Yes. There are lots of possible solutions. For example, a square of 4 x 4 has an area of 16. Adjust the angles (converting it into a rhombus), and you can lower the area all the way down to zero. Use trigonometry to find the correct angle.
Yes. There are lots of possible solutions. For example, a square of 4 x 4 has an area of 16. Adjust the angles (converting it into a rhombus), and you can lower the area all the way down to zero. Use trigonometry to find the correct angle.
Yes. There are lots of possible solutions. For example, a square of 4 x 4 has an area of 16. Adjust the angles (converting it into a rhombus), and you can lower the area all the way down to zero. Use trigonometry to find the correct angle.
Yes. There are lots of possible solutions. For example, a square of 4 x 4 has an area of 16. Adjust the angles (converting it into a rhombus), and you can lower the area all the way down to zero. Use trigonometry to find the correct angle.
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Yes. There are lots of possible solutions. For example, a square of 4 x 4 has an area of 16. Adjust the angles (converting it into a rhombus), and you can lower the area all the way down to zero. Use trigonometry to find the correct angle.
4x4 square: perimeter - 16 area - 16 6x2 rectangle perimeter - 16 area - 12
Oh, what a happy little question! Let's think about shapes that could have a perimeter of 15 and an area of 16. One shape that comes to mind is a rectangle with dimensions 4 by 4. Another possibility is a square with sides of length 4. These shapes show us that there can be different ways to create beautiful combinations of perimeter and area.
The perimeter is all the way around the shape so it would be.. 16 + 16 + 2 + 2 = 36. Therefore th perimeter must be 36cm.
* 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.
The perimeter is 16 feet.