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.
The dimensions of the rectangle are 3 inches by 14 inches
42 square units.
Make it 2 wide and 21 long and you've got it.
28
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.
Largest = 86, Smallest 26
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.
The dimensions of the rectangle are 3 inches by 14 inches
42 square units.
Make it 2 wide and 21 long and you've got it.
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.
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.
That depends how you define the perimeter for a cube. The term perimeter is usually used for plane figures, not for 3D solids.