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This is a rectangle with length 5 units and width 3 units.
The area = length x width = 5 x 3 = 15 sq units.
The perimeter = 2 x (length + width) = 2 x 8 = 16 units.
This is a rectangle with length 5 units and width 3 units.
The area = length x width = 5 x 3 = 15 sq units.
The perimeter = 2 x (length + width) = 2 x 8 = 16 units.
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What shape has a perimeter of 16 and area of 15?
A 5 by 3 rectangle
Is there a shape with the same perimeter and area?
There is a square that has a length of 4. Area: 4 x 4: 16 Perimeter: 4+4+4+4= 16 Yes, there is a shape with the same perimeter and area.
Is it possible to make a shape out of the area of 16 and a perimeter of 15?
Yes but not a square (or rectangle). A quadrilateral with an area of 16 sq units must have sides of at least 4 units and so a perimeter of at least 16 units. However, a circle of perimeter 15 units will enclose an area of 17.905 sq units (to 3 dp) so an ellipse of 15 units' perimeter will meet the requirements.
What is the area that has a 16 cm perimeter?
It depends on the shape. Any area up to 20.37 square cm is possible.
What is the perimeter for 16 tiles that are 1 inch?
The perimeter of each tile is 4 inches. So the perimeter of 16 tiles - separately - is 4*16 = 64 inches. The perimeter of a 16-tile shape depends on the shape.
What shape has a perimeter of 16 and area of 15?
A 5 by 3 rectangle
Is there a shape with the same perimeter and area?
There is a square that has a length of 4. Area: 4 x 4: 16 Perimeter: 4+4+4+4= 16 Yes, there is a shape with the same perimeter and area.
Is it possible to make a shape out of the area of 16 and a perimeter of 15?
Yes but not a square (or rectangle). A quadrilateral with an area of 16 sq units must have sides of at least 4 units and so a perimeter of at least 16 units. However, a circle of perimeter 15 units will enclose an area of 17.905 sq units (to 3 dp) so an ellipse of 15 units' perimeter will meet the requirements.
What shapes have a perimeter of 15 and an area of 16?
5x3
What shape has an area of 12 and a perimeter of 16?
4x3?
What is the area that has a 16 cm perimeter?
It depends on the shape. Any area up to 20.37 square cm is possible.
How can the perimeter stay the same and the area change?
4x4 square: perimeter - 16 area - 16 6x2 rectangle perimeter - 16 area - 12
What is the perimeter for 16 tiles that are 1 inch?
The perimeter of each tile is 4 inches. So the perimeter of 16 tiles - separately - is 4*16 = 64 inches. The perimeter of a 16-tile shape depends on the shape.
Is it possible to make a shape of an area of 12 and a perimeter of 16?
Yes a 2 by 6 rectangle for example.
Is it sometimes always or never true that the perimeter of a rectangle is numerically greater than its area?
Sometimes. Experiment with a small square and with a large square (though any shape rectangle will do). A square of 4 x 4 has a perimeter of 16, and an area of 16. A smaller square has more perimeter than area. A larger square has more area than perimeter.
What shape has the same perimeter and area?
A square 4 squares in the middle and 4 lines out so it has to be a square!!! actually a square does not have the same perimeter and area a 4x4 area square has 8 perimeter so no squares do not but i have made a shape that has 32 perimeter and 32 area so it is possible and we don't want to tell you what the shape is. i already gave you too much information with the 32.... << inccorect A 4x4 square does have the same perimeter and area 4x4=16 4+4+4+4=16 But there are also other shapes such as a circle witha diameter of 2 A right angled triangle with sides of 6, and 8 and the hypotenuse being 10 has the same area as perimeter.
What is the perimeter and area of a shape that has equal sides of 4m?
A shape is a nebulous term. It may mean a triangle, a square, or a pentagon in 2D, or a block or pyramid in 3D etc. The perimeter and area cannot be found without defining the shape first. How many dimensions? How many sides? For example, if the shape is a square, which has 4 edges, its perimeter = 4 * 4 [m] = 16 [m] and its area = 4 [m] * 4 [m] = 16 [m^2].
