Suppose a circle is inscribed in an equilateral triangle with side length two units.
Another circle is inscribed in the upper corner.
It touches two sides of the triangle and the circle.
Find the area A between the smaller circle and the upper corner of the triangle.
The radius of the large circle is 1/sqrt(3), so its diameter is 2/sqrt(3). The height of the given triangle is sqrt(3), so the bottom of the smaller circle is sqrt(3) - 2/sqrt(3) from the apex.
That means that the small circle is inside a small triangle of height sqrt(3) - 2/sqrt(3), which is a miniature version of the large triangle of height sqrt(3), so any area of that triangle or corresponding parts of it is in the ratio ((sqrt(3) - 2/sqrt(3))/sqrt(3))^2 = 1/9.
So if we find the corresponding area above the large circle we can apply this ratio and get the desired area above the small circle.
The area of the large triangle is sqrt(3). The area of the large circle is pi*(1/sqrt(3))^2 = pi/3, so the three "triangular" areas around it have a total area of sqrt(3) - pi/3; each is therefore
(sqrt(3) - pi/3) / 3
For the miniature triangle in which the small circle lies the area ratio needs to be applied:
((sqrt(3) - pi/3) / 3) / 9 = (sqrt(3) - pi/3) / 27 =~ 0.0253649354211955.
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Posted by Charlie
on 2023-07-25 12:02:56 |
same answer
