[Sumit kumar]

A solid trapizium cylinder having height 6 cm and the sum of area of both end is twice of lateral surface area then find the radius of base?

[Chic_Bowdrie]

A solid trapizium cylinder having height 6 cm and the sum of area of both end is twice of lateral surface area then find the radius of base?

There are infinite solutions depending on the difference (d) between the radius of the top and bottom. The area of the lateral surface of the cylinder is pi*s*(r + R) where s is the hypotenuse of triangle with base = d and height = 6cm at either side of a cross-section of the trapezium cut perpendicular to the base. So, the equation to solve is pi*r2 + pi*R2 = 2*pi*s*(r + R) or r2 + R2 = 2*s*(r + R). If you substitute for r = R – d, you can rearrange to get R2 - (d + 2s)*R + d2/2 +s*d = 0. This is a quadratic equation with one real solution R = d/2 + s + (s2 - d2/4)0.5 where s2 = 36 + d2.

It is interesting that r has a minimum around 11.563 cm when R = 13.343 cm. It makes sense that when the trapizoidal cylinder is close to a true cylinder of length 6 cm, r would be less than 12cm and R would be greater than 12cm, but if d > 4cm both r and R are greater than 12.