CK-12 Geometry Concepts

11.6. Cones http://www.ck12.org

36 π=πr^2 +πr( 5 ) Because every term hasπ,we can cancel it out.

36 =r^2 + 5 r Set one side equal to zero, and this becomes a factoring problem.

r^2 + 5 r− 36 = 0

(r− 4 )(r+ 9 ) = 0 The possible answers forrare 4 and− 9 .The radius must be positive,

so our answer is 4.

Example C

Find the volume of the cone.

To find the volume, we need the height, so we have to use the Pythagorean Theorem.

52 +h^2 = 152

h^2 = 200

h= 10

√

2

Now, we can find the volume.

V=

1

3

( 52 )

(

10

√

2

)

π≈ 370. 24

Watch this video for help with the Examples above.

MEDIA

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CK-12 Foundation: Chapter11ConesB

Concept Problem Revisited

The standard cone has a surface area ofπ+ 6 π= 7 π≈ 21. 99 in^2. The “king size” cone has a surface area of
4 π+ 16 π= 20 π≈ 62 .83, almost three times as large as the standard cone.

Vocabulary

Aconeis a solid with a circular base and sides that taper up towards a vertex. A cone has aslant height.

Surface areais a two-dimensional measurement that is the total area of all surfaces that bound a solid.Volumeis a
three-dimensional measurement that is a measure of how much three-dimensional space a solid occupies.