Asked by kayla
A rectangle has an area of 60 square inches and a perimeter of 38 inches. Find the length and width if the rectangle.
Answers
Answered by
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
Answered by
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.
Answered by
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.
Answered by
Henry
NO, SUE DID NOT MAKE THE ERROR THAT
I MADE!
I MADE!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.