i am doing my homework right now and I am stuck on that problem
3*27 = 81 and 3+3+27+27 = a perimeter of 60 inches
There are infinitely many possible rectangles. Suppose A >= 2.25 cm is the length of the rectangle. and B = 4.5 - A cm is the width. Then perimeter = 2*(A + B) = 2*(A + 4.5 - A) = 2*4.5 = 9 cm Also, it is easy to show that A >=B so that A and B cannot swap places. For each of the infinitely many values of A, you have a rectangle with perimeter 9 cm.
3
Yes.
You can't tell the perimeter from knowing the area.There are an infinite number of rectangles with different dimensions that all have the same area.Here are a few examples.1 x 176 . . . perimeter = 3542 x 88 . . . perimeter = 1804 x 44 . . . perimeter = 968 x 22 . . . perimeter = 6016 x 11 . . . perimeter = 5413.266 x 13.266 . . . perimeter = 53.066All of these have an area of 176, but they all have different perimeters.The last one on the list is a square. That's the rectangle with the shortest possible perimeterthat has the area you want.The perimeter of this square is 53.066. You can make a rectangle with any perimetermore than that, and an area of 176.
There is no relationship between the perimeter and area of a rectangle. Knowing the perimeter, it's not possible to find the area. If you pick a number for the perimeter, there are an infinite number of rectangles with different areas that all have that perimeter. Knowing the area, it's not possible to find the perimeter. If you pick a number for the area, there are an infinite number of rectangles with different perimeters that all have that area.
3*27 = 81 and 3+3+27+27 = a perimeter of 60 inches
There would be an infinite number of rectangles possible
The answer is, you can draw a rectangle with these measurements: 6cm and 9cm 5cm and 10cm 7cm and 8cm
There are infinitely many possible rectangles. Let A be ANY number in the range (0,6] and let B = 12-A. Then a rectangle with width A and length B will have a perimeter of 2*(A+B) = 2*12 = 24 units. Since A is ANY number in the interval (0,6], there are infinitely many possible values for A and so infinitely many answers to the question.
* It is unclear if the question is asking about two rectangles, each with a perimeter of 16, or two rectangles whose perimeters sum to 16. This answer assumes the former.Other than the 4x4 square, which coincidentally has both a perimeter and area of 16, some examples would be:1 x 7 rectangle : perimeter 16 in. , area 7 sq. in2 x 6 rectangle : perimeter 16 in., area 12 sq. in3 x 5 rectangle: perimeter 16 in., area 15 sq. inYou can calculate that for a given perimeter, the largest area is found in the square with a side measurement of P/4, i.e. the length and the width are the same.
There are infinitely many possible rectangles. Suppose A >= 2.25 cm is the length of the rectangle. and B = 4.5 - A cm is the width. Then perimeter = 2*(A + B) = 2*(A + 4.5 - A) = 2*4.5 = 9 cm Also, it is easy to show that A >=B so that A and B cannot swap places. For each of the infinitely many values of A, you have a rectangle with perimeter 9 cm.
The perimeter of a rectangle does not provide enough information to determine its dimensions.Let 0
3
(p/4)2, where p is the perimeter.
52 ft
Yes.