The greatest possible area of a rectangle is simply the area of a square, which is a special type of rectangle.
in order to find the area of that square:
4s=96 (4s=4 sides)
s=24
A=lw
A=24*24
A=576
so the area of that rectangle would be 576 ft...
Greatest area is achieved by a square whose sides are 72/4 = 18 cm. The area, in that case, is 18*18 = 324 cm2
56 cm Perimeter of a rectangle is given by 2(length + breadth). So, perimeter of given rectangle = 2(18 + 10) = 56
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.
The answer is, you can draw a rectangle with these measurements: 6cm and 9cm 5cm and 10cm 7cm and 8cm
Think about two numbers whose multiplication is 9 or simply think about the factors of 9. Factors of 9 are 3x3 Now see perimeter of rectangle is 2 times sum of length and breadth. Perimeter of rectangle is 4 times side. 2 and 4 are of the factors of 9. So the figure cannot be rectangle or square. Perimeter of triangle= Sum of all sides 9 = 1+3+5 =9 Hence it is possible to have a triangle with perimeter 9cm as there are various numbers whose addition could be 9. So the figure could be triangle.
The perimeter of a rectangle whose length is 5cm wit a width of 6cm is 22cm
Greatest area is achieved by a square whose sides are 72/4 = 18 cm. The area, in that case, is 18*18 = 324 cm2
You forgot to put in the length of the rectangle's perimeter.
56 cm Perimeter of a rectangle is given by 2(length + breadth). So, perimeter of given rectangle = 2(18 + 10) = 56
Perimeter = 5+5+3.5+3.5 = 17 feet
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That depends on the rectangle! You can have different rectangles with the same area, but with different perimeters.
5x2 + 3x2 = 16
6 feet
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.