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1. A sphere of radius 10 inches is inscribed in a cone with a base of radius 15  inches. In cubic inches, what is the volume of the cone?

 

2. A right square pyramid with base edges of length 8sqrt(2) ts each and slant edges of length 10 units each is cut by a plane that is parallel to its base and which sits units above its base. What is the volume, in cubic units, of the new pyramid that is cut off by this plane?

 

                         ^

                   3 units

 Apr 6, 2019
 #1
avatar+101150 
0

First one

 

Looking at  the cross-section of the cone....half of the measure of the angle formed by the intersection of the side of the cone and its base is given by arctan (10/ 15)   =  arctan(2/3)

 

 

So.....again, looking at the cross-section of the cone...twiice this measure will be the angle formed by the intersection of the side and the base =  2arctan(2/3)   

 

So....we can find the height of the cone, h, as :

 

tan (2  arctan(2/3)) =    h/15 ⇒   h =   15 * tan (2 arctan(2/3))  =  15*(2.4) = 36

 

So....the volume of the cone is given by :

 

(1/3) pi * radius^2 * height  =

 

(1/3) pi *15^2 * 36 = 

 

2700 pi units^3 

 

 

cool cool cool

 Apr 6, 2019
 #2
avatar+101150 
0

The second one cannot be solved unless we know how many units the parallel plane to the base sits above that base.

 

cool cool cool

 Apr 6, 2019
 #3
avatar+101150 
0

Here is the first one without using any trig

Look at the following diagram :

 

 

 

Again.....imagining that we have a cross-section.....

Let EF  be the radius of the sphere drawn to tangent DB

Let  AD be the cone height

We have triangles DAB and DEF

Angle DAB  = angle DFE = 90°

And angle ADF = angle EDF

So....by AA congruency....triangle ADB  is similar to triangle FDE

So.....DE / DB   = FE/AB

Let DB = the slant height of the cone = S

Let DE = cone height - 10  =  H - 10

FE = 10

AB = 15

 

So we have that

 

H-10    =   10

____        __

 S             15

 

H - 10           2

_____   =    ____

  S                 3

 

H - 10   = (2/3)S

H = (2/3)S + 10

 

So......using the Pythagorean Theorem

 

S^2 - [ (2/3)S + 10 ]^2  = 15^2

S^2 - [ (4/9)S^2 + (40/3)S + 100] =  225

S^2 - (4/9)S^2 -(40/3)S^2 - 100  =  225

(5/9)S^2  - (40/3)S - 325   = 0       multiply through by 9/5

S^2  - 24S  - 585  = 0       factor as

(S - 39) (S + 15) =  0

Since S is positive....then setting the first factor to 0  and solving for S  gives that S = 39

So....the height of the cone  = (2/3)(39) + 10   =  26 + 10   = 36

 

So....the volume of the cone is

 

(1/3)(15)^2* 36 * pi =

 

2700 pi  units^3

 

 

cool cool cool

 Apr 7, 2019

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