The area of a rectangle is 66 ft^2 and the length of the rectangle is 7 feet less than three times the width find the dimensions
of the rectangle
2 answers:
<h2>Length = 11</h2><h2>Width = 6</h2>
The area of a rectangle is 66 ft^2:
L * W = 66
Length of the rectangle is 7 feet less than three times the width:
L = 3W-7
Substitute L in terms of W:
(3W-7) * W = 66
Factorise the equation:
3W^2 -7W = 66
3W^2 - 7W - 66 = 0
Factors of 66 =
1 66, 2 33, 3 22, 6 11
(3W + 11) (W - 6) = 0
Solve for W:
3W + 11 = 0
3W = 11
W = 3/11
W - 6 = 0
W = 0 + 6
W = 6
Using the original equation, find L:
L = 3W-7
L = 3(6)-7
L = 18-7
L = 11
L * W = 66
11 * 6 = 66
Answer:
Step-by-step explanation:
Length * width =Area
3(x-7)*x=66
x^2-7x-22=0
Solve for 'x' using quadratic formula
Now width=9.35234995ft
Length=7.05704985ft
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