refreshment wrote:cjb wrote:
Draw a second equilateral triangle, with corners at the centres of the three circles. This has sides of length four. Draw smaller 30-60-90 triangles including the corners of the internal and external triangles. The perpendicular distance from the internal triangle to the external triangle is one radius, or 2. You should see that the length of a side of the original triangle is 4 + 4(sqrt(3)).
Then you can use
area = 0.5 * base * height
to solve.
I get 24+16(sqrt(3))
Could you please explain how did you draw the smaller 30-60-90 triangles? and how does it include the internal and external triangle corners???
Don't have a diagramming tool, but here's my best text effort:
From one of the corners of the internal triangle, draw two lines. The first line is direct to the nearest corner of the larger, external triangle. The second line is perpendicular to one of the two near edges of the outer triangle. This will be a radius of the circle that the internal corner is in, so it will have a length of two.
Those two lines, plus the section of the outer triangle that lies between the points at which they touch it, form a new triangle. The angle between the line formed by a radius of the circle and the outer triangle is 90 degrees, and the line straight to the corner of the outer triangle bisects the angle there, meaning that there's a 30 degree angle inside our new triangle. The last angle must therefore be 60 degrees, so we know the ratio of the sides and the length of one of them.
