Asked by K
The flask consist of a cylindrical part and a frustum of a cone. The diameter of the base is 10 cm while that of neck is 2 cm. The vertical height of the flask is 12 cm.
(a) the slant height of the frustum part;
(b) the slant height of the smaller cone that was cut off to make the frustum part.
(c) the external surface area of the flask. (Take pi = 3.142)
(a) the slant height of the frustum part;
(b) the slant height of the smaller cone that was cut off to make the frustum part.
(c) the external surface area of the flask. (Take pi = 3.142)
Answers
Answered by
Steve
Consider the cut-off part of the cone. If the height of the frustrum is 4x, then since the radius shrinks from 5 to 1, using similar triangles, the missing top part has height x.
thus, the volume of the missing part is 1π/3 * 1^2 x = π/3 x
The volume of the frustrum is thus π/3 (5^2*5x - 1^2 x) = 8πx
(a) s1^2 = (4x)^2+(5-1)^2
(b) s2^2 = x^2+1^2
(c) a = π*5^2*(12-x) + 2π*5*s1 - 2π*1*s2
You did not provide any information on how tall the cylinder or frustrum is. All we have is 12 for the total height.
thus, the volume of the missing part is 1π/3 * 1^2 x = π/3 x
The volume of the frustrum is thus π/3 (5^2*5x - 1^2 x) = 8πx
(a) s1^2 = (4x)^2+(5-1)^2
(b) s2^2 = x^2+1^2
(c) a = π*5^2*(12-x) + 2π*5*s1 - 2π*1*s2
You did not provide any information on how tall the cylinder or frustrum is. All we have is 12 for the total height.
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