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Puzzles Alley Hoop! 
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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!
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Comments | terveloc |  21:40:03, 17 Nov 07 |  Newbie
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| bubba's answer is correct. Jackson, in yours, you somehow ended up with a stray three. The actual answer is (20*sqrt(6))/(sqrt(3) + 2*sqrt(2))
This simplifies to 16*sqrt(3) + 12*sqrt(2), which agrees with Bubba.
Try drawing a scale diagram with a ruler. You will clearly see that the intersection is over 10 feet from the ground.
| | | | Jackson | 17:27:46, 09 Oct 07 | Newbie
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| ok, we know from properties of advanced algebra that the intersection of two such lines is half the harmonic mean (2ab)/(a+b) but in our case it is one half so it is ab/(a+b), so all we need is the heights.
We can find the height of the 20 ft with our knowledge of 60-30-90 triangles.so the height is 10(root3)
and for the 30 ft, just Pyth. therom.
root(900 - 100)
root(800)
simplify down with factors, 4*100*2=800, so we can take the 400 out of 800 leaving us with
20root(2)
equation time!
(a*b)/a+b
so (10(root3)*20(root2))/(30(root3) + 20(root2))
so we get: 200(root6) all over 30(root3) + 2o(root2)
we can simplify by taking out a ten from all of everything, leaving us with:
20(root6)/(3(root3)+2(root2))
(the bad version of this is 6.104967191)
Bubba, I think when you did the f(x) and g(x) you started down a path that would lead you to an un-exact answer. And I think that given rounding errors you could have had it at x=6.202 and just gotten variables swapped.
| | | | bubba | 17:00:19, 09 Oct 07 | Newbie
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| First I made 2 equations. I figured out that they are f(x)=1.7321x and g(x)= -2.8284x + 28.28427. I got those with the Pythagorean theorem and then derived the equations. I set f(x)=g(x) and solved and then set it equal to zero. They intersect at x=6.202. Then I plugged that into either of the first two equations and got the height of the hoop at about 10.742 feet high. There may be an easier way. If there is, please tell.
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