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:

P = 0.637

Step-by-step explanation:

By a small online search, i've found that the image is a circle of radius R = 1in, and a square inside of the circle, such that the distance between the center of the square and the one vertex of the square is one inch.

To solve this problem, we need to find the area of each figure.

The area of a circle of radius R is:

A = 3.14*R^2

Then the area of our circle is:

A = 3.14*(1in)^2 = 3.14 in^2

Now for the square.

If the distance between the center and one vertex is 1 inch, then the distance between two opposite vertex is twice that: 2 inches.

Now we can think of a triangle rectangle, where both catheti are equal to L, the side length of the square, and the hypotenuse is 2 inches.

Using the Pythagoreans theorem, we get:

L^2 + L^2 = (2in)^2

2*L^2 = (2in)^2

L^2 = (2in)^2/2 = 2 in^2

L = √(2 in^2) = √2 in

And the area of a square of side length L, the area is:

A = L^2

Then the area of our square is:

A = (√2 in) = 2 in^2

Then:

Area of the circle = 3.14 in^2

Area of the square = 2 in^2

The area of the circle represents the 100% (if a point is chosen at random in the circle, the probability that the point will be inside the circle is P = 1)

The probability that the point will be inside the square is equal to the quotient between the area of the square and the area of the circle, this is:

P = (2 in^2)/(3.14 in^2) = 0.637

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