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.
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
:)👍
+128566 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
