SOLUTION: I have tried this one and cannot figure it out. I have tried versions of the Pythagorean Theorem. Please Help! Two posts, one 12 meters high and the other 21 meters high, are 30

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Question 51300: I have tried this one and cannot figure it out. I have tried versions of the Pythagorean Theorem. Please Help! Two posts, one 12 meters high and the other 21 meters high, are 30 meters apart. A single wire runs from the top of the first post to the ground and from the ground to the top of the second post. The wire is attached to the ground at a distance of x units from the first post. What is a function representation that states the total length of the wire?
Found 2 solutions by venugopalramana, rapaljer:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
I have tried this one and cannot figure it out. I have tried versions of the Pythagorean Theorem. Please Help! Two posts, one 12 meters high and the other 21 meters high, are 30 meters apart. A single wire runs from the top of the first post to the ground and from the ground to the top of the second post. The wire is attached to the ground at a distance of x units from the first post. What is a function representation that states the total length of the wire?
MAKE A SKETCH AS FOLLOWS
LET OX REPRESENT GROUND....LET AB BE ONE POST WITH A ON GROUND (OX LINE) AND B AT THE TOP...AB=12 M......LET CD BE THE SECOND POST WITH C ON GROUND (OX LINE) AND D AT THE TOP....CD=21 M....AC=30 M
TAKE A POINT P ON THE GROUND LINE BETWEEN A AND C ,WITH AP=X
NOW WE HAVE 2 RIGHT ANGLED TRIANGLES ABP AND PCD....PB AND PD ARE THE HYPOTENUSES WHICH TOGETHER REPRESENT THE WIRE LENGTH....L= PB+PD
BUT PB =SQRT{PA^2+AB^2}=SQRT(X^2+12^2)=SQRT(X^2+144)
AND PD =SQRT(PC^2+CD^2)=SQRT{(AC-PA)^2+CD^2}=SQRT{(30-X)^2+21^2}
HENCE THE REQUIRED FUNCTION IS
L=SQRT(X^2+144)+SQRT{(30-X)^2+441}

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
You have two right triangles, in which the two poles are the vertical legs of the triangles, and the length of the wire is the sum of the two hypotenuses (or would that be hypotenii???). Since the distance between the poles is 30 meters, let x= distance from first pole to the point where the wire touches the ground, and 30-x = distance from this point on the ground to the base of the second pole.

By Theorem of Pythagoras, the length of the wire from the top of the first pole to the ground is sqrt%2812%5E2%2Bx%5E2%29+, and the distance from the top of the second pole to the ground is sqrt%2821%5E2+%2B+%2830-x%29%5E2%29+. The total length of the wire, is the sum of these radicals
sqrt%28144%2Bx%5E2%29+%2B+sqrt%28441%2B900-60x%2Bx%5E2%29+

It doesn't simplify much: sqrt%28144%2Bx%5E2%29+%2B+sqrt%281341-60x%2Bx%5E2%29+

R^2 at SCC