SOLUTION: Hello tutors, I am currently working on a Math project, and am on the final instruction. I have had to design a building structure, with office blocks (2.5 m in height) in the

Question 578088: Hello tutors,
I am currently working on a Math project, and am on the final instruction. I have had to design a building structure, with office blocks (2.5 m in height) in the building (so each floor is 2.5 m in height). Now the final task is to find the volume of wasted space of the structure (shown in the link below, the wasted space are the triagles by the office blocks).
The equation of the outer curve (that the structure is based on) is:

I have asked my math teacher and he has told me there's an equation to work out the length of each individual floor. If I known how to find the length of each floor I'd know how to be able to find the volume afterwards. I have included a second image of just a single floor if this in any way helps.

Thank you very much! :)
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
It appears as if no tutor has commented to this problem yet. So before it drops
down the line so far that it disappears from the list, I'll offer you some input
that might be useful to you. However, before that, I'm really impressed with the visual
that you provided with your problem. That was quite spectacular and made your statement
of the problem considerably easier to understand. Good job!!!
.
Now to your problem. My first comment is, "Are you sure that the equation:
.

.
describes the parabolic shape of the building?" The reason I ask this comes from
the following analysis:
.
Suppose y equals zero. That should cause the equation to represent the two places
where the graph intersects the horizontal axis on which the bottom of the building
rests. But if you set y equal to zero, the equation becomes:
.

.
Solve this by factoring the right side of the equation as below:
.

.
Notice that the right side of the equation will equal zero also if either of the factors
are zero. So either:
.
or
.
To solve for the value of x that will make the second factor equal to zero, let's first
multiply both sides (all terms) by 36 to get rid of the denominator. This multiplication
results in:
.
which becomes:
.
When you solve this you get the second answer for the intercept as being
.
So the equation you gave for the parabolic arch should intercept the horizontal axis at 0,
as it does, and also at +108, but it doesn't. Instead of 108 it appears to show the right
hand intercept at 150. Is the equation in error or the drawing? Or perhaps I'm do not
understand the problem as well as I should before I comment to it.
.
Regardless of that, here's an idea of how you might determine the lengths that you need.
Look at the drawing you provided of just the single first layer. You need to get the
length of the top horizontal line. Note that it intersects with the parabolic graph on
both ends. Therefore, the endpoints of the line must satisfy the equation for the parabola.
In other words, the endpoints can be solved for x if the value of y is known in the equation
for the parabola. For this first level, what is the value of y for the top horizontal
line? It is 2.5, is it not? So for the given equation you could write the equation as:
.

.
This equation can be solved by first multiplying both sides (all terms) by 36 to get rid
of the denominator. As a result of this multiplication the equation becomes:
.

.
Add x^2 - 108x to both sides and the equation then becomes:
.

.
This is in the standard quadratic form of:
.

.
and therefore it can be solved for the two values of x at the endpoints by applying the
quadratic formula:
.

.
In which you substitute 1 for a, -108 for b, and 90 for c to get:
.

.
This simplifies to:
.

.
and the terms in the radical combine to give:
.

.
Taking the square root results in the two answers of:
.
and
.
and this leads to the two answers for x of 0.839865 and 107.160135.
.
(This is correct for the given equation results in the ground level intercepts are
at 0 and 108, not 0 and 150.)
.
For each following set of x values, all you have to do is increase the value of y by 2.5.
So for the next level up, you would solve the equation:
.

.
This would lead to the quadratic form:
.

.
And in general, you could say that the quadratic solution is:
.

.
which by multiplying out the third term becomes:
.

.
where n is 1, 2, 3, 4, ... up to the ceiling level for the top floor.
.
by applying the quadratic formula you get:
.

.
and by multiplying out the terms you get:
.

.
And then you can divide by 2 to get:
.

.
Now to get the values of x for the multiple levels of the building, just substitute
for n the values of 1, 2, 3, 4, 5, and so on and calculate the pairs of x at each ceiling
level for the entire building. That should enable you to find the triangles of unused space
at each level.
.
I hope this helps you with your problem. Check my math and logic to see if I made some
dumb mistake in doing this analysis rather quickly. If my interpretation of the problem
doesn't help you, please post your problem again and some other tutor may be able to
solve it in a better way than I did.
.

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