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A figure with an area of 6 and a perimeter of 12? - Answers

The figure could be a rectangle with length 4.73 and width 1.27(2dp). Area (A) of a rectangle = Length (L) x Width (W) = 6 : Then W = 6/L Perimeter = 2 (L + W) = 12 : substituting for W then :- 2(L + 6/L) = 12 : 2L + 12/L = 12 : 2L2 - 12L + 12 = 0 Solving for a quadratic equation gives L = (12 ± √48) ÷ 4 = 3 ± √3 = 3 ± 1.73 (2dp)



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A figure with an area of 6 and a perimeter of 12? - Answers

https://math.answers.com/geometry/A_figure_with_an_area_of_6_and_a_perimeter_of_12

The figure could be a rectangle with length 4.73 and width 1.27(2dp). Area (A) of a rectangle = Length (L) x Width (W) = 6 : Then W = 6/L Perimeter = 2 (L + W) = 12 : substituting for W then :- 2(L + 6/L) = 12 : 2L + 12/L = 12 : 2L2 - 12L + 12 = 0 Solving for a quadratic equation gives L = (12 ± √48) ÷ 4 = 3 ± √3 = 3 ± 1.73 (2dp)



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https://math.answers.com/geometry/A_figure_with_an_area_of_6_and_a_perimeter_of_12

A figure with an area of 6 and a perimeter of 12? - Answers

The figure could be a rectangle with length 4.73 and width 1.27(2dp). Area (A) of a rectangle = Length (L) x Width (W) = 6 : Then W = 6/L Perimeter = 2 (L + W) = 12 : substituting for W then :- 2(L + 6/L) = 12 : 2L + 12/L = 12 : 2L2 - 12L + 12 = 0 Solving for a quadratic equation gives L = (12 ± √48) ÷ 4 = 3 ± √3 = 3 ± 1.73 (2dp)

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      The figure could be a rectangle with length 4.73 and width 1.27(2dp). Area (A) of a rectangle = Length (L) x Width (W) = 6 : Then W = 6/L Perimeter = 2 (L + W) = 12 : substituting for W then :- 2(L + 6/L) = 12 : 2L + 12/L = 12 : 2L2 - 12L + 12 = 0 Solving for a quadratic equation gives L = (12 ± √48) ÷ 4 = 3 ± √3 = 3 ± 1.73 (2dp)

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