Four congruent circles are arranged, so that three of them are symmetric and tangent to the fourth circle. An equilateral triangle is drawn so that each side is tangent to two of the circles. If the radius of each circle is 1, then find the area of the triangle.
+128475
From the center of the small circle on the bottom left draw a radius to the tangent formed by the side of the equilateral triangle and draw a segment to the bottom left vertex of the equilateral triangle....call this x
This will form a 30-60-90 right triangle
The vertex angle of the equilateral triangle will be bisected = 30°
The side opposite of this = radius of a small circle = 1
So....x = twice the radius = 2
So.....the distance from the bottom left vertex to the center of the middle circle = x + 2 = 2 + 2 = 4
And we can form another triangle with two sides of 4 and an included angle of 120°
And the side of the equilateral triangle ,S, wil be opposite this angle
So.....using the Law of Cosines
S^2 = 4^2 + 4^2 - 2 (16) cos (120°)
S^2 = 32 - 32 (-1/2)
S^2 = 48
S = sqrt (48)
So....the area of the equilateral triangle =
(1/2) (sqrt (48))^2 sin (60°) =
(1/2) (48) sqrt (3) /2 =
12 sqrt (3) units ^2
