According to question, we have
Circumference of circle = Perimeter of rhombus
We know that the circumference (C) of a circle having radius r is given by C = 2 π r
$$ \Rightarrow 2 \pi r = 8.8 $$ $$ \Rightarrow r = \frac {8.8 \times 7}{22 \times 2} $$ $$ \Rightarrow r = 1.4 \text { m} $$Hence, the radius of the circle is 1.4 m
Sol.
Given length of side of equilateral triangle }=6.6 cm
Therefore, length of the wire = Perimeter of equilateral triangle
=3 × 6.6 =19.8 cm
Let’s assume that diameter of new formed circle be d cm.
Then, Circumference of circle (C) } = Perimeter of the rectangle =19.8 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
$$ \Rightarrow d = \frac {C}{\pi} $$ $$ \Rightarrow d =\frac{19.8 \times 7}{22} = 6.3 \text { cm} $$Q8. The diameter of a wheel of a car is 63 cm. Find the distance travelled by the car during the period, the wheel makes 1000 revolutions.
Sol. Given diameter of the wheel of the car =63 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
$$ \Rightarrow C =\frac{22 \times 63}{7} = 198 \text { cm} $$ $$ \therefore \text {Car cover 198 cm in one revolution.} $$ $$ \Rightarrow \text {The distance covered by car in 1000 revolutions } = 1000 \times 198 $$ = 198000 cm
$$ = 198000 \times \frac {1}{100} \text { m} \quad \quad [\because 1 \text { cm} = \frac {1}{100} \text { m}] $$ = 1980 m
Q9. The diameter of a wheel of a car is 98 cm. How many revolutions will it make to travel 6160 metres.
Sol. Given diameter of the wheel of the car = 98 cm
We know that the circumference (C) of a circle having diameter d is given by C = π d
$$ \Rightarrow C =\frac{22 \times 98}{7} = 308 \text { cm} $$ $$ \therefore \text {Car cover 308 cm in one revolution.} $$Given that total distance covered by car = 6160 m
$$ =6160 \times 100 \text { cm} \quad \quad [\because 1 \text { m} = 100 \text { cm}] $$ = 616000 cm
$$ \therefore \text {Total no. of revolutions } = \frac {\text {Total distance travelled by car}}{\text {Distance covered in 1 revolution}} $$ $$ = \frac {616000}{308} $$ = 2000
Q10. The moon is about 384400 km from the earth and its path around the earth is nearly circular. Find the circumference of the path described by the moon in lunar month.
Sol. The radius of the path of the moon around the earth =384400 km
$$ \therefore \text {The circumference of the path of the moon (C)}=2 \pi r $$ $$ =2 \times \frac{22}{7} \times 384400 \text { km} $$ =2416228.57 km
Q11. How long will John take to make a round of a circular field of radius 21 m cycling at the speed of 8 km/hr
Sol. Given radius (r) of the circular field =21 m
We know that the circumference (C) of a circle having radius r is given by C = 2 π r
$$ \therefore \text {Circumference of the circular field }=2 \times \frac{22}{7} \times 21 =132 \text { m} $$ Given that John cyclying speed is 8 km/hr
$$ \Rightarrow \text { Distance travel by John in 1 hr } = 8 \text { km} $$ $$ =8 \times 1000 \text { km} \quad \quad [\because 1 \text { mm} = 1000 \text { m}] $$ =8000 m
$$ \Rightarrow \text {Time taken by John to travel 8000 m } = 1 \text { hr} $$ $$ \Rightarrow \text {Time taken by John to travel 1 m } = \frac {1}{8000} \text { hr} $$ $$ \Rightarrow \text {Time taken by John to travel 138 m } = 138 \times \frac {1}{8000} \text { hr} $$ $$ = 138 \times \frac {1}{8000} \times 3600 \text { sec} \quad \quad [\because 1 \text { hr} = 3600 \text { sec}] $$ = 59.4 sec
Q12. The hour and minute hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days.
Sol. We know that the circumference (C) of a circle having radius r is given by C = 2 π r
$$ \text {The radius } r_{1}\text { of the path of hour hand } = \text {Length of the hour hand } =4 \text { cm} $$ $$ \text {The radius } r_{2}\text { of the path of minute hand }= \text {Length of the minute hand }=6 \text { cm} $$ $$ \text {Distance covered by the hour hand in 1 revolution } = \text {Circumference } C_{1}\text { of the path of hour hand } $$ $$ =2 \pi r_{1} =2 \times \frac{22}{7} \times 4=\frac{176}{7} \text { cm} $$Hour hand makes 2 revolutions in one day.
$$ \therefore \text {Distance covered by the hour hand in 2 days }=\frac{176}{7} \times 2 \times 2=100.57 \text { cm} $$$$ \text {Now, Distance covered by the minute hand in 1 revolution } $$ $$ = \text {Circumference } C_{2}\text { of the path of minute hand } $$ $$ =2 \pi r_{2} =2 \times \frac{22}{7} \times 6=\frac{264}{7} \text { cm} $$Minute hand makes 24 revolutions in one day.
$$ \therefore \text {Distance covered by the minute hand in 2 days } $$ $$=\frac{176}{7} \times 2 \times 24=1810.28 \text { cm} $$Sum of the distances travelled by the hour and minute hands in 2 days = 100.57 + 1810.28 =1910.85 cm
Q13. A rhombus has the same perimeter as the circumference of a circle. If the side of the rhombus is 2.2 m, find the radius of the circle.
Sol. Given that the side of a rhombus =2.2 m
We know that perimeter of rhombus = 4 × (Side)
=4 × 2.2 = 8.8 m
According to question, we have
Circumference of circle = Perimeter of rhombus
We know that the circumference (C) of a circle having radius r is given by C = 2 π r
$$ \Rightarrow 2 \pi r = 8.8 $$ $$ \Rightarrow r = \frac {8.8 \times 7}{22 \times 2} $$ $$ \Rightarrow r = 1.4 \text { m} $$Hence, the radius of the circle is 1.4 m