Surface Area of a Conical Frustum (Truncated Cone)
A truncated cone or conical frustum is a solid shape based on a cone, except that the pointed tip has been sliced off somewhere below the point. In a standard conical frustum, the slice is parallel to the base, so it has two parallel circular faces for the top and bottom as with a cylinder. But unlike a cylinder, the circular faces of a conical frustum are not equal sizes.
To find the surface area of a conical frustum, you must know the height of the object, the radius of the base, and the radius of the top. The formula, its derivation, and some examples are explained here.
Surface Area Formula for Truncated Cone
If you exclude the area of the bases, then the lateral surface area of a conical frustum is
π(a + b)sqrt[(b - a)^2 + h^2]
where "a" is the radius of the top, "b" is the radius of the bottom, and "h" is the height of the object. If you include the top and bottom in your calculation, then the total surface area of this solid shape is
π(a + b)sqrt[(b - a)^2 + h^2] + π(a^2 + b^2).
Derivation of the Formula
The lateral surface area of a conical frustum is the difference in surface area between the larger cone and the part that is cut off.
For a conical frustum with dimensions b, a, and h, the dimensions of the larger cone from which it is cut are
base radius = b
height = bh/(b-a)
The dimensions of the smaller cone that has been cut off the top are
base radius = a
height = ah/(b-a)
The lateral surface area of a cone with a radius of R and a height of Y is
πR*sqrt(R^2 + Y^2).
If we plug the larger and smaller cone dimensions into this expression and subtract, we will get the lateral surface area of a conical frustum.
πb*sqrt[ b^2 + (bh/(b-a))^2 ] - πa*sqrt[ a^2 + (ah/(b-a))^2 ]
= (πb^2)sqrt[ 1 + (h/(b-a))^2 ] - (πa^2)sqrt[ 1 + (h/(b-a))^2 ]
= π(b^2 - a^2)sqrt[ 1 + (h/(b-a))^2 ]
= π(b + a)sqrt[ (b-a)^2 + h^2 ].
A conical frustum has a base radius of 20 cm, a top radius of 11 cm, and a height of 10 cm. This means
a = 11
b = 20
h = 10
Plugging these values into the lateral surface area formula gives you
π(20+11)sqrt[ (20-11)^2 + 10^2 ]
≈ 1310.24 square centimeters.
A truncated cone has a surface area of 2500 square centimeters and a height of 10 cm. If the base has twice the radius of the top, what is the radius of the base?
Here we have h = 10 and b = 2a. If we plug these into the formula and set it it equal to 2500, we get
2500 = π(2a + a)*sqrt[ (2a-a)^2 + 10^2 ]
2500 = 3πa*sqrt(a^2 + 100)
2500^2 = (9π^2)(a^2)(a^2 + 100)
9π^2*a^4 + 900π^2*a^2 - 6250000 = 0
a^2 = [-900π^2 + sqrt(810000π^4 + 225000000π^2)]/(18π^2)
a^2 ≈ 219.9294965
a ≈ 14.83
Therefore the radius of the top is about 14.83 cm, and the radius of the base is about 2*14.83 = 29.66 cm.
A conical frustum as integer height and radii. If its total surface area is 378π, what could be its dimensions?
For this problem, we first set up the surface area equation
378π = π(a + b)sqrt[(b-a)^2 + h^2] + π(a^2 + b^2)
Dividing both sides by π gives us
378 = (a + b)sqrt[(b-a)^2 + h^2] + a^2 + b^2
So the real problem is finding h, a, and b such that (b-a)^2 + h^2 is a perfect square. If we solve the above equation for h, we get
h = 2sqrt[(189 - a^2)(189 - b^2)]/(a + b)
At this point, we just need to use trial-and-error or guess-and-check to find pairs (a, b) such the expression on the right side equals an integer. One condition to make the process easier is that a^2 and b^2 must both be less than 189 or both greater than 189. Testing integer values of a and b gives us some possible solutions
a = 3, b = 12, h = 12
a = 6, b = 11, h = 12
a = 9, b = 9, h= 12
The last one is actually a cylinder since a = b.
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