It might depend on how you draw it, but I don't think it is possible. Draw the figure, and look at the points where lines meet. Count how many of those points are "odd", meaning that an odd number of lines meets there. Each of these points has to be either the starting point, or the end point, and you can only have one starting point and one end point.
It is not. If you draw yourself a square then inscribe a circle with a radius of half the length of a side of the square, the circle will fit inside the square but the corners of the square will be outside the circle. Thus by inspection the area of the square is larger than the area of the circle.
Perpendicular bisector.
if area of square is Z area of largest circle just fit inside
Assuming that the 14 inch box is square, you could fit a circle inside with a 14 inch diameter.
If there is a square field fit in a circle,then the middle point of the hypotenuse of the square is the the centre point of the circle. So,the half of hypotenuse is the radius of the circle. So Hypotenuse of square = √{(5)²+(5)²} cm ==5√2 cm So, the radius of the circle = (5√2 )/2 =5/√2 cm.
you don't
The answer depends on their relative size: is the circle inside the square, the square inside the circle or something else?
The diagonal of the square.
You find the area of the whole square first. Then you find the area of the circle inside of it And then subtract the area of the circle from the area of the square and then you get the shaded area of the square
The is not stated that the circle inside the square was the greatest possible circle, so all one can say is 8pi at most.
If the circle is inside the square, four.
The same as half the side of the square, as the radius of the circle is half its diameter, and the diameter of the circle is equal to the side of the square.
Yes
If the circle inscribes the square, the diameter equals the square's side length. In this case, 16mm.
2.5
It is not. If you draw yourself a square then inscribe a circle with a radius of half the length of a side of the square, the circle will fit inside the square but the corners of the square will be outside the circle. Thus by inspection the area of the square is larger than the area of the circle.
Well, since the circle is inside the square, the edges of the square define the limits of the circle. Since we have a square, the length of each side is the square root of 800ft. The radius of the circle is half the length of the sides. Therefore, the answer is half of (the square root of 800), which makes for an ugly radius value of "14.142135623730950488016887242097" ft