math problem?

its difficult to describe without a diagram but i'll try my best

firrst imagine the triangle inscribed in the circle.

join two vertices to the centre

these are the radii ( say r cm)

now the angle between them will be double the angle at the other vertex

sincle the triangle is eqilateral all angles are 60

so the angle is 120

in the triangle formed by the 2 radii the other two angles will be equal to 30 each as it is isosceles.

draw the perpendicular from the centre cutting the base of the triangle

cos30 =(sqrt3)/2

so base of the small triangle is (sqrt3r) / 2

so the length of the base of the big triangle is (sqrt3)r

so the side of the equilateral triangle is (sqrt)3r

area = (sqrt3)/4 x side^2

=3(sqqrt3r^2) / 4

Now consider the equilateral triangle inscribed around the circle

draw the perpendicular from the centre onto the triangle

join one vertice on this line to the centre .

the angle at the vertice is 1/2(60) = 30

tan30 = 1 / sqrt3

so base of small triangle = (sqrt 3)r

so side of equilateral triangle = 2r (sqrt 3)

so area by formula = 3(sqrt3)r^2

difference of areas = 9 (sqrt 3) r^2 / 4

equating it to 25cm^2

r = 10 / 9sqrt(sqrt 3)

Well no I have 8 digit phone number !