a rectangular security area is enclosed with 200 ft of fencing. if this rectangular area is arranged so that the ratio of the length to the width is 5/3, find the dimensions.
In a rectangle, the parallel sides are equal. Thus,
x = width
5x/3 = lenght
2x + 2(5x/3) = 200
6x/3 + 10x/3 = 200
16x/3 = 200
x=37.5
Now, you have the short side. To find the long one, multiply by ratio:
37.5 * 5/3 = 62.5
Control:
2*37.5 + 2*62.5 = 200
200 = 200
5L = 3W
L = 3/5W
2L + 2W = 200
Substitute 3/5W for L in last equation and solve for W. Insert that value into the first equation and solve for L. Check by inserting both values into the last equation.
To find the dimensions of the rectangular security area, we'll set up an equation based on the given information.
Let's assume the length of the rectangular area is 5x, and the width is 3x (since the ratio of length to width is 5/3).
The perimeter of a rectangle is given by the formula: Perimeter = 2(length + width)
Given that the perimeter of the security area is 200 ft, we can set up the equation:
200 = 2(5x + 3x)
Simplifying:
200 = 2(8x)
Divide both sides of the equation by 2:
100 = 8x
Now divide both sides by 8 to isolate x:
x = 100/8
x = 12.5
So the value of x is 12.5.
Now, we can find the dimensions of the rectangular security area:
Length = 5x = 5(12.5) = 62.5 ft
Width = 3x = 3(12.5) = 37.5 ft
Therefore, the dimensions of the rectangular security area are 62.5 ft by 37.5 ft.