Answer:
[tex]\fbox{\begin{minipage}{5.8 em}x =5.97 cm\end{minipage}}[/tex]
Step-by-step explanation:
(1) As definition, the area ([tex]A[/tex]) of a circle is calculated by:
[tex]A = \pi[/tex] x [tex]r^{2}[/tex] with [tex]r[/tex] is radius of circle.
(2) The relation between radius ([tex]r[/tex]) and diameter ([tex]d_{1}[/tex]) of a circle is:
[tex]d_{1}[/tex] = 2 x [tex]r[/tex]
(3) Applying the Pythagorean theorem for a square with sides of length [tex]x[/tex],
the relation between side [tex]x[/tex] and diagonal [tex]d_{2}[/tex] is:
[tex]x^{2} +x^{2} = d^{2} \\or\\2x^{2} = d^{2}[/tex]
=> The diagonal ([tex]d_{2}[/tex]) of that square is calculated by:
[tex]d_{2}[/tex] = [tex]x\sqrt{2}[/tex]
(4) The square fits exactly inside a circle, with each of its vertices being on the circumference of the circle.
Applying the theorem of inscribed angle in a circle, the measure of arc BD is twice as the measure of inscribed angle BAD.
In other words, the measure of arc AC = 2 x 90 = 180 deg, or half of circle.
It means that the diagonal BD is coincide with a diameter of circle.
=> [tex]d_{1} = d_{2}[/tex]
or [tex]2[/tex] x [tex]r[/tex] = [tex]x\sqrt{2}[/tex]
or [tex]r[/tex] = [tex]x\frac{\sqrt{2} }{2}[/tex]
Substitute [tex]r[/tex] back into original formula used for calculating the area of circle:
=> A = [tex]\pi[/tex] x [tex](x\frac{\sqrt{2}}{2})^{2}[/tex] = [tex]\pi[/tex] x [tex]\frac{x^{2}}{2}[/tex]
=> [tex]x^{2} = A[/tex] x [tex]\frac{2}{\pi }[/tex]
=> [tex]x = \sqrt{A}[/tex] x [tex]\sqrt{\frac{2}{\pi } }[/tex] = [tex]\sqrt{56}[/tex] x [tex]\sqrt{\frac{2}{\pi } }[/tex] = 5.97 cm
Hope this helps!
:)
