SOLUTION: A piece of wire is shaped to enclose an equilateral triangle whose area is 27.71cm^2. It is then reshaped to enclose a rectangle whose length is 9cm., find the area of the rectangl
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-> SOLUTION: A piece of wire is shaped to enclose an equilateral triangle whose area is 27.71cm^2. It is then reshaped to enclose a rectangle whose length is 9cm., find the area of the rectangl
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Question 1124653: A piece of wire is shaped to enclose an equilateral triangle whose area is 27.71cm^2. It is then reshaped to enclose a rectangle whose length is 9cm., find the area of the rectangle Found 2 solutions by rothauserc, Theo:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! let s be the length of a side of the equilateral triangle
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area of equilateral triangle in terms of s is
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Area = s^2 * square root(3)/4
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Note this formula comes from the fact that the base is s/2 and height = s * square root(3)/2 for an equilateral triangle
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27.71 = s^2 * square root(3)/4
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s^2 = (4 * 27.27)/square root(3) = 62.9774
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s = square root(62.9774) is approximately 7.94 cm
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Perimeter of the equilateral triangle is 3 * 7.94 = 23.82 cm
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Perimeter of rectangle is 2l + 2w where l is length and w is width
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we are given that l=9
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2(9) +2w = 23.82
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2w = 23.82 -18 = 5.82
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w = 2.91 cm
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Area of rectangle(A) = l * w
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A = 9 * 2.91 = 26.19 cm^2
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