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radius of sector

by Shubham
(Allahabad, uttar pradesh, india)











































if the radius of sector of circle increases 10% then how much decreases it angle, so that area be remain constant.

Comments for radius of sector

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Dec 20, 2010
Circle - Area of Sector
by: Staff

The question:

by Shubham
(Allahabad, Uttar Pradesh, India)


if the radius of sector of circle increases 10% then how much decreases it angle, so that area be remain constant.


The answer:

Begin with the formula for the area of the entire circle

A = area of circle
r = radius of circle

A = Pi*r^2

The formula for the area of a sector of the circle is:

Area of Sector = [(Central Angle of sector in degrees)/360]* Pi*r^2

Your problem:

The area of the sector of your first circle is:

Angle1 = central angle of sector of first circle
Area1 = area of the sector of your first circle

Area1 = [(Angle1)/360]* Pi*r^2

The area of the sector of your second circle is:

Since the radius of your second circle increases 10%, the radius will actually be = r + .1r

Angle2 = central angle of sector of second circle
Area2 = area of the sector of your second circle

Area2 = [(Angle2)/360]* Pi*(r + .1r)^2

Solving your problem

Since the areas of the sectors of both circles is equal:

Area1 = Area2

Substitute the formulas for Area1 and Area2

Area1 = Area2

[(Angle1)/360]* Pi*r^2 = [(Angle2)/360]* Pi*(r + .1r)^2

Eliminate the 360 on each side of the equation by multiplying each side of the equation by 360

[(Angle1)/360]* Pi*r^2 = [(Angle2)/360]* Pi*(r + .1r)^2

360*[(Angle1)/360]* Pi*r^2 = 360*[(Angle2)/360]* Pi*(r + .1r)^2

(360/360)*(Angle1)* Pi*r^2 = (360/360)*(Angle2)* Pi*(r + .1r)^2

1*(Angle1)* Pi*r^2 = 1*(Angle2)* Pi*(r + .1r)^2

(Angle1)* Pi*r^2 = (Angle2)* Pi*(r + .1r)^2

Eliminate the Pi on each side of the equation by dividing each side of the equation by Pi

(Angle1)* Pi*r^2 = (Angle2)* Pi*(r + .1r)^2

[(Angle1)* Pi*r^2]/Pi = [(Angle2)* Pi*(r + .1r)^2]/Pi

[(Angle1)*r^2]*(Pi/Pi) = [(Angle2)*(r + .1r)^2]*(Pi/Pi)

[(Angle1)*r^2]*1 = [(Angle2)*(r + .1r)^2]*1

(Angle1)*r^2 = (Angle2)*(r + .1r)^2

Since you asked how much the Angle of the circle would be reduced when the radius in increased, we will solve the equation for Angle2:

(Angle1)*r^2 = (Angle2)*(r + .1r)^2

Divide each side of the equation by (r + .1r)^2

[(Angle1)*r^2]/(r + .1r)^2 = [(Angle2)*(r + .1r)^2]/(r + .1r)^2

[(Angle1)*r^2]/(r + .1r)^2 = (Angle2)*[(r + .1r)^2/(r + .1r)^2]

[(Angle1)*r^2]/(r + .1r)^2 = (Angle2)*1

[(Angle1)*r^2]/(r + .1r)^2 = (Angle2)

Angle2, the angle of the circle with the increased radius is:

Angle2 = [(Angle1)*r^2]/(r + .1r)^2
Angle2 = [(Angle1)*r^2]/(1.1r)^2
Angle2 = [(Angle1)*r^2]/[1.21*r^2]
Angle2 = [(Angle1)/1.21]*(r^2/r^2)
Angle2 = [(Angle1)/1.21]*1
Angle2 = (Angle1)/1.21
Angle2 = (1/1.21)*(Angle1)
Angle2 = (0.8264)*(Angle1)
Angle2 = 82.64% of Angle1

The final answer: Angle2 is 82.64% of Angle1. That means that Angle2 is 17.36% smaller than Angle1.



Thanks for writing.


Staff
www.solving-math-problems.com

Jan 10, 2011
Spot for more math stuff
by: Anonymous

Here is some good stuff. http://volumeformula.org

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