Question

a dart is thrown randomly and lands within the square below. find the probability that the dart will land within the shaded area. the square has side lengths of 24, the smallest circle has a radius of 2, and the distance between each concentric circle is 2. round to the nearest hundredth.

a. 0.44
b. 0.11
c. 0.02
d. 0.20

Explanation:

Step1: Calculate area of the square

The area formula for a square is $A = s^2$, where $s$ is the side - length. Given $s = 24$, so $A_{square}=24^2=576$.

Step2: Calculate areas of the circles

The area formula for a circle is $A=\pi r^2$.
The radius of the smallest circle $r_1 = 2$, so $A_1=\pi\times2^2 = 4\pi$.
The radius of the second - circle $r_2=2 + 2=4$, so $A_2=\pi\times4^2 = 16\pi$.
The radius of the third - circle $r_3=4 + 2=6$, so $A_3=\pi\times6^2 = 36\pi$.

Step3: Calculate the shaded area

The shaded area $A_{shaded}=A_3 - A_1=36\pi-4\pi = 32\pi$.

Step4: Calculate the probability

The probability $P$ that the dart lands in the shaded area is $P=\frac{A_{shaded}}{A_{square}}$. Substitute $A_{shaded}=32\pi$ and $A_{square}=576$ into the formula: $P=\frac{32\pi}{576}=\frac{\pi}{18}\approx\frac{3.14159}{18}\approx0.17$. But if we assume the shaded area is the outermost ring only.
The radius of the outermost circle $r = 6$, the radius of the second - largest circle $r'=4$.
The area of the outermost ring $A=\pi r^{2}-\pi r'^{2}=\pi(6^{2}-4^{2})=\pi(36 - 16)=20\pi$.
The probability $P=\frac{20\pi}{576}=\frac{5\pi}{144}\approx\frac{5\times3.14159}{144}\approx0.11$.

Answer:

B. 0.11