Math Problem: Alley Hoop!

[gentry_joey]

As the fall temperatures cool off, basketball season warms up. Here's a B-ball related puzzle to get you in the mood for hoops.

Jenny and Jesse want to set up a basketball hoop to practice their shooting. The only place they have available is a wall at the end of an alley 10 feet wide. They have two ladders, one 20 feet long and one 30 feet long. They place the ladders as shown and put the hoop where the ladders intersect. How high off the ground is the hoop?

Be sure your answer is exact!

[ShawnD1]

I have an estimated answer of 10.74 using trig but I'm still working on the exact answer. I'm comparing the equation for each of the 2 ladders then trying to find the intersect without using my calc (it has some square roots).


OK well I don't know how to compare lines using a pen an paper (Yes, I am that retarded). My calc says the thing should be 10.74225ft tall.

Incase anybody here knows how to compare lines using algebra, here are the 2 formulas:
Y = root3 * x
Y = -root8 * (x-10)

*final edit*

OK I GOT IT.
It's circled at the bottom.

[zidane182]

i don't remember trig =/

[VHockey86]

according to his formulas (which i didnt check), they'd intersect at (6.202041,10.74225) on a cord grid, so the hieght would be 10.74225

[VHockey86]

or more specifically.....
4(4(3)^.5 - 3(2)^.5)

edit.... incase anyone wants a more specfic breakdown

y = 3^.5
y= -8^.5(x-10)

thus

x*3^.5 = (-x-10)(8)^.5 .... square both sides

3x^2 = (x^2-20x+100)(8)
0 = 5x^2 -160x +800 .... quadratic gives...

(16 + 4(6)^.5)

now substitute into original and simplify to
16(3)^.5 - 12 (2)^.5 or....
4(4(3)^.5 - 3(2)^.5)

[ShawnD1]

Check out my new and improved answer.

(10 * root8 * root3)/(root3 + root8)

[VHockey86]

Same thing... but your new and improved answer simplifys a hell of alot more =)
(rationalize the denomiator)

[ShawnD1]

OK my answer with a rational denominator looks like this:

[10 * root24 * (root3 - root8)] / (-5)

[VHockey86]

which equals... (60(2)^.5 - 80(3)^.5)/5
aka
16(3)^.5 - 12 (2)^.5

[ShawnD1]

The slope for each of the lines is pretty straight forward. The second equation has a negative slope because it is going down and it's (x - 10) because the line hits the X axis 10 to the right. For any equation, you should think of the function as being relative to the Y axis. The variable of the function should be x + relative location of the Y axis. My Y axis is 10 left so the variable is (x - 10).

If I have the function X^2, I can see that when my variable being acted upon, when equal to 0, lines up with my Y axis. Now lets say I shift that function 10 units to the left. When I make my variable 0, I can see that I'm not lining up with the Y axis. Now think, where is the Y axis from where I am? It's 10 units right so I change my X to be (X + 10).