Draw a rectangle that has a perimeter that has of12 uints and an area of 8 square units
With a perimeter of 12 units, the sides must be
1 by 5
2 by 4
3 by 3
Which of those dimensions will give you an area of 8 square units?
To draw a rectangle with a perimeter of 12 units and an area of 8 square units, we need to find the dimensions of the rectangle.
Let's assume the length of the rectangle is L units and the width is W units.
The formula for calculating the perimeter of a rectangle is: Perimeter = 2(L + W).
Given that the perimeter is 12 units, we can write the equation as:
12 = 2(L + W)
Now, let's find the formula for calculating the area of a rectangle: Area = L * W.
Given that the area is 8 square units, we can write the equation as:
8 = L * W
We now have a system of equations:
12 = 2(L + W)
8 = L * W
To solve this system of equations, we can use substitution or elimination. Let's use substitution:
Rearrange the first equation to solve for L:
L = (12 - 2W) / 2
Substitute this value of L into the second equation:
8 = [(12 - 2W) / 2] * W
Simplify the equation:
16 = 12W - 2W^2
Rearrange the equation:
2W^2 - 12W + 16 = 0
Now, we can solve this quadratic equation to find the values of W. We can use factoring, completing the square, or the quadratic formula. By factoring, we find:
(2W - 4)(W - 4) = 0
Setting each factor equal to zero, we have:
2W - 4 = 0 or W - 4 = 0
Solving these equations, we find:
W = 2 or W = 4
If W = 2, substituting this value into the first equation:
12 = 2(L + 2)
12 = 2L + 4
2L = 8
L = 4
So, when W = 2, L = 4.
If W = 4, substituting this value into the first equation:
12 = 2(L + 4)
12 = 2L + 8
2L = 4
L = 2
So, when W = 4, L = 2.
Therefore, we have two possible rectangles: one with a length of 4 units and a width of 2 units, and the other with a length of 2 units and a width of 4 units.
Now, you can draw a rectangle with either dimension and label the sides according to the given values.