The radius length r of the inscribed circle equals to one half of the length side of the square, 10 cm.
The area A of the inscribed circle: A = pir2 = 102pi ≈ 314 cm2
The radius length r of the circumscribed circle equals to one half of the length diagonal of the square.
Since the diagonals of the square are congruent and perpendicular to each other, and bisect the angles of the square, we have
sin 45⁰ = length of one half of the diagonal/length of the square side
sin 45⁰ = r/20 cm
r = (20 cm)(sin 45⁰)
The area A of the circumscribed circle: A = pir2 = [(20 cm)(sin 45⁰)]2pi ≈ 628 cm2.
Depends on the size of the circles and the square.
the side of the square
An inscribed square is a regular polygon with four sides such that each of its vertices is on the boundary of some other shape which lies wholly outside the square.
The circumscribing square has sides of length 155 cm. The inscribed square has diagonals of 155 cm and so has sides of 155/sqrt(2) cm. The sides of a circumscribing square is always larger than those of the inscribed square by sqrt(2) = 1.4142 (approx). The area of a circumscribing square is always larger twice as large as that of the inscribed square.
The sides of the Square.
If we denote the measure of the length side of the circumscribed square with a, then the vertexes of the inscribed square will point at the midpoint of the side, a, of the circumscribed square.The area of the circumscribed square is a^2The square measure of the length of the inscribed square, which is also the area of this square, will be equal to [(a/2)^2 + (a/2)^2]. Let's find it:[(a/2)^2 + (a/2)^2]= (a^2/4 + a^2/4)= 2(a^2)/4= a^2/2Thus their ratio is:a^2/(a^2/2)=[(a^2)(2)]/a^2 Simplify;= 2
He circumscribed the square with a circle
A circle with a diameter of 2 is the guiding cynosure when Pi is the square of all possible circles: If the square root of Pi defines the side of a square and that square can be inscribed within a circle or enclose a circle, then the diameters of all possible circles between the largest and smallest include the circle of which Pi is its perfect square (a diameter of 2).
If I understand your question correctly, you would need to subtract the area of the inscribed circle from the circumscribed circle. Which would approximately be 78.60cm squared.
It could mean: 202 -4*pi*52 = 86 square mm rounded to nearest integer
Two circles with a square on top
Yes.
the side of the square
Depends on the size of the circles and the square.
You cannot circumscribe a "true rhombus". The opposite angles of a circumscribed quadrilateral must be supplementary whereas the opposite angles of a rhombus must be equal. That means a circumscribed rhombus is really a square.
An inscribed square is a regular polygon with four sides such that each of its vertices is on the boundary of some other shape which lies wholly outside the square.
The question asks about the "following". In those circumstances would it be too much to expect that you make sure that there is something that is following?