SOLUTION: Good day! Can you please help me with this problem? The lateral edge of a regular hexagonal pyramid is two times the length of the base edge. If the apothem of the base is 8

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Question 1005912: Good day! Can you please help me with this problem?

The lateral edge of a regular hexagonal pyramid is two times the length of the base edge. If the apothem of the base is 8 cm, find the altitude and volume of the cone inscribed in a pyramid.

Thank you!

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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The lateral edge of a regular hexagonal pyramid is two times the length of the base edge.
If the apothem of the base is 8 cm, find the altitude and volume of the cone inscribed in a pyramid.
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Wikipedia says:

"The apothem of a regular polygon is a line segment from the center to the midpoint of one of its sides. 
Equivalently, it is the line drawn from the center of the polygon that is perpendicular to one of its sides. 
The word "apothem" can also refer to the length of that line segment. Regular polygons are the only polygons 
that have apothems. Because of this, all the apothems in a polygon are be congruent".


The Figure shows a regular hexagonal pyramid,  an apothem of the regular hexagon in its base        
(OP),  the pyramid's height  (RO)  and its slant height  (RP).

The length of the apothem  OP  is given:  |OP| = 8 cm.
We also know that the length of the lateral edge is twice the base edge length:  |AR| = 2*|AB|.

The apothem  OP  is the height of the regular triangle  DELTAOAB.
Therefore,  |OP| = sqrt%283%29%2F2*|AB| = sqrt%283%29%2F2*|OA|.  Hence,  |OA| = |AB| = 2%2Fsqrt%283%29*|OP| = 2%2Fsqrt%283%29*8 = 16%2Fsqrt%283%29.

Since the length of the lateral edge is twice the base edge length, we have
|AR| = 2*|AB| = 2*16%2Fsqrt%283%29.

Now from the right-angled triangle  DELTAAOR  we have for the height of the pyramid
|OR| = sqrt%28abs%28AR%29%5E2+-+abs%28OA%29%5E2%29 = sqrt%28%282%2A16%2Fsqrt%283%29%29%5E2+-+%2816%2Fsqrt%283%29%29%5E2%29%29 = sqrt%28+%284%2A16%5E2%29%2F3+-+%2816%5E2%29%2F3%29+%29 = sqrt%28+%283%2A16%5E2%29%2F3+%29 = 16 cm.



  Regular hexagonal pyramid,
its height and a slant height.

Thus for the inscribed cone you just know the radius of its base |OP| = 8 cm and its height |OR| = 16 cm.

It means that you can easily calculate the volume of the inscribed cone V = 1%2F3.pi%2Aabs%28OA%29%5E2%2Aabs%28OR%29 = 1%2F3.pi%2A8%5E2%2A16.