We have an isosceles triangle, apex at the center of a circle, sides #r#, #r#, and #8#, central angle #theta = pi/6-pi/8=pi/24.#
Law of Cosines:
# 8^2 = r^2 + r^2 - 2 (r)(r) cos theta = 2(r^2-1) cos theta#
#r^2 = 1 + 32/cos (pi/24) #
Our answer is #pi# times that. Are we done?
No, #pi/24 = 7.5^circ# is constructible so its cosine has an "exact" answer using square roots of positive numbers. Let's work it out.
#cos 15^circ = cos (45^circ - 30 ^circ) = cos 45 cos 30 + sin 45 sin 30 = sqrt{2}/2(sqrt{3}/2+1/2)=1/4(sqrt{6}+sqrt{2})#
Now the half angle formula; we choose the positive square root.
#cos 7.5^circ = sqrt{1/ 2 ( 1 + cos 15^circ ) } = sqrt{ 1/2(1 + 1/4(sqrt{6}+sqrt{2}))} = 1/2 sqrt(2 + sqrt(2 + sqrt(3))) #
#text{Area}=pi r^2 = pi( 1 + 32/ ( 1/2 sqrt(2 + sqrt(2 + sqrt(3))) )) #
#text{Area}= pi( 1 + 64/ ( sqrt(2 + sqrt(2 + sqrt(3))) )) #
Let's leave it there.
