Perimeter = 2*(L + B)So 30 = 2*(9 + B)
15 = 9 + B so that B = 6
And therefore, Area = L*B = 9*6 = 54 sq metres.
Perimeter = 2*(L + B)So 30 = 2*(9 + B)
15 = 9 + B so that B = 6
And therefore, Area = L*B = 9*6 = 54 sq metres.
Perimeter = 2*(L + B)So 30 = 2*(9 + B)
15 = 9 + B so that B = 6
And therefore, Area = L*B = 9*6 = 54 sq metres.
Perimeter = 2*(L + B)So 30 = 2*(9 + B)
15 = 9 + B so that B = 6
And therefore, Area = L*B = 9*6 = 54 sq metres.
Perimeter = 2*(L + B)So 30 = 2*(9 + B)
15 = 9 + B so that B = 6
And therefore, Area = L*B = 9*6 = 54 sq metres.
A square of side 22 has an area of 484. Rectangle 23 x 21 has an area of 483...
The perimeter of any rectangle is [ 2 x (length + width) ]. Since the length and width of a square are equal, the perimeter of a square is also [ 2 x (side + side) ] = (4 x side).
nope because if u have a square with a side length of 4 then the perimeter is 16 and the area is 16 and say if u have a rectangle with side lengths of 2 and 6 then the perimeter is 16 but the area is 12 so the answer is no
A rectangle by definition has two pairs of sides with equal length. Since perimeter equals the length of all the sides. The equation for the perimeter of a rectangle could be thought of as: 2L + 2W = P Where L represents the length of one side of the rectangle and W represents the length of the adjacent (next to) side of the rectangle. If you know the length of one side and the perimeter, plug those values in as L and P and then solve for W. That will give you L and W which are the dimensions of the rectangle.
Both the side lengths and the perimeter are linear measurements, therefore they are proportional. In other words, twice the side length results in twice the perimeter.
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.
You can find the perimeter of a rectangle if you know its area and the length of one side. Divide the area by the length of the known side and the quotient will be the length of a side perpendicular to the known side, and then multiply the sum of the two sides by two to find the perimeter.
28
It is: 12 units in length
A square of side 22 has an area of 484. Rectangle 23 x 21 has an area of 483...
The maximum area for a rectangle of fixed perimeter is that of the square that can be formed with the given perimeter. 136/4 = 34, so that the side of such a square will be 34 and its area 342 = 1156.
10
24 cm is.
No it is not possible the dimensions are 200 by 1/2
perimeter of a rectangle is 20 on each side
Area: 15ab Perimeter: 2(3a+5b) or as 6a+10b
Simple the Perimeter you just add the outside numbers. For area. When I look at an octagon I see a rectangle and two Trapezoids on either side of the rectangle. So find the area of each object then add the areas up and voila you got the area of a rectangle.