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In triangle ABC, AB=5, AC=6, and BC=7. 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.

 Feb 8, 2018
 #1
avatar+9489 
+4

There might be a better way, but here is one way.

 

Let the radius of circle A be  a , the radius of circle B be  b , and the radius of circle C be  c .

 

From the given information, we can make these three equations:

 

a + b  =  5        solve for  a  to get        a  =  5 - b

a + c  =  6        solve for  a  to get        a  =  6 - c

b + c  =  7        solve for  b  to get        b  =  7 - c

 

5 - b   =   6 - c

                                 Substitute  7 - c  in for  b .

5 - (7 - c)   =   6 - c

                                 Distribute the  -1  to the terms in parenthesees.

5 - 7 + c   =   6 - c

                                 Rearrange

c + c   =   6 - 5 + 7

 

2c   =   8

 

c   =   4       Use this value of  c  to find  a  and  b .

 

a   =   6 - c   =   6 - 4   =   2

 

b   =   7 - c   =   7 - 4   =   3

 

area of circle A  +  area of circle B  +  area of circle C   =

 

22π   +   32π   +   42π   =

 

4π   +   9π   +   16π   =

 

29π    sq. units

 Feb 8, 2018
 #2
avatar+26399 
+3

In triangle
ABC,¯AB=5,¯AC=6,
and
¯BC=7.
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.

 

 

Let ra= radius of circle A Let rb= radius of circle B Let rc= radius of circle C 

 

(1)ra+rb=5(2)rb+rc=7(3)rc+ra=6(1)+(2)+(3):2(ra+rb+rc)=5+6+72(ra+rb+rc)=18|:2ra+rb+rc=9

 

rc= ?

ra+rb=5+rc=95+rc=9rc=95rc=4

 

ra= ?

ra+rb+rc=7=9ra+7=9ra=97ra=2

 

rb= ?

ra+rb+rc=9|ra+rc=66+rb=9rb=96rb=3

 

 

The sum of the areas of the circles:

πr2a+πr2b+πr2c=π(r2a+r2b+r2c)=π(22+32+42)=π(4+9+16)=29π

 

 

laugh

 Feb 8, 2018

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