5x ≥ 60
x ≥ 12
No, or at least not always.
square
a square * * * * * It need not be a square. It is simply a right angled quadrangle! It can be a trapezium, kite, arrowhead, rectangle or square. It cannot be a parallelogram or rhombus - other than a rectangle or square respectively.
The least perimeter is attained when the rectangle is, in fact a square. A square with an area of 32 square feet will have sides of sqrt(32) = 4*sqrt(2) ft. So the perimeter of the square will by 4*4*sqrt(2) = 16*sqrt(2) = 22.63 feet, approx.
A trapezoid (or trapezium) has at least one pair of parallel sides.If you asked, "Name all the types of quadrilaterals which have at least one pair of parallel sides?", I would answer: Trapezoid, rhombus, parallelogram, rectangle and square.
1 foot = 12 inches thus 2 feet = 2 * 12 = 24 inches. Therefore: The square has an area of 24 * 24 = 576 square inches. The rectangles each have an area of 2 * 3 = 6 square inches. The exact number required (which is of course also the least) is 576/6 = 96.
No, or at least not always.
square
A rectangle (or square).
a heart, square, and rectangle
47
Oh honey, let's break it down for you. The greatest area of the rectangle occurs when it's a square, so the sides are 4.5 inches each, giving an area of 20.25 square inches. The least area happens when it's a skinny little strip, like 1 inch by 8 inches, giving an area of 8 square inches. So the difference between the greatest and least possible areas is 12.25 square inches. Honey, you're welcome.
A rectangle or a square.
a square * * * * * It need not be a square. It is simply a right angled quadrangle! It can be a trapezium, kite, arrowhead, rectangle or square. It cannot be a parallelogram or rhombus - other than a rectangle or square respectively.
5.5 inches
You cannot. First of all, an area cannot be 6 inches since that is a measure of distance, not of an area. So a rectangle with an area of 36 inches is impossible. In any case, assuming the area is 36 SQUARE inches and that the length of the rectangle is at least as great as the width, all that you can say is that the width is at most 6 inches. Suppose the length is L inches where L >= 6 inches. and let W = 36/L inches. Then Area = L*W = L*36/L = 36 sq inches as required. But W is indeterminate.
A square, rhombus, rectangle, hexagon, and octagon.