Answer:
The probability that a randomly selected point within the circle falls in the white area is approximately 36.3%
Step-by-step explanation:
The given parameters of the area are;
The length of the radius of the circle, r = 4 cm
The length of the side of the inscribed quadrilateral, s = 4·√2 cm
The diagonals of the inscribed quadrilateral = 2 × The radius of the circle
∴ The diagonals of the inscribed quadrilateral are equal and given that the sides of the quadrilateral are equal, the quadrilateral is a square
The probability that a randomly selected point within the circle is white is given by the ratio of the white area to the brown square area as follows;
The white area = The total area - The area brown square
The total area, A = The area of the circle with radius, r = π·r²
∴ A = π·(4 cm)² = 16·π cm²
The area of the brown square, Asq = s² = 4·√2 cm × 4·√2 cm = 32 cm²
The white area, Aw = 16·π cm² - 32 cm² = 16·(π - 2) cm²
The probability that a randomly selected point within the circle falls in the white area is therefore;
[tex]P(W) = \dfrac{A_W}{T_A} = \dfrac{16 \cdot (\pi - 2) \, cm^2}{16 \cdot \pi \, cm^2 } \times 100= \left (1 - \dfrac{2}{\pi}\right) \times 100 \approx 36.338022763241865692 \%[/tex]
By rounding to the nearest tenth of a percent, we have;
The probability that a randomly selected point within the circle falls in the white area, P(W) ≈ 38.3%.
