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This question has no unique answer. A (3 x 2) rectangle has a perimeter = 10, its area = 6 A (4 x 1) rectangle also has a perimeter = 10, but its area = 4 A (4.5 x 0.5) rectangle also has a perimeter = 10, but its area = 2.25. The greatest possible area for a rectangle with perimeter=10 occurs if the rectangle is a square, with all sides = 2.5. Then the area = 6.25. You can keep the same perimeter = 10 and make the area anything you want between zero and 6.25, by picking different lengths and widths, just as long as (length+width)=5.
The formula for the perimeter of a rectangle is 2(L + W), where L is the length and W is the width. Then 16 = 2(L + W) : 8 = L + W. And L = W - 8, or W = L - 8 So, if you were dealing integers, you can choose any two numbers between 1 and 7 so that the numbers when added equal 8.
The perimeter of a rectangle is not enough to determine its shape. Let B < 5.5 be the breadth of the rectangle and let L = (11 - B) be its length. Then perimeter = 2*(L + B) = 2*(11 - B + B) = 2*11 = 22. The choice of B was arbitrarily any number between 0 and 5.5. Therefore there are infinitely many shapes that will meet the requirements.
To find the area of a rectangle, when you only know the perimeter, you can just just break the sides down into two pairs of lengths. For example; if the known perimeter is 100, you can call the two short sides, we'll call those the width, 10 (x2=20) and the two long sides, we'll call those length, 40 (x2=80) 20+80=100... Now you know all the lengths of the sides, so the area formula is L x W = area or 10 x 40 = 400. Any rectangle with a perimeter of 100 will have an area of 400, no matter what the lengths are and the process works for any rectangle, you just have to break the perimeter length down into two pairs of lengths.+++That starts off right but is NOT the full method.'You need to know the ratio between its length and breadth. Then apply that ratio to half of the perimeter to find the length and breadth.'Try it:L = 4, B= 6 so P = 20 but Area = 4 X 6 = 24 square units.L = 2, B = 8 so P = 20 again BUT Area now = 2 X 8 = 16 sq. units.For any given rectangle perimeter, there is an infinite number of possible areas.For a quadrilateral, Perimeter-only works only for the Square, for which A = [P/4]^2 = [(P^2)/16]
To find the area of a rectangle with a given perimeter, we need to use the formula for the perimeter of a rectangle, which is 2(length + width) = 16cm. Since the perimeter is given as 16cm, we have 2(length + width) = 16. If we divide by 2, we get length + width = 8. Without knowing the specific dimensions of the rectangle, we cannot determine the exact area.
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17. A 4x4 rectangle has a perimeter of 16 units and an area of 16 square units. A 3x6 rectangle has a perimeter of 18 units and an area of 18 square units. The number 17 is halfway between 16 and 18.
perimeter is the measure around the figure; area is the measure within the figure formula: perimeter: length+length+width+width=perimeter (for square or rectangle) area: length times width= area ( for square or rectangle)
This question has no unique answer. A (3 x 2) rectangle has a perimeter = 10, its area = 6 A (4 x 1) rectangle also has a perimeter = 10, but its area = 4 A (4.5 x 0.5) rectangle also has a perimeter = 10, but its area = 2.25. The greatest possible area for a rectangle with perimeter=10 occurs if the rectangle is a square, with all sides = 2.5. Then the area = 6.25. You can keep the same perimeter = 10 and make the area anything you want between zero and 6.25, by picking different lengths and widths, just as long as (length+width)=5.
The formula for the perimeter of a rectangle is 2(L + W), where L is the length and W is the width. Then 16 = 2(L + W) : 8 = L + W. And L = W - 8, or W = L - 8 So, if you were dealing integers, you can choose any two numbers between 1 and 7 so that the numbers when added equal 8.
Yes, there is. The area of a rectangle sets a lower limit on its perimeter.If the area is A, then the quadrilateral shape with the smallest perimeter has sides of length sqrt(A). Therefore the minimum perimeter is 4*sqrt(A). The perimeter can have any value grater than that since the area of the rectangle can be maintained while making it thinner and longer and thus increasing its perimeter with out any upper limit.
Any number between 0 and 12 that you like. There is no need for the sides of the rectangle to be whole numbers!
Length is the distance between two different points. But perimeter is the distance between the same point. Really interesting. Isn't it? Perimeter of sqaure is 4 x side Perimeter of rectangle is 2 (length + breadth) Perimeter of triangle is a + b + c Perimeter of a circle is specially named as circumference.
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.
No. Different rectangles, all with the same area, may have a different perimeter. Example:* A rectangle of 4 x 1 has an area of 4 square units, and a perimeter of 2(4+1) = 10. * A rectangle of 2 x 2 has an area of 4 square units, and a perimeter of 2(2+2) = 8. * A rectangle of 8 x 1/2 has an area of 4 square units, and a perimeter of 2(8 + 1/2) = 17. In fact, for any given area, you can make the perimeter arbitrarily large. On the other hand, you get the lowest perimeter if your rectangle is a square.
Picture a rectangle. The area is the space inside the lines. The perimeter is the distance around the lines.
The perimeter of a rectangle is not enough to determine its shape. Let B < 5.5 be the breadth of the rectangle and let L = (11 - B) be its length. Then perimeter = 2*(L + B) = 2*(11 - B + B) = 2*11 = 22. The choice of B was arbitrarily any number between 0 and 5.5. Therefore there are infinitely many shapes that will meet the requirements.