SOLUTION: The length of a rectangle is 1cm. more than twice its height. If the area of the rectangle is 351 sq cm, what is the perimeter of the rectangle?
I already tried to solve using t
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I already tried to solve using t
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Question 744996: The length of a rectangle is 1cm. more than twice its height. If the area of the rectangle is 351 sq cm, what is the perimeter of the rectangle?
I already tried to solve using these
L=2h+1
I got H=-1/3
P=2l+2h
I don't know what to do from there...? Found 2 solutions by stanbon, algebrahouse.com:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 1cm. more than twice its height. If the area of the rectangle is 351 sq cm, what is the perimeter of the rectangle?
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Let height be "x" cm.
Then length is 2x+1 cm
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Area = length*height = x(2x+1) = 2x^2+x
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Equation:
2x^2 +x = 351
2x^2 + x -351 = 0
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(x-13)(2x+27) = 0
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height = x = 13 cm
length = 2x+1 = 27 cm
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Perimeter = 2(height + length) = 2(40) = 80 cm
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Cheers,
Stan H.
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x(2x + 1) = 351 {multiplied length x height and set equal to area, 351}
2x² + x = 351
2x² + x - 351 = 0
2x² - 26x + 27x - 351 = 0 {split x into -26x and 27x}
2x(x - 13) + 27(x - 13) = 0 {factored 2x out of 1st two terms and 27 out of last two terms - factoring by grouping}
(2x + 27)(x - 13) = 0 {factored (x - 13) out of the two terms}
2x + 27 = 0 or x - 13 = 0 {set each factor equal to 0}
2x = -27 or x = 13 {subtracted 27 in 1st and added 13 in 2nd}
x = -13.5 or x = 13 {solved each equation for x}
x = 13 {height of a rectangle cannot be negative}
2x + 1 = 27 {substituted 13, in for x, into 2x + 1}
height is 13 cm and length is 27 cm
Perimeter of a rectangle is 2(height) + 2(length)
P = 2(13) + 2(27) {substituted 13 for height and 27 for length, into perimeter formula}
= 26 + 54 {multiplied}
= 80 {added}
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