SOLUTION: This was under the conic section of my book but I am not sure if it is a review or if it is really a conic section problem. It seems really simple but after I looked at it for a w

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: This was under the conic section of my book but I am not sure if it is a review or if it is really a conic section problem. It seems really simple but after I looked at it for a w      Log On


   

Question 10295: This was under the conic section of my book but I am not sure if it is a review or if it is really a conic section problem. It seems really simple but after I looked at it for a while I realized I didn't know what to do.
Find the dimentions of a rectangle that has the area of 10 and a dioganal length of 5.

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = width of the rectangle
y = length of the rectangle
5 = diagonal of the rectangle

By the Theorem of Pythagoras,
+x%5E2+%2B+y%5E2+=+25 By the way, this equation would be the equation of a circle, which is probably why this problem is in your conic section section of the book!!

Solve for y:
y%5E2+=+25+-+x%5E2
y+=+sqrt%2825+-+x%5E2%29+

Area = xy = 10
+x%2Asqrt%2825-+x%5E2%29+=+10

Square both sides:
x%5E2+%2A+%2825-x%5E2%29+=+100+

Set the equation equal to zero by taking everything to the right side of the equation, in order to get the x^4 to have a positive coefficient.
25x%5E2+-+x%5E4+=+100+
0+=+x%5E4+-+25x%5E2+%2B+100

It doesn't always happen, but it sure feels good when it does--that math comes out even! This does indeed factor!!
0+=+%28x%5E2+-+20%29%28x%5E2+-+5%29+
x%5E2+=+20
x+=+0%2B-sqrt%2820%29
x=+2%2Asqrt%285%29+or+-2%2Asqrt%285%29

x%5E2+=+5
x+=+sqrt%285%29+or+-sqrt%285%29+

It turns out that if x+=+sqrt%285%29, then y+=+2sqrt%285%29 and if x=2sqrt%285%29, then y+=sqrt%285%29. So there is actually only one solution. The rectangle is sqrt%285%29 by 2sqrt%285%29.

R^2 at SCC