Quadradic Equation Solving by using square roots

tklopfstein

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Jul 16, 2005
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I really need some help if you please:

I have the following problem:

A Tree is hit by lightning. The trunk of the tree breaks and the top of the tree touches the ground 20 ft. from the base of the tree. If the top of the tree is twice as long as the bottom part. approximately how tall was the tree before the tree was hit by lightning?
The book gave this hint:
In a right triangle: a + b = c, where a and b are the legs of the triangle and c is the hypotenuse. So 20 + x= (2x).

Can you show me how to set this problem up and solve it please?[/code][/u][/list][/quote]
 
Hello, tklopfstein!

A tree is hit by lightning.
The trunk of the tree breaks and the top of the tree touches the ground 20 ft. from the base of the tree.
If the top of the tree is twice as long as the bottom part,
approximately how tall was the tree before the tree was hit by lightning?

The book gave this hint: In a right triangle: a + b = c,
where a and b are the legs of the triangle and c is the hypotenuse.
So: 20 + x= (2x).

Can you show me how to set this problem up and solve it please?
Sorry, but I have to say this . . . The problem is already set up!
They already gave you the equation for the problem . . . and you can't solve it?
Did you try to make a sketch?
Code:
    *
    |                      This is the tree 
    | 
 2x |                    before the lightning.
    |
    |
    + <=== lightning struck here
    |
  x |
    |
  - + - - - - - - - -



    + 
    | \                    This is the tree after
  x |   \2x
    |     \                 the lightning struck.
  - + - - - *
      20
Pythagorus says: . (hyp)<sup>2</sup> .= .(leg)<sup>2</sup> + (leg)<sup>2</sup>

So we have: . (2x)<sup>2</sup> .= .x<sup>2</sup> + 20<sup>2</sup>

. . . . . . . . . . . . .4x<sup>2</sup> .= .x<sup>2</sup> + 400

. . . . . . . . . . . . .3x<sup>2</sup> .= .400

. . . . . . . . . . . . . .x<sup>2</sup> .= .400/3
. . . . . . . . . . . . . . . . . . . . _____ . . . . . _
. . . . . . . . . . . . . . . x .= .√400/3 .= .20√3/3 .= .11.54700538 . .11' 6.5"

Therefore, the height of the tree . = . 3x . . 34' 7.5"
 
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