A rectangle has an area of 60 square inches and a perimeter of 38 inches. Find the length and width if the rectangle.
The dimensions for the area must be either
10 * 6
or
12 * 5
or
15 * 4
Which of those sets of dimensions will give you a perimeter of 38 inches?
Which
OOPS!! I used 36 for the perimeter.
It should be 38. Sue made the same error.
L * W = 60 IN^2, W = 60/L
2L + 2W = 38 IN
2L + 2 *60/L = 38
2L + 120/L = 38
2L^2 + 120 = 38L
L^2 + 60 = 19L
L^2 - 19L + 60 = 0
(L - 4) (L - 15) = 0
L = 4, and 15.
2 solutions: L = 4, W = 15.
L = 15 , W = 4.
To find the length and width of the rectangle, we can set up a system of equations based on the given information.
Let's assume that the length of the rectangle is represented by "L" and the width is represented by "W".
According to the problem, the area of the rectangle is 60 square inches. The formula for the area of a rectangle is A = L * W, so we can write the equation:
L * W = 60 Equation 1
Additionally, the problem states that the perimeter of the rectangle is 38 inches. The formula for the perimeter of a rectangle is P = 2L + 2W, so we can write the equation:
2L + 2W = 38 Equation 2
Now, we have a system of two equations (Equation 1 and Equation 2) with two variables (L and W) that we can solve simultaneously to find their values.
To solve the system of equations, we can use various methods such as substitution, elimination, or graphing. In this case, we will use the method of substitution.
We can solve Equation 1 for either L or W and substitute it into Equation 2.
Solving Equation 1 for L:
L = 60 / W
Substituting this value of L into Equation 2:
2(60 / W) + 2W = 38
Now we can simplify and solve for W.
Multiplying through by W to eliminate the denominator:
2(60) + 2W^2 = 38W
120 + 2W^2 = 38W
Rearranging and simplifying:
2W^2 - 38W + 120 = 0
Now we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. In this case, we can factor the equation as follows:
2(W^2 - 19W + 60) = 0
2(W - 4)(W - 15) = 0
Setting each factor equal to zero and solving for W:
W - 4 = 0 or W - 15 = 0
W = 4 or W = 15
Since the width cannot be negative, we discard W = 15 as an extraneous solution. Therefore, the width of the rectangle is 4 inches.
Now that we have the width, we can substitute it back into Equation 1 to solve for the length:
L * 4 = 60
L = 60 / 4
L = 15
So, the length of the rectangle is 15 inches.
Therefore, the length of the rectangle is 15 inches, and the width is 4 inches.
L * W = 60 sq in
W = 60/L
2L + 2W = 36 IN
Substitute 60/l for w:
2L + 2*60/L = 36
2L + 120/L = 36
Multiply both sides by L:
2L^2 + 120 = 36L
L^2 + 60 = 18L
L^2 - 18L + 60 = 0
Use Quadratic formula to find l:
L = 9 +- sqrt(21) = 9 +- 4.58
L = 13.58, and 4.42.
13.58 * W = 60, W = 4.42 IN
4.42 * W = 60, W = 13.57 in
2 Solutions: L * W =13.58 * 4.42 and
L * W = 4.42 * 13.58.
NO, SUE DID NOT MAKE THE ERROR THAT
I MADE!