Power of a point rule: O * W = O * W ( outside * whole = outside * whole )
3rd problem: ST * SU = SV * SW
The length of a chord YZ = 17.889
Let the side lengths of the triangle ABC be: a=1, c=1, and b= ac/2 = 0.5
cos(A) /a + cos(C)/c=(cos75.522°/1)+(cos75.522°/1) = 0.25 + 0.25 = 0.5
Is it possible to inscribe the ellipse in that shaded rhombus, and what would be the area of it?
I misunderstood the question.
The diameter of the semi circle is: D = 2{ sin60° * sqrt(4) }
My answer is somewhat different. The area GEFC = 0.75 cm2
Besides those 4 small circles, it's possible to inscribe 24 more corcles.
The area of a larger square is: 6 + 3*sqrt(3) = ≈ 11.196 u2 side = ≈3.346
The area of a smaller square is: sin15° * 3.346 = 0.866
0.866 * 2 = 1.732
1.7322 = 3.0 u2
The ratio of the shaded triangle to the square is 1 : 10
