Radius = 35 cm
By converting units ,
Radius = 0.35 m
Circumference of wheel = 2πr
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = 2 \times \dfrac{22}{7} \times 0.35 \\ \\ \sf{} = 2 \times 22 \times 0.5 \\ \\ \sf{} = 2.2 \: m[/tex]
Therefore , In one revolution wheel covers 2.2 m .Therefore ,365 m is covered in revolutions ;
[tex] \boxed {\sf{}No \: of \: Revolution \: = \dfrac{total \: distance}{distance \: covered \: in \: 1 \: revolution} }[/tex]
Therefore,
[tex] \sf{}No. \: of \: Revolution \: = (\dfrac{356}{2.2} ) \\ \\ \sf{}No. \: of \: Revolution= 161.8[/tex]
Therefore,wheel takes 161.8 revolution to cover 356 m of distance .
This Figure is the combination of a Rectangle and two semi circles . Therefore ,
Area of the figure = Area of rectangle + Area of 2 semi circles.
Area of rectangle = Length × Breadth
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = (133 \times 46) \: {cm}^{2} \\ \\ \sf{} = 6118 \: {cm}^{2} [/tex]
Now area of 2 semi circles = 2 ( 1/2 πr²)
Diameter = 46 m
Radius = 46/2 = 23 m
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = 2( \dfrac{1}{2} \times \dfrac{22}{7} \times 23 \times 23) \\ \\ \sf{} \: = \frac{22}{7} \times 23 \times 23 \\ \\ \sf{ }= 1662.5 \: {cm}^{2} [/tex]
Now , Area of given figure ;
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = (6118 + 1662.5) \: {cm}^{2} \\ \\ \sf{} = 7780.5 \: {cm}^{2} [/tex]
Therefore,area of given figure is equals to 7780.5 cm².
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Verified answer
Solution ( 9 ) :
Radius = 35 cm
By converting units ,
Radius = 0.35 m
Circumference of wheel = 2πr
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = 2 \times \dfrac{22}{7} \times 0.35 \\ \\ \sf{} = 2 \times 22 \times 0.5 \\ \\ \sf{} = 2.2 \: m[/tex]
Therefore , In one revolution wheel covers 2.2 m .Therefore ,365 m is covered in revolutions ;
[tex] \boxed {\sf{}No \: of \: Revolution \: = \dfrac{total \: distance}{distance \: covered \: in \: 1 \: revolution} }[/tex]
Therefore,
[tex] \sf{}No. \: of \: Revolution \: = (\dfrac{356}{2.2} ) \\ \\ \sf{}No. \: of \: Revolution= 161.8[/tex]
Therefore,wheel takes 161.8 revolution to cover 356 m of distance .
Solution ( 10 ) :
This Figure is the combination of a Rectangle and two semi circles . Therefore ,
Area of the figure = Area of rectangle + Area of 2 semi circles.
Area of rectangle = Length × Breadth
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = (133 \times 46) \: {cm}^{2} \\ \\ \sf{} = 6118 \: {cm}^{2} [/tex]
Now area of 2 semi circles = 2 ( 1/2 πr²)
Diameter = 46 m
Radius = 46/2 = 23 m
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = 2( \dfrac{1}{2} \times \dfrac{22}{7} \times 23 \times 23) \\ \\ \sf{} \: = \frac{22}{7} \times 23 \times 23 \\ \\ \sf{ }= 1662.5 \: {cm}^{2} [/tex]
Now , Area of given figure ;
[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{} = (6118 + 1662.5) \: {cm}^{2} \\ \\ \sf{} = 7780.5 \: {cm}^{2} [/tex]
Therefore,area of given figure is equals to 7780.5 cm².