Question 375207: find the sides of a rectangle whose perimeter is 40 and diagonal is 18? Found 2 solutions by unlockmath, mananth:Answer by unlockmath(1688) (Show Source):
You can put this solution on YOUR website! Hello,
Lets have x be the width and y be the length. We can set up two equations as:
2x+2y=40
x^2+y^2=18^2
With the first equation lets do this:
y=20-x
Now substitute this into the second equation:
x^2+(-x+20)^2=18^2
This can be rewritten as:
x^2+x^2-40x+400=324
Subtract 324 from both sides and combine like terms to get:
2x^2-40x+76=0
Rewritten as:
2(x^2-20x+38)=0
From here you can use the quadratic formula to solve for x and then you can solve for y.
Got it?
RJ
www.math-unlock.com
You can put this solution on YOUR website! perimeter = 40
2(L+W)=40
L+W=20
L= 20-W
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Diagonal = 18
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width^2 + length^2 = diagonal ^2
W^2+(20-W)^2=diagonal^2
W^2+400-40W+W^2=D^2
2W^2-40W+400 = 18^2
2W^2-40W+400=324
2W^2-40w+76=0
/2
W^2-20W+38=0
..
find the roots x1 & x2 by using quadratic formula
discriminant b^2-4ac=248
x1=((20+sqrt(248))/2
x1= 17.87
x2=((20-sqrt(248))/2
2.13
the other side will be 20-17.87 = 2.13
..
m.ananth@hotmail.ca
