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If the degree measure of an arc of a circle is increased by 20% and the radius of the circle is increased by 25% , by what percent does the length of the arc increase?

 Mar 3, 2021
 #1
avatar+485 
+1

lets say the length radius is r, and the degree measure is d. we know that the length of the arc is 2$\pi$r$\cdot$d/360=d$\pi$r/180, so increasing the degree measure of the arc makes 2$\pi$r$\cdot$(6d/5)/360, and increasing the radius by 25%, we get 5$\pi$r/2$\cdot$(6d/5)/360 and we get:

$\frac{5\pi r \cdot \frac{6d}5}{720}=\frac{d\pi r}{120}$

$\frac{\frac{d\pi r}{180}}{\frac{d\pi r}{120}}=\frac32$

$\frac{3}{2} - 1=\frac{1}{2}=\boxed{50\%}$

 Mar 3, 2021
 #2
avatar+118069 
+2

Call  the original  arc length, S

Call the original radius, R

Call the orignal arc measure , T

 

So

 

S =  RT

 

And  when the arc measure  increases by 20%  and the  radius increases by 25%  we have

 

(1.20)R * ( 1.25) T  =

 

1.5  RT

 

So .....  the  arc length  increases  by   (1.5 -  1) * 100%  =  .5 * 100%  =   50%

 

cool cool cool

 Mar 3, 2021
 #3
avatar+31281 
+2

Original arc length would be    x / 360   pi  2r  

 

 now multiply  x by 1.2   and   r by 1.25

 

1.2x / 360   pi  2 r * 1.25   =      1/2 * 1.25  x/360 pi 2r

                                                   = 1.5  *  original                        so 50% longer

 Mar 3, 2021
 #4
avatar+118069 
0

By unanimous consent......50%  it is    !!!!!

 

 

 

cool cool cool

 Mar 3, 2021

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