1. ## Pyramid Math Problem.

Kay so we had this problem today that we're sposed to do tonight. Goes something like this:

There's a pyramid 150m tall, with a square base. Each side of the base is 230m. Find the area of the horizontal cross section 50m above the base, using 3 different methods.

So first off, I did a right triangle to get the height of one of the 4 triangles forming the pyramid. So the triangle would have a base of 115 m, and the height would be 150. Pythagorean theorem says the hypotenuse is 189 .

Then I cut off a part of the triangle at the 50m mark, and form another triangle with that, so I have two right triangles and a rectangle. I need to find the length of the rectangle to get the area of the horizontal cross section 50m up. So I use sin on the original triangle to get the base angle, which turns out to be 52.53. We also know that the height of the newly formed base triangle is 50 because that's where I need to find the horizontal cross section.

So I use the sin function again and find out that the base of that triangle is 38.33m. Subtract that from the original base of 115m, and I get about 76.67. Now I have to remember that this is only half of the length of the horizontal cross section, so I multiply that by two, then use the area of a square formula and get the answer, which happens to be around 23, 515m^2.

Now this is great and all, but the question requires me to solve this using 3 different methods. So I'm looking for helping figuring out what the other two methods are. Thanks! =D

2. Here is a second for you.

The sides of a pyramid are all triangles. You can draw a horizontal line across and it will be proportional to the base based on how far up you are in terms of the height of the triangle. For example, one side of the pyramid is an equilateral triangle with base 230 and height 150.

Let's make X a fraction.
H will be the height of the triangle
W will be the width of the base

For a base Z that is
X*H above the original base, it will have a length of (1-X) * W

A base 50 m(1/3) above the original base will have 2/3(1-X) times the original width of the base.
2/3 * 230 = 153.3~
That is the length of one side of your square cross section.
Square it
23511.1~

Alright, just need one more now =P

3. Originally Posted by Crysolice
Kay so we had this problem today that we're sposed to do tonight. Goes something like this:

There's a pyramid 150m tall, with a square base. Each side of the base is 230m. Find the area of the horizontal cross section 50m above the base, using 3 different methods.

...
With the given values you can calculate the volume of the complete pyramid:

$V_{cp} = \dfrac13\cdot a^2 \cdot h$

If you cut off the small pyramid at the top you know its height (=100 m) and you know that all other measures has changed by the same factor. That means:

$V_{sp}=\left(\dfrac23\right)^3\cdot V_{cp}$

Let x denote the side of the cross sectional square then you have to solve for x:

$\dfrac13\cdot x^2 \cdot 100=\left(\dfrac23\right)^3\cdot V_{cp}$