
Прямоугольный треугольник с катетами 9 и 12 вписан в окружность найдите длину окружности и площадь
круга

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Problem Analysis
We are given a right triangle with the lengths of its legs as 9 and 12. This triangle is inscribed in a circle. We need to find the circumference of the circle and the area of the circle.Solution
To find the circumference of the circle, we can use the formula `C = 2πr`, where `C` is the circumference and `r` is the radius of the circle. In this case, the radius of the circle is equal to half the length of the hypotenuse of the right triangle.To find the area of the circle, we can use the formula `A = πr^2`, where `A` is the area and `r` is the radius of the circle.
Let's calculate the values step by step.
Calculating the Radius
The hypotenuse of the right triangle can be found using the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.Using the given lengths of the legs (9 and 12), we can calculate the length of the hypotenuse as follows:
Hypotenuse^2 = 9^2 + 12^2
Hypotenuse^2 = 81 + 144
Hypotenuse^2 = 225
Taking the square root of both sides, we get:
Hypotenuse = √225
Hypotenuse = 15
Therefore, the length of the hypotenuse is 15.
Since the radius of the circle is half the length of the hypotenuse, the radius is:
Radius = 15/2
Radius = 7.5
Calculating the Circumference
Using the formula for the circumference of a circle, we can calculate the circumference as follows:Circumference = 2π * Radius
Circumference = 2 * 3.14159 * 7.5
Circumference ≈ 47.12385
Therefore, the length of the circumference of the circle is approximately 47.12385.
Calculating the Area
Using the formula for the area of a circle, we can calculate the area as follows:Area = π * Radius^2
Area = 3.14159 * 7.5^2
Area ≈ 176.71459
Therefore, the area of the circle is approximately 176.71459.
Answer
The length of the circumference of the circle is approximately 47.12385, and the area of the circle is approximately 176.71459.Note: The calculations are approximate due to rounding.


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