Consider first the vertical plane that includes the vertical axis of each shape. The intersection of this plane with these shapes is a circle and triangle. We next need to cut the circle and triangle with a horizontal line at a height that traverses the same interior width in each 2d shape. The problem just got much simpler. That width is a diameter, but let's rather focus on the radius.

For the triangle:

r(h) = 5 (1-h/10) (at the base: h=0, the half width is 5, and at the top: h=10, it is 0)

For the circle:

r(h)=sqrt {100- (10-h)^2} (since x^2 + y^2 =10^2 and r = x along this line).

Equating these gives:

h^2 - 20 h + 20 = 0

h = 10 (+/-) 4 sqrt (5) = 1.0557, 18.944 (choose the first root, since the second root gives a negative radius)

h = 1.0557 cm

radius = 4.472 (this is sqrt 20) and area = 62.83 cm^2

*Edited on ***September 8, 2018, 4:22 am**