16 and 12
It is not possible to answer the question because you have not defined what 32 measures: is it the area of the rectangle, its perimeter, its diagonal or some other property?
No, it is not possible for a rectangle to have a perimeter of 46 and an area of 42 simultaneously. For a rectangle, the perimeter ( P ) is given by ( P = 2(l + w) ), and the area ( A ) is ( A = l \times w ), where ( l ) is the length and ( w ) is the width. Solving these equations shows that the dimensions needed for these values are inconsistent, meaning no such rectangle exists.
7 x 19 cm
If the dimensions are restricted to whole numbers, then the only possibilities are 1 x 4 and 2 x 3.
i think that it si 4x3 and it eqals 12
The dimensions of the rectangle will then work out as 14 cm by 10 cm because the perimeter is 14+10+14+10 = 48 cm
It is not possible to answer the question because you have not defined what 32 measures: is it the area of the rectangle, its perimeter, its diagonal or some other property?
No, it is not possible for a rectangle to have a perimeter of 46 and an area of 42 simultaneously. For a rectangle, the perimeter ( P ) is given by ( P = 2(l + w) ), and the area ( A ) is ( A = l \times w ), where ( l ) is the length and ( w ) is the width. Solving these equations shows that the dimensions needed for these values are inconsistent, meaning no such rectangle exists.
7 x 19 cm
L + W >= 24, L = W + 4, Possible dimensions 14 and 10 giving an area of 140 sq cm.
The greatest area that a rectangle can have is, in fact, attained when it is a square. A square with perimeter of 16 cm must have sides of 4 cm and so an area of 4*4 = 16 cm2.
If the dimensions are restricted to whole numbers, then the only possibilities are 1 x 4 and 2 x 3.
No it is not possible the dimensions are 200 by 1/2
Your question cannot be answered. An area of 14 acres is equivalent to 609,840 square feet, but a rectangle with that area could be 609,840 feet by 1 foot, or it could be 1089 feet by 560 feet, or many other possible dimensions, each of which would have a different perimeter.
i think that it si 4x3 and it eqals 12
There are often multiple 'correct' dimensions for these problems. The most straight forward way to solve it is to list all the factors that, when multiplied, equal the area. Then from this list, cross out the factors that DON'T equal your perimeter. The remaining factors are your possible dimensions.
Select any number, W, between 0 and 4 units.Consider a rectangle with width W and length, L = 8 - W units.Then the rectangle with dimensions L and W will have a perimeter of 16 units.Since there are infinitely many possible values for W, there are infinitely many possible answers to the question.