SOLUTION: The area of a rectangle is 12sq. units and the diagonal of the rectangle has a length of 2 then the square root and underneath that is a 10. Find the perimeter of the rectangle?

Algebra ->  Rectangles -> SOLUTION: The area of a rectangle is 12sq. units and the diagonal of the rectangle has a length of 2 then the square root and underneath that is a 10. Find the perimeter of the rectangle?      Log On


   

Question 273089: The area of a rectangle is 12sq. units and the diagonal of the rectangle has a length of 2 then the square root and underneath that is a 10. Find the perimeter of the rectangle?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The area of a rectangle is 12sq. units and the diagonal of the rectangle has a length of 2 then the square root and underneath that is a 10.
Find the perimeter of the rectangle?
:
Area = 12 sq/units; Diagonal = 2sqrt%2810%29
:
Find legs of the right triangle; a^2 + b^2 = c^2
sqrt%28L%5E2+%2B+W%5E2%29 = 2sqrt%2810%29
sqrt%28L%5E2+%2B+W%5E2%29 = sqrt%284%2A10%29
sqrt%28L%5E2+%2B+W%5E2%29 = sqrt%2840%29
square both sides
L^2 + W^2 = 40
:
The area equation
L*W = 12
L = 12%2FW; for substitution below
:
%2812%2FW%29%5E2 + W^2 = 40
%28144%2FW%5E2%29 + W^2 = 40
multiply equation by W^2
144 + W^4 = 40W^2
:
A quadratic equation
W^4 - 40W^2 + 144 = 0
:
this will factor
(W^2 - 36)(W^2 - 4) = 0
:
Two solutions
W^2 = 36
W = 6, call this one the length
and
W^2 = 4
W = 2, the width
:
P = 2(6) + 2(2)
P = 12 + 4
P = 16 units is the perimeter
:
A = 6*2 = 12 confirm using the area