How does one solve frustum problems?

Question

I recently encountered one in my SAT prep, but I have never formally learned a method of for solving frustum problems successfully. The problem in question involved a picture, but I will describe all measurements to the best of my ability.

 

Whent eh top of a pyramid or a cone is cut off, the remaining bottom part is called a frustum. Suppose the third (based on height) of the square pyramid described below is cut off and discarded. What will be the volume, in cubic meters, of the remaining frustum?

 

height from base to tip: 24 meters

side length of base: 26 meters

slant length: 30 meters

base length of inner right triangle: 18 meters

Answer 1

The top cut off and the original pyramid are similar three dimensional figures.

The linear scale is 1/3, therefore the ratio of the volume of top to whole is (1/3)³, or 1/27.

If the top is 1/27 the volume of the whole, the frustum is 26/27 of the whole.

V_frustum = (26/27)(1/3)bh, where b is the area of the base_pyramid, and h is the height_pyramid.

Substitute values for b and h and solve for V.

 

 

Answer 2

Pari,

 

The easiest way to do this is to fins the volume of the pyramid before it the top is lopped off, then subtract the volume of the top portion which is removed.

 

What is the formula for the volume of a pyramid?

 

V = 1/3 Bh,

 

where B is the area of the base and h is the height.

 

In order to use this formula using what you were given, you need to find the height and base areas from the given information.

 

Based upon what you stated, the area of the whole pyramid is

 

V = 1/3 (26)² (24) = 5408 m³.

 

Since you are cutting off the top third, which has height 24/3 = 8 m, the only problem remaining is to find the length of the base of the removed portion. This can be done using similar triangles.

 

Let s be half the length of the base of the lopped off pyramid. Then you can draw two similar triangles using a right triangle with base 13, height 24, and hypotenuse √745 (I do not know what the slant length means here, so I am ignoring it entirely. I also do not know what the last measurement indicates, so I will also ignore that as well.) At the top of this right triangle is a smaller one with height 8. You can then solve for s using the equation

 

s/13 = 8/24,

 

which has solution

 

s = 13/3.

 

Next, use the same pyramid volume formula to find the volume removed:

 

V_r = 1/3 (13/3 · 2)² (8) = 200096/999 = 5408/27.

 

Finally, you can subtract the two volumes to get the final answer:

 

V − V_r = 5408 − 5408/27 = 5408(26/27) = 140608/27 m³ ≈ 5207.70 m³.

 

Let me know if you have any questions. There may be a quicker way to do this, but I was in a hurry to get you a solution.