SOLUTION: At a certain time of day a tree that is X meters tall cast a shadow that is X -3 m long if the distance from the top of the tree to the end of the shadow is X +1 m what is the heig

Question 1148735: At a certain time of day a tree that is X meters tall cast a shadow that is X -3 m long if the distance from the top of the tree to the end of the shadow is X +1 m what is the height X of the tree
Found 3 solutions by ikleyn, ankor@dixie-net.com, William_KO:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.

You have a right angled triangle with the legs X and X-3 and the hypotenuse (X+1) meters. + = = = 0 Use the quadratic formula = = = = Only the root X = makes sense. ANSWER. The height of the tree is X= = 6.828 meters.

Solved.

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Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
At a certain time of day a tree that is X meters tall cast a shadow that is X -3 m long
if the distance from the top of the tree to the end of the shadow is X +1 m, what is the height X of the tree
:
This forms a right triangle: a^2 + b^2 = c^2; where
a = x
b = (x-3)
c = (x+1)
Using Pythagoras we have
x^2 + (x-3)^2 = (x+1)^2
FOIL
x^2 + x^2 - 6x + 9 = x^2 + 2x + 1
Combine like terms on the left
x^2 + x^2 - x^2 - 6x - 2x + 9 - 1 = 0
x^2 - 8x + 8 = 0
Use the quadratic formula a=1; b=-8; c=8
the two solutions
x = 1.17
and
x = 6.83, this is the only solution that makes sense (x-3)
therefore the tree is 6.83 m high
:
:
Check: find the hypotenuse using 6.83 and 3.83
h =
h = 7.83 which is x+1

Answer by William_KO(1)   (Show Source): You can put this solution on YOUR website!
(x+1)^2=(x-3)^2+x^2
(x^2+2x+1)=(x^2-6x+9)+x^2 square them
x^2+2x+1=2x^2-6x+9 move everything to the right side
0=x^2-8x+8 then use quadratic formula

x= 4+2 root 2 or 4-2 root 2
Hope this helps.

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