A point is chosen at random in the circle. What percent of the time will the point be in the square? Round to the nearest tenth of a percent.

Answer:
[tex]63.7\%[/tex]
Step-by-step explanation:
we know that
To find the percent divide the area of the square by the area of the circle
step 1
Find the area of the circle
The area of the circle is
[tex]A=\pi r^{2}[/tex]
we have
[tex]r=1\ in[/tex]
substitute the values
[tex]A=(3.14)(1)^{2}=3.14\ in^{2}[/tex]
step 2
Find the area of the square
The area of the square is
[tex]A=b^{2}[/tex]
where
b is the length side of the square
we have
[tex]D=2\ in[/tex] ---> the diagonal of the square is equal to the diameter of the circle
Applying Pythagoras Theorem
[tex]D^{2}=b^{2}+b^{2}[/tex]
substitute the values
[tex]2^{2}=2b^{2}[/tex]
[tex]4=2b^{2}[/tex]
[tex]b^{2}=2\ in^{2}[/tex] ------> the area of the square
step 3
Find the percent
[tex]\frac{2}{3.14}= 0.6369[/tex]
Convert to percent
[tex]0.6369*100=63.69\%[/tex]
Round to the nearest tenth of a percent
[tex]63.7\%[/tex]