Question

1. a circle is inscribed in a square. a point in the figure is selected at random. find the probability that the point will be in the part that a circle covers about 10% about 20% about 25% about 5%

Explanation:

Step1: Assume side length of square

Let the side length of the square be \(s\). Then the area of the square \(A_{square}=s^{2}\).

Step2: Find radius of circle

Since the circle is inscribed in the square, the diameter of the circle \(d = s\), and the radius \(r=\frac{s}{2}\). The area of the circle \(A_{circle}=\pi r^{2}=\pi(\frac{s}{2})^{2}=\frac{\pi s^{2}}{4}\).

Step3: Calculate probability

The probability \(P\) that a randomly - selected point is in the circle is \(P=\frac{A_{circle}}{A_{square}}\). Substitute \(A_{circle}=\frac{\pi s^{2}}{4}\) and \(A_{square}=s^{2}\) into the formula, we get \(P = \frac{\frac{\pi s^{2}}{4}}{s^{2}}=\frac{\pi}{4}\approx\frac{3.14}{4}=0.785\) (This is wrong. Wait, no, the problem is to find the probability that the point is in the part that is not shaded? No, no, wait, the problem says "Find the probability that the point will be in the part that is not shaded". Wait, no, no, the formula for probability of a point being in a region is \(P=\frac{\text{Area of the region}}{\text{Total area}}\). If the circle is shaded, and we want the non - shaded part (the area of the square minus the area of the circle). \(A = s^{2}-\frac{\pi s^{2}}{4}=s^{2}(1 - \frac{\pi}{4})\). Then \(P=\frac{s^{2}(1 - \frac{\pi}{4})}{s^{2}}=1-\frac{\pi}{4}\approx1 - 0.785=0.215\approx20\%\)

Answer:

about \(20\%\)