SOLUTION: During a hurricane, a telegraph pole was broken in such a way that the top struck the level ground at a distance of 20 feet from the base of the pole. It was replaced by an identic
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Question 901768: During a hurricane, a telegraph pole was broken in such a way that the top struck the level ground at a distance of 20 feet from the base of the pole. It was replaced by an identical pole which was broken by another gale at a point 5 feet lower and the top struck the ground a distance of 30 feet from the base. What was the original height of the poles? Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! i grapahed the equations and then i solved them algebraically for you.
they both point to the same solution which is a good sign.
the solution is:
the base of the pole from the original break was 21 feet and the piece of the pole that fell to the ground was 29 feet for a total of 50 feet in length.
the base of the pole from the replacement break was 16 feet and the piece of the pole that fell to the ground was 34 feet for a total of 50 feet again.
the equations were set up as follows:
two right triangles were formed.
The first case triangle had the following characteristics.
x = vertical leg of the triangle = bottom of the pole that is still connected to the base.
y = hypotenuse of the triangle = top of the pole that fell over.
20 = horizontal leg of the triangle = the distance from the top of the pole that is touching the ground to the base of the pole.
the second case triangle had the following characteristics.
x-5 = vertical leg of the triangle = bottom of the pole that is still connected to the base.
y+5 = hypotenuse of the triangle = top of the pole that fell over.
30 = horizontal leg of the triangle = the distance from the top of the pole that is touching the ground to the base of the pole.
the pythagorean formula was used on both triangles.
first case triangle is x^2 + 20^2 = y^2
second case triangle is (x-5)^2 + 30^2 = (y+5)^2
these 2 equations were graphed and the intersection of the two graphs is the solution.
that solution is that x = 21 and y = 29 as shown on the graph.
the graphical solution is shown below:
the algebraic solution was solved as follows:
both equations needed to be solved simultaneously since the same solution for x and y had to be common to both equations.
start with:
x^2 + 20^2 = y^2
(x-5)^2 + 30^2 = (y+5)^2
the vertical leg of the replacement pole was 5 feet shorter than the vertical leg of the original pole which is the reason for (x-5) as the vertical leg.
since the vertical leg was 5 feet shorter, the upper part that fell had to be 5 feet longer which is the reason for (y+5) as the hypotenuse.
simplify these equations to get:
x^2 + 400 = y^2
(x-5)^2 + 900 = (y+5)^2
solve for y^2 and y from the first equation to get:
y^2 = x^2 + 400
y = sqrt(x^2 + 400)
simplify the second case equation of (x-5)^2 + 900 = (y+5)^2 to get:
x^2 - 10x + 25 + 900 = y^2 + 10y + 25
replace y^2 with x^2 + 400 and replace y with sqrt(x^2 + 400) to get:
subtract x^2 and subtract 25 from both sides of the equation and subtract 400 from both sides of the equation to get:
-10x + 500 = 10*sqrt(x^2 + 400)
divide both sides of the equation by 10 to get:
-x + 50 = sqrt(x^2 + 400)
rearrange the terms on the left side of the equation to get:
50 - x = sqrt(x^2 + 400)
square both sides of the equation to get:
2500 - 100x + x^2 = x^2 + 400
subtract x^2 from both sides of the equation and subtract 400 from both sides of the equation and add 100x to both sides of the equation to get:
2100 = 100x
divide both sides of the equation by 100 to get:
21 = x
that's the length of the vertical leg of the first case triangle.
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using pythagorus again, you get:
21^2 + 20^2 = y^2
solve for y^2 to get:
y^2 = 841
solve for y to get:
y = 29
the original pole was broken 21 feet above the ground and the top of the pole that fell over was 29 feet in length for a total length of 50 feet.
the replacement pole was broken 5 feet further down, so the replacement pole was broken 16 feet above the ground the the top of the pole was 34 feet in length for a total length of 50 feet again.
since x = 21 and y = 29, then (x-5) = 16 and (y+5) = 34.
21^2 + 20^2 = 29^2
16^2 + 30^2 = 34^2
solving these equations gets:
841 = 841
1156 = 1156
this confirms the solution is correct with x = 21 and y = 29
