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.
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,¯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=9−5rc=4
ra= ?
ra+rb+rc⏟=7=9ra+7=9ra=9−7ra=2
rb= ?
ra+rb+rc=9|ra+rc=66+rb=9rb=9−6rb=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