SOLUTION: 6076 is the area of a rectangle and 320 is the perimeter. What are the dimensions

Question 928504: 6076 is the area of a rectangle and 320 is the perimeter. What are the dimensions

Found 2 solutions by TimothyLamb, MathTherapy:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
LW = 6076
L = 6076/W
2L + 2W = 320
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2L + 2W = 320
2(6076/W) + 2W = 320
2*6076/W = 320 - 2W
12152 = W(320 - 2W)
12152 = 320W - 2WW
2WW - 320W + 12152 = 0
---
the above quadratic equation is in standard form, with a=2, b=-320 and c=12152
---
to solve the quadratic equation, by using the quadratic formula, copy and paste this:
2 -320 12152
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
---
the quadratic has two real roots at:
---
W = 98
W = 62
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either root will work for W, but because W is typically less than L, let's use W = 62:
---
L = 6076/W
L = 6076/62
L = 98
---
answer:
W = 62
L = 98
---
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Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

6076 is the area of a rectangle and 320 is the perimeter. What are the dimensions

Let length and width, be L and W, respectively
Then: LW = 6,076 ------- eq (i)
Also, 2(L + W) = 320_____2(L + W) = 2(160)_____L + W = 160_____W = 160 - L ------- eq (ii)
L(160 � L) = 6,076 -------- Substituting 160 � L for W in eq (i)

(L - 98)(L - 62) = 0
L, or length = 98 OR L = 62
If length = 98, then width = 62
If length = 62, then width = 98
Dimensions:
You can do the check!!
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