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 externall

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

In triangle
\(ABC, \overline{AB}=5, \overline{AC}=6,\)
and
\(\overline{BC}=7.\)
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.

\(\text{Let $r_a =$ radius of circle $A$ } \\ \text{Let $r_b =$ radius of circle $B$ } \\ \text{Let $r_c =$ radius of circle $C$ }\)

\(\begin{array}{|lrcll|} \hline (1) & r_a + r_b &=& 5 \\ (2) & r_b + r_c &=& 7 \\ (3) & r_c + r_a &=& 6 \\ \hline (1)+(2)+(3): \\ & 2(r_a+r_b+r_c) &=& 5+6+7 \\ & 2(r_a+r_b+r_c) &=& 18 \quad & | \quad : 2 \\ & r_a+r_b+r_c &=& 9 \\ \hline \end{array}\)

\(\mathbf{r_c = \ ?}\)

\(\begin{array}{|rcll|} \hline \underbrace{r_a+r_b}_{=5}+r_c &=& 9 \\ 5+r_c &=& 9 \\ r_c &=& 9-5 \\ \mathbf{r_c} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

\(\mathbf{r_a = \ ?}\)

\(\begin{array}{|rcll|} \hline r_a+\underbrace{r_b+r_c}_{=7} &=& 9 \\ r_a+7 &=& 9 \\ r_a &=& 9-7 \\ \mathbf{r_a} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

\(\mathbf{r_b = \ ?}\)

\(\begin{array}{|rcll|} \hline r_a+r_b+r_c &=& 9 \quad & | \quad r_a+r_c = 6\\ 6 + r_b &=& 9 \\ r_b &=& 9-6 \\ \mathbf{r_b} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}\)

The sum of the areas of the circles:

\(\begin{array}{|rcll|} \hline && \pi r_a^2 + \pi r_b^2 + \pi r_c^2 \\ &=& \pi \cdot( r_a^2 + r_b^2 + r_c^2 ) \\ &=& \pi \cdot( 2^2 + 3^2 + 4^2 ) \\ &=& \pi \cdot( 4+9+16 ) \\ &\mathbf{=}& \mathbf{29\pi} \\ \hline \end{array} \)