Question

Вписане коло прямокутного трикутника ABC дотикаеться до гіпотенузи AB у точці М. Знайдіть сторони трикутника, якщо радіуе кола дорівнюе 2 см i AM менше від ВМ на 7 см.

Answer 1

Problem Analysis

We are given a right triangle ABC, where a circle is inscribed and touches the hypotenuse AB at point M. We need to find the sides of the triangle, given that the radius of the circle is 2 cm and AM is 7 cm less than BM.

Solution

Let's denote the sides of the triangle as follows: - AC = a - BC = b - AB = c

We know that the radius of the inscribed circle is 2 cm, so the distance from the center of the circle to the point of tangency (M) is also 2 cm. Let's denote the distance from the center of the circle to point M as r.

Since the circle is inscribed, we can use the property that the distance from the center of the circle to any tangent line is perpendicular to that line. Therefore, we can draw a perpendicular line from the center of the circle to the hypotenuse AB, which will pass through point M.

Let's denote the length of AM as x. According to the problem statement, AM is 7 cm less than BM, so BM = x + 7.

Now, we can use the Pythagorean theorem to find the lengths of the sides of the triangle. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Using the Pythagorean theorem, we can write the following equations: - (a + r)^2 + (b + r)^2 = c^2 (Equation 1) - x + (x + 7) = c (Equation 2)

We have two equations with two unknowns (a and b), so we can solve them simultaneously to find the values of a and b.

Let's solve the equations step by step:

1. Expand Equation 1: - a^2 + 2ar + r^2 + b^2 + 2br + r^2 = c^2 - a^2 + b^2 + 2ar + 2br + 2r^2 = c^2 (Equation 3)

2. Substitute Equation 2 into Equation 3: - a^2 + b^2 + 2r(a + b) + 2r^2 = (x + (x + 7))^2 - a^2 + b^2 + 2r(a + b) + 2r^2 = (2x + 7)^2 - a^2 + b^2 + 2r(a + b) + 2r^2 = 4x^2 + 28x + 49 (Equation 4)

3. Simplify Equation 4: - a^2 + b^2 + 2r(a + b) + 2r^2 - 4x^2 - 28x - 49 = 0 (Equation 5)

Now, we have a quadratic equation in terms of a and b. We can solve this equation to find the values of a and b.

Let's solve the quadratic equation using the quadratic formula: - a = (-b ± √(b^2 - 4ac)) / (2a)

In our case, the coefficients of a and b are both 1, and the constant term is -4x^2 - 28x - 49. Substituting these values into the quadratic formula, we get:

- a = (-(2r) ± √((2r)^2 - 4(1)(2r^2 - 4x^2 - 28x - 49))) / (2(1)) - a = (-2r ± √(4r^2 - 8r^2 + 32x^2 + 224x + 392)) / 2 - a = (-2r ± √(-4r^2 + 32x^2 + 224x + 392)) / 2 - a = -r ± √(8x^2 + 56x + 98) (Equation 6)

Similarly, we can find the value of b using Equation 6.

Now, we have the values of a and b in terms of r, x, and the square root term. We can substitute the given values of r and x into Equation 6 to find the values of a and b.

Let's substitute the given values and solve for a and b:

- r = 2 cm - x = BM - 7 = (2x + 7) - 7 = 2x

Substituting these values into Equation 6, we get:

- a = -2 ± √(8(2x)^2 + 56(2x) + 98) - a = -2 ± √(32x^2 + 224x + 392) - a = -2 ± √(32(x^2 + 7x + 14))

Similarly, we can find the value of b using the same equation.

Now, we have the values of a and b in terms of x. To find the specific values of a and b, we need to know the value of x.

Unfortunately, the problem statement does not provide the value of x, so we cannot determine the specific values of a and b. However, we have derived the general equations to find a and b in terms of x.

In summary, the sides of the triangle ABC can be found using the equations: - a = -2 ± √(32(x^2 + 7x + 14)) - b = -2 ± √(32(x^2 + 7x + 14)) - c = 2x + 7

Please provide the value of x, and I can help you find the specific values of a and b.