Question

Will give brainliest

Answer 1

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;

$P(W) = \dfrac{A_W}{T_A} = \dfrac{16 \cdot (\pi - 2) \, cm^2}{16 \cdot \pi \, cm^2 } \times 100= \left (1 - \dfrac{2}{\pi}\right) \times 100 \approx 36.338022763241865692 \%$

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%.