SOLUTION: A large pipe is placed next to a wall and a 1 ft high block is placed 5 ft from the wall to keep the pipe in place as show in the figure. What is the radius of the pipe ? The answe

Algebra ->  Circles -> SOLUTION: A large pipe is placed next to a wall and a 1 ft high block is placed 5 ft from the wall to keep the pipe in place as show in the figure. What is the radius of the pipe ? The answe      Log On


   

Question 349269: A large pipe is placed next to a wall and a 1 ft high block is placed 5 ft from the wall to keep the pipe in place as show in the figure. What is the radius of the pipe ? The answer in the book says 6-square root 10, but I'm not sure how they got that answer. I have a link to the image from my textbook.
http://img237.imageshack.us/img237/2158/dsc00343ts.jpg
Found 2 solutions by Alan3354, Fombitz:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Extend a line from the top of the block to the vertical line thru the center of the pipe.
You get a right triangle with r as the hypotenuse.
r%5E2+=+%285-r%29%5E2+%2B+%28r-1%29%5E2
If you solve that you'll get 6 ± sqrt(10)
6 - sqrt(10) is the answer.
6 + sqrt(10) puts the block on the other side of the pipe, between the pipe and the wall.
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btw, that's not a "large" pipe, less than 3 ft OD.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!

You can find the triangle with the green leg xsince the triangle is a right triangle with the other leg R-1 and hypotenuse of R.
R%5E2=%28R-1%29%5E2%2Bx%5E2
.
.
x%2BR=5
x=5-R
x%5E2=25-10R%2BR%5E2
.
.
Substituting,
R%5E2=R%5E2-2R%2B1%2B%2825-10R%2BR%5E2%29
-2R%2B1%2B25-10R%2BR%5E2=0
R%5E2-12R%2B26=0
Complete the square,
R%5E2-12R%2B36-10=0
%28R-6%29%5E2=10
R-6=0+%2B-+sqrt%2810%29
R=6+%2B-+sqrt%2810%29
Since x%2BR=5, then R must be less than 5.
So,
highlight%28R=6-+sqrt%2810%29%29