28
The area is about 127.3 square inches.
Perimeter = 168 inches so 4*length of side = 168 inches Length of side = 168/4 = 42 inches So area = 42*42 = 1764 square inches.
Even if you knew how many sides the polygon has, you stillcould not calculate its perimeter with that much information.Examples:-- An equilateral triangle with area of 20 has perimeter of 20.3885 .-- A square with area of 20 has perimeter of 17.889(rounded).-- A rectangle with area of 20 can have any perimeter more than 17.889 .4 by 5 . . . . area = 20, perimeter = 182 by 10 . . . area = 20, perimeter = 241 by 20 . . . area = 20, perimeter = 42..etc.
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.
When the linear dimensions of a plane figure are quadrupled, its perimeter is quadrupled, and its area is multiplied by 42 = 16 .
Assuming that it's a rectangle then:- Area = 42*14 = 588 square cm Perimeter = 42+42+14+14 = 112 cm
Area 42 cm2, perimeter 26 cm.
The area is about 127.3 square inches.
Perimeter = 168 inches so 4*length of side = 168 inches Length of side = 168/4 = 42 inches So area = 42*42 = 1764 square inches.
For a given perimeter, the greatest possible area is enclosed by a circle.A circle with a circumference of 18 has a diameter of (18/pi) and a radius of (9/pi).Its area is (pi R2) = (pi 92/pi2) = 81/pi = 25.78 (rounded)So an area of 42 cannot be enclosed by a perimeter of 18.
That depends how you define the perimeter for a cube. The term perimeter is usually used for plane figures, not for 3D solids.
The 2 lengths that you described are diagonals. The area of a rhombus when you know the diagonals is half the product of the diagonals: Area = (1/2) * ( 12 * 7) = 42.
Even if you knew how many sides the polygon has, you stillcould not calculate its perimeter with that much information.Examples:-- An equilateral triangle with area of 20 has perimeter of 20.3885 .-- A square with area of 20 has perimeter of 17.889(rounded).-- A rectangle with area of 20 can have any perimeter more than 17.889 .4 by 5 . . . . area = 20, perimeter = 182 by 10 . . . area = 20, perimeter = 241 by 20 . . . area = 20, perimeter = 42..etc.
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.
Largest = 86, Smallest 26
When the linear dimensions of a plane figure are quadrupled, its perimeter is quadrupled, and its area is multiplied by 42 = 16 .
The perimeter of a rectangle is 42. Meters. The length of the rectangle is threemeter less than twice the width.Mar