Glencoe Math: Course 3, Volume 2
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Glencoe Math: Course 3, Volume 2 View details
5. Surface Area of Cones
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Exercise 12 Page 636

Remember the formula for the surface area of a cone.

Example Solution:
Practice makes perfect

We are asked to draw a cone with a surface area that is between 50 and 100 square units.

Cone
Before we do that, let's remember the formula for the surface area of a cone. The surface area S.A. of a cone with slant height l and radius r is given as follows. S.A. = π r l + π r^2We should choose such values of r and l that the expression on the right-hand side has a value between 50 and 100. 50 < π r l + π r^2 <100 There are infinitely many pairs of numbers that satisfy the inequality. We should choose just one. Since it is easier to focus on just one variable, let's say that the slant height will always be twice the radius. l =2r We can now substitute 2r for l in the expression and simplify.
50 < π r l + π r^2 <100
50 < π r ( 2r) + π r^2 <100
50 < 2 * π * r * r + π r^2 <100
50 < 2 * π * r ^2 + π r^2 <100
50 < 2π r ^2 + π r^2 <100
50 < 3π r ^2 <100
Now there is just one variable for us to consider. See that we need to choose a value of r that satisfies the last inequality. 50 < 3π r ^2 <100 Notice that r=3 satisfies the condition. We can check that using a calculator. Since the radius of our cone is 3 units, this means that the slant height is 2(3)=6. Let's now find the surface area of the cone.
S.A. = π r l + π r^2
S.A. = π( 3)( 6)+ π( 3)^2
S.A. = π(3)(6)+π(9)
S.A. = 18π +9π
S.A. = 27 π
S.A. = 84.823001 ...
S.A. ≈ 84.82
The volume of the cone is about 84.82 units^3. Since this more than 50 and less than 100, the dimensions are correct. Last, let's draw the cone.
Dimensions

Note that this is just a sample solution. The slant height does not need to be twice the radius. Also, the radius does not have to be a natural number. We could choose any two numbers r and l that satisfy the inequalities.