SOLUTION: the perimeter of a square and a rectangle are equal. The length of the rectangle is 11cm and area of the square is 4cm^2 more than the area of the rectangle.find the side of the sq

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: the perimeter of a square and a rectangle are equal. The length of the rectangle is 11cm and area of the square is 4cm^2 more than the area of the rectangle.find the side of the sq      Log On


   

Question 165927: the perimeter of a square and a rectangle are equal. The length of the rectangle is 11cm and area of the square is 4cm^2 more than the area of the rectangle.find the side of the square?
Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= square's side
Let y=length of the rectangle
Let z=width of the rectangle
the perimeter of a square and a rectangle are equal
then 4x=2(y+z)
The length of the rectangle is 11cm
then y=11
area of the square is 4cm^2 more than the area of the rectangle
x%5E2=yz%2B4
so the system to solve is:
4x=2%2811%2Bz%29
x%5E2=11z%2B4
then
2x=11%2Bz
then
z=2x-11
then
x%5E2=11%282x-11%29%2B4
x%5E2-22x%2B117=0
x+=+%2822+%2B-+sqrt%28+%28-22%29%5E2-4%2A1%2A117+%29%29%2F%282%29+ then

x=13 or x=9
if x=9 then z=18-11=7
if x=13 then z=26-11=15
then z=7 or z=15 but z Answer: x=9,z=7 and y=11