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.

Guest Dec 3, 2020

#1**+1 **

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

:)👍

Guest Dec 3, 2020

#2**+1 **

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

CPhill Dec 3, 2020