SOLUTION: please help me how to solve this. because i can't get it.
1) the width of the rectangle is one-forth of its lenght. if the perimeter is 40 cm, find the dimensions of the rectang
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Question 208023: please help me how to solve this. because i can't get it.
1) the width of the rectangle is one-forth of its lenght. if the perimeter is 40 cm, find the dimensions of the rectangles?
2) the perimeter of a standard basketball court is 288ft. find the dimensions of the basketball court if its lenght is 44ft longer than its width.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
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PROBLEM 1
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1) the width of the rectangle is one-forth of its lenght. if the perimeter is 40 cm, find the dimensions of the rectangles?
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let W = width
let L = length
L = (4 * W) which is the same as (W = L / 4)
P = 40
= (2 * L) + (2 * W)
= (2 * (4 * W)) + (2 * W) after replacing L with 4*W
= (8 * W) + (2 * W)
= (10 * W)
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40 = (10 * W)
W = (40 / 10) = 4 after dividing both sides of the equation by 10.
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so far you have W = 4
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L = (4 * W) = (4 * 4) = 16
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now you also have L = 16
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W = 4
L = 16
W = (L / 4) = (16 / 4) = 4 = correct based on the problem statement.
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(2 * L) + (2 * W) = 40
= (2 * 16) + (2 * 4) = 40
= (32 + 8) = 40
40 = 40 = correct with W = 4 and L = 16.
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your answer is:
the width of the rectangle is 4 and the length of the rectangle is 16.
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PROBLEM 2
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2) the perimeter of a standard basketball court is 288ft. find the dimensions of the basketball court if its lenght is 44ft longer than its width.
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P = 288 feet.
L = (W + 44)
(2 * L) + (2 * W) = 288
= (2 * (W + 44)) + (2 * W) = 288 after replacing L with W + 44.
= (2 * W) + (2 * 44) + (2 * W) = 288
= (4 * W) + 88 = 288
= (4 * W) = 200
= W = (200 / 4) = 50 after dividing both sides of the equation by 4.
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so far you have W = 50
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L = (W + 44) = (50 + 44) = 94
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now you also have L = 94
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(2 * L) + (2 * W) = 288
(2 * 94) + (2 * 50) = 288
(188 + 100) = 288 = correct based on W = 50 and L = 94.
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your answer is:
length = 94 feet
width = 50 feet
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