1. ## Solving quadratic word problems - Help

Area of a Rectangle P=2l+2w
x=width,1.6x=the length of the rectangle
2(1.6x)+2x=9000m^2
3.2x+2x=9000m^2
5.2x=9000m^2
5.2x/5.2=(9000m^2)/5.2

I am not sure what to do at this point. Any suggestions?

2. Originally Posted by mcruz65
Area of a Rectangle P=2l+2w
x=width,1.6x=the length of the rectangle
2(1.6x)+2x=9000m^2
3.2x+2x=9000m^2
5.2x=9000m^2
5.2x/5.2=(9000m^2)/5.2

I am not sure what to do at this point. Any suggestions?
... area of a rectangle is A = LW

P = 2L + 2W is the perimeter.

3. ## Solving quadratic word problems - Help

Here is the actual problem. Golden Rectangle. The so-called golden rectangle is said to be extremely pleasing visually and was used often by ancient Greek and Roman architects. The length of a golden rectangle is approximately 1.6 times the width. Find the dimensions of a golden rectangle if its area is 9000 m^2.

w = the width of the rectangle and 1.6w = the length

Area = length x width

w x 1.6w = 9000 m^2

1.6w^2 = 9000 m^2

sqrt(1.6w^2) = sqrt(9000 m^2)

1.6w = 9000m

1.6w/1.6 = 9000m/1.6

w = 5625

Is this correct??

4. Originally Posted by mcruz65
Here is the actual problem. Golden Rectangle. The so-called golden rectangle is said to be extremely pleasing visually and was used often by ancient Greek and Roman architects. The length of a golden rectangle is approximately 1.6 times the width. Find the dimensions of a golden rectangle if its area is 9000 m^2.

w = the width of the rectangle and 1.6w = the length

Area = length x width

w x 1.6w = 9000 m^2

1.6w^2 = 9000 m^2

sqrt(1.6w^2) = sqrt(9000 m^2)

1.6w = 9000m

1.6w/1.6 = 9000m/1.6

w = 5625

Is this correct??

$\displaystyle 1.6w^2 = 9000$

$\displaystyle w^2 = \frac{9000}{1.6}$

$\displaystyle w = \sqrt{\frac{9000}{1.6}}$