Seven identical circles are arranged like the diagram below so that each one is tangent to one another. A hexagon is then drawn around the circles. The radius of each circle is 3. Find the area of the space inside of the hexagon but outside of the circles.
+128475 The only hard part here is trying to determine the side length of the hexagon
The vertex angle of a hexagon = (2/3)(180) = 120°
Looking at the top right circle...if we draw a segment connnecting the center of this circle with the top right vertex of the hexagon....this segment will bisect the vertex angle
Next from the center of ths circle draw a radius perpendicular to the top edge of the hexagon....this distance = 3
So....we have a 30- 60 -90 right triangle...call the distance from the top right vertex of the hexagon to the point where the radial line meets the top edge = x
And we can find this distance as
3/sin 60 = x/sin 30
x = 3 sin (30)/ sin (60)
x = 3 (1/2)/ [√ 3/2]
x = 3(1/2) * (2/√ 3]
x = √ 3
So using symmetry.....the side length of the hexagon = x + 3 + 3 + x = 2( 3 + x) = 2 (3 + √ 3) = 6 + √ 12
And the area of the hexagon = 3√ 3 side^2 / 2 = 3√ 3 [ 6 + √ 12]^2 / 2 =
3√ 3 [ 36 + 12√ 12 + 12] / 2 =
3√ 3 [ 48 + 12√ 12] / 2 =
3√ 3 [ 24 + 6√ 12] =
72√ 3 + 18 √ 36 =
72√ 3 + 108
The area of the circles is easy = 7pi 3^2 = 63 pi
Then the area inside the hexagon but outside the circles =
72√ 3 + 108 - 63 pi ≈ 34.8 units^2
