SOLUTION: the radius and the height of a cone are in the ratio 4:3 . the area of the base is 154 cm^2. find the area of the curved surface. (please answer this step by step)
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Question 842137: the radius and the height of a cone are in the ratio 4:3 . the area of the base is 154 cm^2. find the area of the curved surface. (please answer this step by step)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the radius and the height of the cone form a triangle which is a right triangle that is in the ratio of 3:4:5.
3 is the height of the cone.
4 is the radius of the cone.
5 is the slant height of the cone.
if we let x equal the common multiple of this ratio, then we get:
3x is the height of the cone.
4x is the radius of the cone.
5x is the slant height of the cone.
if x is equal to 1, then the ratio is 3:4:5.
if x is equal to 2, then the ratio is 6:8:10, etc.
the formula for the total surface area of the cone is equal to pi * r * s + pi * r^2
pi * r * s is the area of the curved surface.
pi * r^2 is the area of the base.
we know the area of the base is equal to 154 cm^2.
we can use that fact to solve for x.
the formula for the area of the base is pi * r^2 = 154
since r is equal to 4x, this means that the formula for the base becomes:
pi * (4x)^2 = 154 which can be simplified to:
pi * 16x^2 = 154
we solve for x in this formula to get:
x = sqrt (154 / (16 * pi) which makes x = 1.750352152
the area of the base is equal to (4x)^2 * pi which is equal to (4 * 1.750352152)^2 * pi which is equal to 154.
the area of the curved surface of the cone is equal to pi * (4x) * (5x) which is equal to pi * (4 * 1.750352152 * 5 * 1.750352152) which is equal to (pi * 4 * 5 * (.750352152)^2) which is equal to 192.5 cm^2.
that's your answer.
the area of the base is equal to 154 cm^2.
the area of the curved surface area is equal to 192.5 cm^2.
the total surface area is equal to 346.5 cm^2.
here a link to where i found the formula for the surface area of a cone.
http://math.about.com/od/formulas/ss/surfaceareavol_2.htm
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