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Could a rectangle have a perimeter 20cm and area 20cm2? - Answers

Yes. The maximum area for a rectangle with a given perimeter is the square (in this case, 5 cm x 5 cm = 25 cm2), and any area between that and 0 is possible by slowly stretching out the square, increasing its length while decreasing its width. If you'd like to figure out the dimensions of the rectangle with a perimeter of 20 and an area of 20 (I'll leave off the units for simplicity), we can use algebra: Let x and y be the length and width. We know two things: Area: xy = 20 Perimeter: 2x + 2y = 20 Solve the second equation for y and we get y = 10 - x. Substituting that in the first equation we get x(10 - x) = 20 which turns into the quadratic equation x2 - 10x + 20 = 0 Applying the quadratic equation, we get x = 5 +/- sqrt(5), which gives us our two dimensions: 5 + sqrt(5) and 5 - sqrt(5)



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Could a rectangle have a perimeter 20cm and area 20cm2? - Answers

https://math.answers.com/geometry/Could_a_rectangle_have_a_perimeter_20cm_and_area_20cm2

Yes. The maximum area for a rectangle with a given perimeter is the square (in this case, 5 cm x 5 cm = 25 cm2), and any area between that and 0 is possible by slowly stretching out the square, increasing its length while decreasing its width. If you'd like to figure out the dimensions of the rectangle with a perimeter of 20 and an area of 20 (I'll leave off the units for simplicity), we can use algebra: Let x and y be the length and width. We know two things: Area: xy = 20 Perimeter: 2x + 2y = 20 Solve the second equation for y and we get y = 10 - x. Substituting that in the first equation we get x(10 - x) = 20 which turns into the quadratic equation x2 - 10x + 20 = 0 Applying the quadratic equation, we get x = 5 +/- sqrt(5), which gives us our two dimensions: 5 + sqrt(5) and 5 - sqrt(5)



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https://math.answers.com/geometry/Could_a_rectangle_have_a_perimeter_20cm_and_area_20cm2

Could a rectangle have a perimeter 20cm and area 20cm2? - Answers

Yes. The maximum area for a rectangle with a given perimeter is the square (in this case, 5 cm x 5 cm = 25 cm2), and any area between that and 0 is possible by slowly stretching out the square, increasing its length while decreasing its width. If you'd like to figure out the dimensions of the rectangle with a perimeter of 20 and an area of 20 (I'll leave off the units for simplicity), we can use algebra: Let x and y be the length and width. We know two things: Area: xy = 20 Perimeter: 2x + 2y = 20 Solve the second equation for y and we get y = 10 - x. Substituting that in the first equation we get x(10 - x) = 20 which turns into the quadratic equation x2 - 10x + 20 = 0 Applying the quadratic equation, we get x = 5 +/- sqrt(5), which gives us our two dimensions: 5 + sqrt(5) and 5 - sqrt(5)

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      Yes. The maximum area for a rectangle with a given perimeter is the square (in this case, 5 cm x 5 cm = 25 cm2), and any area between that and 0 is possible by slowly stretching out the square, increasing its length while decreasing its width. If you'd like to figure out the dimensions of the rectangle with a perimeter of 20 and an area of 20 (I'll leave off the units for simplicity), we can use algebra: Let x and y be the length and width. We know two things: Area: xy = 20 Perimeter: 2x + 2y = 20 Solve the second equation for y and we get y = 10 - x. Substituting that in the first equation we get x(10 - x) = 20 which turns into the quadratic equation x2 - 10x + 20 = 0 Applying the quadratic equation, we get x = 5 +/- sqrt(5), which gives us our two dimensions: 5 + sqrt(5) and 5 - sqrt(5)
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