SOLUTION: Dear Tutor, My question involves "Graphic Representation of Radicals of the Second Order" in Algebra 2. I know that the length of the hypotneuse of a right triangle equals the s

Algebra ->  Graphs -> SOLUTION: Dear Tutor, My question involves "Graphic Representation of Radicals of the Second Order" in Algebra 2. I know that the length of the hypotneuse of a right triangle equals the s      Log On


   



Question 577610: Dear Tutor,
My question involves "Graphic Representation of Radicals of the Second Order" in Algebra 2. I know that the length of the hypotneuse of a right triangle equals the square root of the sum of the lengths of the two legs. The problems I'm having trouble with are the square roots of 15, 24, 34. Theres also one with 1/2 in front of the square root of 5. There is a question similar to this on algebra.com, but i do not understand what the tutor is saying about how to solve it. I have tried all posible square root possibilties but nothing adds up to 15, 24, or 34. My dad tried to help me, but the answers given in the book dont make sense. I have posted this question everywhere. I really hope u can help me. This is all Greek to me.
Sincerely,
A Really Confused Algebra 2 Student

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You are really not doing a very good job of explaining what you want to do, but here is my guess. I think you want to create a right triangle with two sides that have integer measures such that the third side meassures

Let represent the measure of the short leg of the triangle, let represent the measure of the other leg of the triangle, and let represent the measure of the hypotenuse.

Pythagoras in its pristine form says:



But that same relationship can be expressed:



Hence



What we need to find is two integers such that the difference of their squares is 15. What if ? Then , then







Ah ha! Two integers.

So we need to construct a right triangle that has sides of , , and and the right angle has to be between the two shorter segments.

On your graph paper, plot the point (1, 0). Draw a heavy line from the origin to this point. Use the scale on your graph paper to set the points of your compass 4 units apart. Put the metal point of your compass on the point (1,0) then strike a short arc that intersects the positive -axis. This point of intersection will be the point .

Hints for the other two: 24 = 25 - 1 and 34 = 25 + 9

John

My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism