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https://math.answers.com/math-and-arithmetic/Can_a_rectangle_have_the_same_perimeter_and_the_same_area

Can a rectangle have the same perimeter and the same area? - Answers

Technically, no because a perimeter is a linear measure and an area is a square measure. However, there are infinitely many rectangles such that the NUMERICAL VALUE of their perimeter is the same as the NUMERICAL value of their area. Select any number y greater than or equal to 4. let x = 2*y/(y-2) Consider the rectangle with dimensions width x and length y. Its area is x*y = [2*y/(y-2)]*y = 2y2/(y-2) square units. Its perimeter is 2(x+y) = 2*[(2y/(y-2) + y] = 2/(y-2)*[2y+y*(y-2)] = 2/(y-2)*[2y+y2-2y] = 2/(y-2)*y2 = 2y2/(y-2) units Since y is an arbitrary number greater than 4, there are infinitely many choices for y giving rise to infinitely many shapes. The one with the smallest y: y = 4 is actually a square - with sides of 4 units and perimeter/area = 16.



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Can a rectangle have the same perimeter and the same area? - Answers

https://math.answers.com/math-and-arithmetic/Can_a_rectangle_have_the_same_perimeter_and_the_same_area

Technically, no because a perimeter is a linear measure and an area is a square measure. However, there are infinitely many rectangles such that the NUMERICAL VALUE of their perimeter is the same as the NUMERICAL value of their area. Select any number y greater than or equal to 4. let x = 2*y/(y-2) Consider the rectangle with dimensions width x and length y. Its area is x*y = [2*y/(y-2)]*y = 2y2/(y-2) square units. Its perimeter is 2(x+y) = 2*[(2y/(y-2) + y] = 2/(y-2)*[2y+y*(y-2)] = 2/(y-2)*[2y+y2-2y] = 2/(y-2)*y2 = 2y2/(y-2) units Since y is an arbitrary number greater than 4, there are infinitely many choices for y giving rise to infinitely many shapes. The one with the smallest y: y = 4 is actually a square - with sides of 4 units and perimeter/area = 16.



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https://math.answers.com/math-and-arithmetic/Can_a_rectangle_have_the_same_perimeter_and_the_same_area

Can a rectangle have the same perimeter and the same area? - Answers

Technically, no because a perimeter is a linear measure and an area is a square measure. However, there are infinitely many rectangles such that the NUMERICAL VALUE of their perimeter is the same as the NUMERICAL value of their area. Select any number y greater than or equal to 4. let x = 2*y/(y-2) Consider the rectangle with dimensions width x and length y. Its area is x*y = [2*y/(y-2)]*y = 2y2/(y-2) square units. Its perimeter is 2(x+y) = 2*[(2y/(y-2) + y] = 2/(y-2)*[2y+y*(y-2)] = 2/(y-2)*[2y+y2-2y] = 2/(y-2)*y2 = 2y2/(y-2) units Since y is an arbitrary number greater than 4, there are infinitely many choices for y giving rise to infinitely many shapes. The one with the smallest y: y = 4 is actually a square - with sides of 4 units and perimeter/area = 16.

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      Technically, no because a perimeter is a linear measure and an area is a square measure. However, there are infinitely many rectangles such that the NUMERICAL VALUE of their perimeter is the same as the NUMERICAL value of their area. Select any number y greater than or equal to 4. let x = 2*y/(y-2) Consider the rectangle with dimensions width x and length y. Its area is x*y = [2*y/(y-2)]*y = 2y2/(y-2) square units. Its perimeter is 2(x+y) = 2*[(2y/(y-2) + y] = 2/(y-2)*[2y+y*(y-2)] = 2/(y-2)*[2y+y2-2y] = 2/(y-2)*y2 = 2y2/(y-2) units Since y is an arbitrary number greater than 4, there are infinitely many choices for y giving rise to infinitely many shapes. The one with the smallest y: y = 4 is actually a square - with sides of 4 units and perimeter/area = 16.

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