Inscribed circles in hexagon
Geometry
Level
3
A regular hexagon with side length 1 is divided into 6 equilateral triangles. What is the total sum of areas of the circles inscribed in these triangles.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
The inscribed circle is tangent to each side of the triangle and its center is the center of the triangle, O.
Let one of the angles of the triangle be A, and the point where the side is tangent to the circle be D.
Then OAD is a 30 - 60 - 90 right triangle with long leg = 0.5.
Therefore shortleg = 0.5 / sqrt(3) and is equal to the radius of the circle.
Area of circle = pi * r^2 = pi / ( .5^2 * sqrt(3)^2) = pi/ 12.
Since there are 6 circles, 6 * pi / 12 = pi / 2.