Alley Hoop!

One way.

Can we assume the corners are right angles?

If so, ...

The short height can be found: (h2)^2 + 10^2 = 20^2

h2 = 10*sqrt(3)

The tall height can be found: (h3)^2 + 10^2 = 30^2

h3 = 20*sqrt(2)

Put an Origin at the lower, lefthand corner. y-axis runs up the tall height. x-axis runs across the bottom.

The equation with positive slope is now obvious: y = x*sqrt(3)
The equation with negative slope is a little trickier. y = sqrt(2)*[20 - 2x]

From there, it's not TOO hard to find the point of intersection. Of course, for this problem, you don't really need the x-value. I get a height of 10.742 -- You tell me what the exact value is.
 
I loved your answer...but what do you mean by short heigh? and Long Height?
 
thesloc said:
...what do you mean by short heigh? and Long Height?
Click to expand...
Short Height - the one on the right at the end of the 20
Tall Height - the one on the left at the end of the 30

Apologies for unclear definitions.

10'+11.5" = 10.9583' -- That seems a little high.
 
That's easy. Which one agrees with me? :)

For starters, I have yet to see an EXACT answer. Wasn't that the requirement? That's the only problem with the bottom one, except for the fact that it appears the software knew the answer, rather than the student.

The top one has a little problem. 49.4712 != 90.0000 :? The angle is labled 49.4712º!!! How can one then assume it is 90º to calculate the opposite leg of a right triangle? Doesn't work for me. :(

Somehow, one must get 120/[4*sqrt(3) + 3*sqrt(2)]. It is easily derived from my equations, but I remain a little stunned that someone hasn't presented a prettier solution. Brute force (my speciality) is very unsatisfying.
 
I think you are still missing the ORIGINAL question. You went to an awful lot of trouble to require the EXACT answer. Your feet and inches are not exact. Otherwise, that looks right.
 
I mean; the equations are perfect

But I don't know what the teacher wants....he wants the exact answer. I assumed that metric answers would be enough...
 
If I asked for an "exact" answer, anything rounded or truncated would be unacceptable. "Perfect" is a little harder to track down, but pretty much means the same thing as "exact".

Well, make up your mind and see how it goes.
 
similar triangles.
Code: 
  B 
  * 
  * * 
  *   * 
  *     * 
  *       *                     D 
  *         *                   * 
  *           *              *  * 
  *             *         *     * 
  *               * E  *        * 
  *                 *           * 
  *              *  * *         * 
  *           *     *   *       * 
  *        *        *     *     * 
  *     *           *       *   * 
  *  *              *         * * 
  ******************************* 
  A                 F           C
We have
AC = 10; AD = 15; BC = 20
Pythagorean Theorem gives:
AB = 10sqrt(3)
CD = 5sqrt(5)

Triangles AEF and ADC are similar, so
EF/EC = AB/AC
EC = EF*(AC/AB)
and
DC/AC = EF/AE
AE = EF*(AC/DC)
Adding them gives
AC = EF(AC/AB + AC/DC)
1/EF = 1/AB + 1/DC
We know AB and CD, so we can determine EF
All that's left is simplification
1/EF = 1/(20sqrt(2)) + 1/(10sqrt(3) =
(10sqrt(3) + 20sqrt(2))/(200sqrt(6)) =
(sqrt(3) + 2sqrt(2))/(20sqrt(6))

EF = 20sqrt(6)/(sqrt(3) + 2sqrt(2)) =
20sqrt(6)*(sqrt(3)-2sqrt(2))/(3-8) =
8sqrt(12)-4sqrt(18) =
16sqrt(3)-12sqrt(2)

Exactly.
 
Sorry 'bout that. Right method, wrong numbers. If you had read the first line...
---------------------
Gene
 
Gene's answer is correct. The incorrect lengths must be only typos.

I gave you a correct answer:

120/(4*sqrt(3)+3*sqrt(2))

Using Gene's MUCH BETTER similar triangles, I get an equivalent expression:

20*sqrt(6)/(2*sqrt(2)+sqrt(3))

Both of these are equivalent to Gene's:

16*sqrt(3) - 12*sqrt(2)

Of course, I only verified it algebraically and checked 15 decimal places numerically, but what do I know?! Maybe it's still wrong.
Maybe it's time to move on to a new problem. It would be a shame to lose too much more sleep over a problem that has been solved in several ways already.
 
Sorry 'bout that...the method is correct, however; How do I plug the new solution into the correct method?
 
EDIT: Okay; I think I'm on to something...
Basically I have to solve the intersection point for your lines?
x*sqrt(3)=qrt(2)*[20 - 2x]

so that will give me=120/[4*sqrt(3) + 3*sqrt(2)]? *or is this the height?*

How do I derive a height from the intersection point?
 
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