Answer:
Part 1) The length of each side of square AQUA is [tex]3.54\ cm[/tex]
Part 2) The area of the shaded region is [tex](486\pi-648)\ units^{2}[/tex]
Step-by-step explanation:
Part 1)
step 1
Find the radius of the circle S
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
we have
[tex]A=25\pi\ cm^{2}[/tex]
substitute in the formula and solve for r
[tex]25\pi=\pi r^{2}[/tex]
simplify
[tex]25=r^{2}[/tex]
[tex]r=5\ cm[/tex]
step 2
Find the length of each side of square SQUA
In the square SQUA
we have that
SQ=QU=UA=AS
[tex]SU=r=5\ cm[/tex]
Let
x------> the length side of the square
Applying the Pythagoras Theorem
[tex]5^{2}=x^{2} +x^{2}[/tex]
[tex]5^{2}=2x^{2}[/tex]
[tex]x^{2}=\frac{25}{2}\\ \\x=\sqrt{\frac{25}{2}}\ cm\\ \\ x=3.54\ cm[/tex]
Part 2) we know that
The area of the shaded region is equal to the area of the larger circle minus the area of the square plus the area of the smaller circle
Find the area of the larger circle
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
we have
[tex]r=AB=18\ units[/tex]
substitute in the formula
[tex]A=\pi (18)^{2}=324\pi\ units^{2}[/tex]
step 2
Find the length of each side of square BCDE
we have that
[tex]AB=18\ units[/tex]
The diagonal DB is equal to
[tex]DB=(2)18=36\ units[/tex]
Let
x------> the length side of the square BCDE
Applying the Pythagoras Theorem
[tex]36^{2}=x^{2} +x^{2}[/tex]
[tex]1,296=2x^{2}[/tex]
[tex]648=x^{2}[/tex]
[tex]x=\sqrt{648}\ units[/tex]
step 3
Find the area of the square BCDE
The area of the square is
[tex]A=(\sqrt{648})^{2}=648\ units^{2}[/tex]
step 4
Find the area of the smaller circle
The area of the circle is equal to
[tex]A=\pi r^{2}[/tex]
we have
[tex]r=(\sqrt{648})/2\ units[/tex]
substitute in the formula
[tex]A=\pi ((\sqrt{648})/2)^{2}=162\pi\ units^{2}[/tex]
step 5
Find the area of the shaded region
[tex]324\pi\ units^{2}-648\ units^{2}+162\pi\ units^{2}=(486\pi-648)\ units^{2}[/tex]
