SOLUTION: A rectangle has a diagonal of length 20 ft. and a perimeter or 56 ft. Find the dimensions of the rectangle.

Algebra ->  Rectangles -> SOLUTION: A rectangle has a diagonal of length 20 ft. and a perimeter or 56 ft. Find the dimensions of the rectangle.      Log On


   



Question 163403: A rectangle has a diagonal of length 20 ft. and a perimeter or 56 ft. Find the dimensions of the rectangle.
Found 2 solutions by checkley77, gonzo:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+y^2=20
2x+2y=56
2x=56-2y
x=(56-2y)/2
x=28-y
(28-y)^2+y^2=20^2
784-56y+y^2+y^2=400
2y^2-56y+784-400=0
2y^2-56y+384=0
2(y^2-28y+192)=0
2(y-16)(y-12)=0
y-16=0
y=16 Answer. Then x=12.
y-12=0
y=12 Answer. Then x=16.
Proofs:
2*16+2*12=56
32+24=56
56=56
16^2=1262=20^2
256+144=400
400=400


Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
diagonal 20 feet.
perimeter 56 feet.
-----
diagonal of the rectangle is the hypotenuse of the 2 triangles created by it.
-----
let x = length of the rectangle
let y = width of the rectangle.
x + y are also legs of the right triangles created.
-----
formula for perimeter of the rectangle is
2*x + 2*y = 56
-----
formula for legs of the right triangle is
x^2 + y^2 = 20^2
-----
since 2*x + 2*y = 56, then
2*y = 56 - 2*x
which becomes
y = 28 - x
-----
substituting for y in the second equation gets
x^2 + (28-x)^ = 20^2
which expands to become
x^2 + 784 - 56*x + x^2 = 400
subtracting 400 from both sides of the equation and combining like terms gets
2*x^2 - 56*x + 384 = 0
dividing both sides of the equation by 2 gets
x^2 - 28*x + 192 = 0
-----
solving this equation (i used the quadratic formula because i wasn't able to see the correct factors) yields
x = 16 or x = 12
-----
both values are good by substituting in the original equation.
-----
if x = 12, then
2*x + 2*y = 56 becomes 24+2*y=56 becomes 2*y = 32 becomes y = 16
-----
if x = 16, then
2*x + 2*y = 56 becomes 32+2*y=56 becomes 2*y = 24 becomes y = 12
----
if x = 12, y = 16
if x = 16, y = 12
-----
substituting either one in the equation for the legs of the triangle gets
x^2 + y^2 = 400 becomes (12)^2 + (16)^2 = 400 becomes 144 + 256 = 400 becomes 400 = 400.
-----
x = 12 and y = 16, or x = 16 and y = 12 works for both the perimeter and for the legs of the triangle so these values are good.
-----
answer is
x = 12 or 16
y = 16 or 12
-----
the rectangle is 12 by 16, or 16 by 12, whichever way you want to look at it.
-----