Volumes Practice Test 4

Concept:Equate three times the sum of the areas of the two circular ends to twice the curved surface area of the cylinder.Explanation:Let the radius be $r$ metres and height $h = 6$ m.Area of one end face $= \pi r^2$.Sum of areas of two end faces $= 2\pi r^2$.Three times this sum $= 3 \times 2\pi r^2 = 6\pi r^2$.Curved surface area $= 2\pi rh = 2\pi r(6) = 12\pi r$.Twice the curved surface area $= 2 \times 12\pi r = 24\pi r$.According to the condition: $6\pi r^2 = 24\pi r$.Cancel $\pi r$ since $r \neq 0$: $6r = 24$.So, $r = \frac{24}{6} = 4$ m.Answer:The radius of the base is $4$ m.