what must be the dimensions of a rectangle to have an area of 125 cm^2 and a perimeter of 60 m?
Solve these two equations simultaneously:
2a + 2b = 60
a*b = 125
Solve by substitution to eliminate one variable:
2a + 250/a = 60
a + 125/a = 30
a^2 -30a +125 = 0
(a -25)(a-5) = 0
a = 5 or 25
If a = 5, b = 25
If a = 25, b = 5
To find the dimensions of a rectangle with a given area and perimeter, you can set up a system of equations using the formulas for area and perimeter of a rectangle.
Let's denote the length of the rectangle as "L" and the width as "W".
1. The formula for the area of a rectangle is: Area = Length * Width.
In this case, the area is given as 125 cm^2, so we can write the equation as:
125 = L * W (Equation 1)
2. The formula for perimeter is: Perimeter = 2 * (Length + Width).
The perimeter is given as 60 m, which needs to be converted to centimeters before setting up the equation.
1 meter = 100 centimeters, so 60 m = 60 * 100 = 6000 cm.
Now, we can write the equation as:
6000 = 2 * (L + W) (Equation 2)
We have two equations with two variables (L and W). We can solve the system by substitution or elimination.
From Equation 1, we can rearrange it to get:
L = 125 / W
Substituting this value of L in Equation 2, we get:
6000 = 2 * ((125 / W) + W)
Simplifying the equation, we have:
6000 = (250 / W) + 2W
To get rid of the fraction, we can multiply both sides by W:
6000W = 250 + 2W^2
Now, let's rearrange the equation to a quadratic form:
2W^2 - 6000W + 250 = 0
To solve this quadratic equation for W, we can apply the quadratic formula:
W = (-b ± √(b^2 - 4ac)) / 2a
Using the values from above, we have:
a = 2, b = -6000, c = 250
Substituting the values in the quadratic formula and solving for W, we get two possible values for W.
Once we find the values of W, we can substitute these values back into Equation 1 to find the corresponding values of L.
Keep in mind that we obtain two solutions since a rectangle has two possible orientations: length greater than width or width greater than length.