I'm confused...

Question

a natural history museum is building a pyrmidal glass structure for its tree snake exhibit. its research team has designed a pyramid with a square base and with a height that is 2 yards more than a side of its base. the volume of the pyramid must be 147 cubic yards. what are the dimensions?

Answer 1

Michelle,

 

Let's do it step by step:

Let x = the side of the pyramid's square base (in yards)

Let h = the height of the pyramid (in yards)

Let V = the volume of the pyramid (in cubic yards)

 

The we know or you can look up or even derive the volume formula for a 3-dimensional square base pyramid:

 

V = (1/3) h x².  But the problem says that h = (x + 2) yards, and that V = 147 cu. yds

 

So  147 = (1/3)(x + 2)x² = [x³ + 2x²]/3 , or   [x³ + 2x²] = 3 (147) = 441.

 

Rather than try to solve this algebraically for x (which isn't easy but can be done), I will approach it numerically since I believe they made x an integer.  

 

To get into the neighborhood of the solution, I would first solve x³ = 441, so x = 7.6 yards approximately.

 

Now let me do trial and error on both sides of that:

If x = 6 yards, [x3 + 2x2] = 216 + 72 = 288 too low!

If x = 7 yards, [x3 + 2x2] = 343 + 98 = 441 Right on!

If x = 8 yards, [x3 + 2x2] = 512 + 128 = 640 too high!

 

So the pyramid is 7 yards square at the bottom and 9 yards high in the center.

 

If I had not hit it exactly, then I would do tenths, x = 6.7, 6.8, 6.9, 7.1, 7.2, 7.3

And then hundreds, each time squeezing the possible solution into a narrower band.

 

 

Answer 2

Michelle, the problem states that the height is 2 yards more than a side of its base.

 

Let x = the length of a side of its base

Then x+2 = the pyramid's height.

 

The formula for the volume of a pyramid is:

 

V=1/3*B*h

 

Where B is its base area and h is its height.

 

Now B = x^2 and h=x+2 and V=147. This implies

 

147=1/3*x^2*(x+2) To simply this problem, I multiply the equation by 3 in which becomes:

 

147*3=x^2*(x+2) In which is:

 

441=x^3+2x^2. Now subtract 441 from both sides of the equation in which it becomes:

 

x^3+2x^2-441=0 Hence, this becomes a third degree polynomial in which you could use your graphing calculater to determine what values x make the equation zero by setting it to y=x^3+2x^2-441.

 

Or you could make a table and guess what valus of x make the equation being zero. Note: from my graphing calculater I got x=7 yards being the distance of a side of the base. 

 

Thus, h=7+2=9 yards.

 

Now, check your answer by plugging in the values in determine the volume of the pyramid.

 

V=1/3*7^2*9=147 yards^3 This checks.