Thread: Babylonian "bull's eye"

There is a shape called the Babylonian bull's eye. Show that it has area A=9/32*a^2 where a is the length of the arc (1/3 the circumference). Assume the Babylonian values of C^2/12 for the area of a circle and 7/4 for the sqrt(3). Then show that the length of the long transversal is 7/8*a and the length of the short transversal is 1/2*a.

I tried to find a picture of this figure online to no avail, but to describe it, it is made of two connected arcs of length a each, and each of the arcs is one third of a circle. It looks like a football with the two arcs being the two sides of the ball, meeting at a point.. There are two transversals drawn in, with the first joining the two sharp points where the two arcs meet and one transversal going from the middle of one arc to the middle of the other. The two transversals meet at a 90 degree angle at the center of the shape. I am very stumped. Please help. thanks.

Originally Posted by zhupolongjoe

There is a shape called the Babylonian bull's eye. Show that it has area A=9/32*a^2 where a is the length of the arc (1/3 the circumference). Assume the Babylonian values of C^2/12 for the area of a circle and 7/4 for the sqrt(3). Then show that the length of the long transversal is 7/8*a and the length of the short transversal is 1/2*a.

I tried to find a picture of this figure online to no avail, but to describe it, it is made of two connected arcs of length a each, and each of the arcs is one third of a circle. It looks like a football with the two arcs being the two sides of the ball, meeting at a point.. There are two transversals drawn in, with the first joining the two sharp points where the two arcs meet and one transversal going from the middle of one arc to the middle of the other. The two transversals meet at a 90 degree angle at the center of the shape. I am very stumped. Please help. thanks.

the shaded area in red would be half the area of the "eye" , correct?

Yes, that looks right to me, but I still don't see exactly how it follows. Thanks.