Geometry***Hexagonal Pyramid****tell me if I'm correct...(please)?

Question

In a regular hexagonal pyramid whose base edges measure 2 sqrt3 in., the apothem of the base measures 3 in. If the slant height of the pyramid is 5 in., find the length of its altitude.

My solution:
l^2 = a^2 + h^2
5^2 = 3^2 + h^2
25 = 9 + h^2
h^2 = 16
h = 4in

But I didn't use the info for the side length...I probably have it wrong... :(

Answer 1

Hi,

Your solution is PERFECT!!

Answer 2

In order to do this, you will need to use the Pythagorean Theorem twice.

Firstly, looking at the base, you have the apothem. What you need is the distance between the center and a corner point. We know that the apothem bisects any side and creates a right angle. So, the distance to a corner point is the hypontenuse of the right triange formed between the apothem and a side.

So, half a side is 2sqrt(3)/2 = sqrt(3)
The apothem is 3
So, c^2 = sqrt(3)^2 + 3^2 (where c is the distance to the corner)

c^2 = 3 + 9 = 12
c=sqrt(12) = 2sqrt(3)

Now, the altitude can be caluclated using the slant height as the hypotnuse of a triange with the corner distance and the base and the altitude as the height.

So, 2sqrt(3) + a^2 = 5^2 or 12 + a^2 = 25
a^2 = 25-12 = 13

So a = sqrt(13)

Answer 3

Your work is right. Actually, you used the info for the side length because the apothem of 3 in was calculated from there.

Answer 4

Your solution is correct. The side length was used previously to determine the apothem.