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Using 3.14 as Pi the area of circle is: 113.03999999999999
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What is the formula for finding the Surface Area of a frustum of a cone?
Lateral Surface Area = π(r1 + r2)s = π(r1 + r2)√((r1 - r2)2 + h2)Top Surface Area = πr12Base Surface Area = πr22Total Surface Area = π(r12 + r22 + (r1 * r2) * s) = π[ r12 + r22 + (r1 * r2) * √((r1 - r2)2 + h2) ]
What is the formula for the surface area of a frustum?
For a conical frustum:Lateral Surface Area = π(r1 + r2)s = π(r1 + r2)√((r1 - r2)2 + h2)Top Surface Area = πr12Base Surface Area = πr22Total Surface Area = π(r12 + r22 + (r1 * r2) * s) = π[ r12 + r22 + (r1 * r2) * √((r1 - r2)2 + h2) ]
How do you find the area for a frustum cone?
V = 1/3πh ( R12 + R22 + R1R2 )where π is 3.14159265..., and R1, R2 are the radii of the two bases.
How could you change the height of a cone so that its volume would remain the same when its radius was tripled?
V1 = (1/3)(pi)(r12)(h1) V2 = (1/3)(pi)(xr12)(h2) V1 = V2 , which means that: (1/3)(pi)(r12)(h1) = (1/3)(pi)(xr12)(h2) Divide both sides by (1/3)(pi) and you get: (r12)(h1) = (xr12)(h2) -> (r12)(h1) = x2(r12)(h2) Divide both sides by (r12) and you get: h1 = x2(h2) -> h2 = (h1)/x2 For example: Cone1: r1 = 10, h1 = 10 Cone2: r2 = 30, h2 = (10/32) = 10/9 = 1.11111111 Then to check: Volume of a cone = (1/3)(pi)(r2)(h) V1 = (1/3)(pi)(102)(10) V1 = 1047.197551 = V2 1047.197551 = (1/3)(pi)(302)(h2) h2 = 1047.197551/((900pi)/3) h2 = 1.111111111 = 10/9
What is the formula for the density of a truncated cone?
Density = Mass/[(1/3*pi*h) * (R12 + R22 + R1*R2)] where h is the height of the frustum, and R1 and R2 are the radii of the two circular sections.
