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In triangle ABC, AB=5, AC=6, and BC=8. Circles are drawn with centers A, B, and C, so that any two circles are externally tangent. Find the sum of the areas of the circles.

 

 Dec 3, 2020
 #1
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As we can see in the picture there are three different radiuses, therefore,

let's name the radius of side of circle A: a

name the radius of circle B: b

and the radius of circle C: c

Next we have a system of 3 equations:

a+b=5

a+c=6

b+c=8
next we solve it and get a=3/2, b=7/2, c=9/2, then we can solve these three values with the area of a circle formula which is pi*r^2

 

you should be able to do the calculations 

:)👍

 Dec 3, 2020
 #2
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Let A  = radius  of  circle A         B  = radius of circle B       C  = radius of  circle C

 

And we have this system of equations

 

A + C   = 6

A + B =  5

B + C =  8

 

Subtract the  first equation from the  second   →   B - C   =  -1

Add this to  the 3rd equation

We get  that 2B  = 7

B  = 7/2  = 3.5

 

And

B + C  = 8

3.5  + C  =8

8 -3.5  = C  = 4.5

 

A + 3.5 =  5

A = 5 -3.5  =   1.5

 

Combined area of all three circles  =   pi  [ 1.5^2  + 3.5^2  +  4.5^2  ]  =  34.75 pi  units^2

 

 

cool cool cool

 Dec 3, 2020
 #3
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Oh my! Me and Cphill's answer are so alike😂

 Dec 3, 2020
 #4
avatar+114592 
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LOL!!!...sorry, guest....didn't mean to steal your thunder.....!!!!   (I didn't see  you working on it at the same time as  I was.....)

 

cool cool cool

CPhill  Dec 3, 2020
 #5
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No problem, never apologize, you a true legend and helper

 Dec 3, 2020

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