Question
A rectangle has an area of 60 square inches and a perimeter of 38 inches. Find the length and width if the rectangle.
Answers
Ms. Sue
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
10 * 6
or
12 * 5
or
15 * 4
Which of those sets of dimensions will give you a perimeter of 38 inches?
Which
Henry
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.
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.
Henry
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.
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.
Henry
NO, SUE DID NOT MAKE THE ERROR THAT
I MADE!
I MADE!