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?

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Originally Posted by 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?

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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.