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Please advise, thanks.

2007-07-05 11:10:43 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

Hi Dr D, can you explain why the ratio is 4:3 please? thanks.

2007-07-05 11:27:53 · update #1

4 answers

If R is the radius of the circumscribed circle, you can calculate the radius of the inscribed circle by drawing a triangle between the circle center and one side of the hexagon, bisecting the side, and solving the triangle (the bisector will pass through the circle center). This will permit an easy solution to the problem.

2007-07-05 11:20:20 · answer #1 · answered by Anonymous · 0 0

4/3

The radius, R of the circumscribed triangle is the "diagonal". The radius, r of the inscribed circle is the distance from the center to the midpoint of an edge.
These form a 30-60-90 triange with hypothenuse R, and
r = R*sqrt(3) / 2

So R/r = 2 / sqrt(3)
R^2 / r^2 = 4 / 3 = ratio of areas

2007-07-05 18:25:16 · answer #2 · answered by Dr D 7 · 0 0

A cirumcircle /A inscribed circle = 4/3
A circumincircle = pi R^2
r = Rsqrt(3)/2
So area incircle = pi3R^2/4
So there ratio is 4/3

2007-07-05 18:35:52 · answer #3 · answered by ironduke8159 7 · 0 0

(cos(30))^2=3/4
or 4/3

2007-07-05 18:18:11 · answer #4 · answered by Alberd 4 · 0 0

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