The perimeter of the square is 16 inches what is the circumference?

Answer 1

The perimeter of a plane figure is the length of its boundary. Thus the perimeter of a square of length L is 4L. So the perimeter of a square of length 4 is 4 x 4 = 16 (4 + 4 + 4 + 4 = 16). The perimeter of a circle is the length of its circumference.

If you are asking for the circumference of the circle circumscribed and inscribed in this square, their circumference will be:

First, we need to find the measure length of their radius. We know that the diagonals of the square form 4 congruent isosceles triangles with the base length equal to the length of the square, and length side equal one half of the diagonal length ( the diagonals of a square are equal in length and bisect each other (and bisect also the angle of the square ), so the center of the circumscribed circle of the square will be the point of their intersection, and its radius will be the one half of the diagonal of the square). We can find the diagonal length by using the Pythagorean theorem. So from the right trianglewhich is formed by drawing one of the diagonals, we find the length of the diagonal which is also the hypotenuse of this right triangle, and which is equal to square root of[2(4^2)]. So the length of the diagonal is equal 4(square root of 2), and its half is 2(square root of 2), which is the length of the radius of the circumscribed circle. So its circumference is equal to (2)(pi)(2(square root of 2)) = 4(square root of 2)pi.

Now, we need to know what is the length of the radius of the inscribed circle, and what is this radius. Let's look at the one of the fourth triangles that are formed by drawing the two diagonals of the square. If we draw the perpendicular from the intersection of the diagonals to the side of the square, this perpendicular is the median of the side of the square and also the altitude of this isosceles triangle. Let's find the measure of its length. Again we can use the Pythagorean theorem. So this measure is equal to the square root of [(2(square root of 2))^2] - 2^2] which is equal to 2. If we extend this perpendicular to the side of the triangle and draw another perpendicular from the point of the intersection of the diagonals to the other sides of the square, their length will be also 2. Since they have the same distance from the point of the intersection of the diagonals, we can say that their length is the length of the radius of the inscribed circle, and the point of the intersection of the diagonals is also its center. So the measure of length of the radius is 2, and the circumference of the inscribed circle is (2)(pi)(r) = (2)(pi)(2) = 4pi.

As a result, we can say that the point of the intersection of the diagonals of a square is the center of its inscribed and circumscribed circle, and the perpendicular lines drawing from this point to the sides of the square bisect each other. (These perpendiculars are parallel and equal in length to the square length, because we know that two lines that are perpendicular respectively to the other two parallel lines, are equal in length and parallel between them). We also can say that in an isosceles triangle with 45 degrees base angle, the median is not only also an altitude, but its length is one half of the length of the base.