The four vertices of the rectangle are
(x, 0), (x, 3-x2), (-x, 3-x2) and (-x,0)
so that the sides are of length 2x and 3-x2
Therefore, the perimeter, P, is 4x + 6 - 2x2
dP/dx = 4 - 4x which is 0 when x = 1
when x = 1, P = 4 + 6 - 2 = 8 units.
d2P/dx2 = -4 < 0 which confirms that x = 1 does give a maximum.
88mm squared
== 25x25== 625 Rectangle sides 5 cm each. ==
The dimensions of the rectangle are 3 inches by 14 inches
perimeter = (5+5+12+12)= 34 area = (5*12) = 60 squared
First divide the perimeter by 2 then subtract the diagonal from this. The number left with must equal two numbers that when squared and added together equals the diagonal when squared (Pythagoras' theorem) These numbers will then be the length and height of the rectangle.
Vertices.
88mm squared
To answer this simply try a few out for yourself. In a 2x1 cm rectangle, the area is 2 cm squared and the perimeter is 6 cm In a 12x10 rectangle, the area is 120 cm squared and the perimeter is 44 cm. In some cases, the perimeter is larger and in others it is smaller. To answer your question, no, the perimeter of a rectangle is NOT always greater than its area.
The perimeter varies, depending on the shape.
The perimeter is 18 feet.
== 25x25== 625 Rectangle sides 5 cm each. ==
The dimensions of the rectangle are 3 inches by 14 inches
the old perimeter is 17 * 2 The old perimeter = 38
perimeter = (5+5+12+12)= 34 area = (5*12) = 60 squared
First divide the perimeter by 2 then subtract the diagonal from this. The number left with must equal two numbers that when squared and added together equals the diagonal when squared (Pythagoras' theorem) These numbers will then be the length and height of the rectangle.
Area = 8*24 =192 square feet Perimeter = 24+24+8+8 = 64 feet
Rectangle with area 28 m2 and width = 4 m => length = 28/4 = 7m. Then, perimeter = 2*(4+7) = 2*11 = 22 metres.