A circle is inscribed in a right-angled triangle. The point of tangency divides the hypotenuse

| March 1, 2021 | education

A circle is inscribed in a right-angled triangle. The point of tangency divides the hypotenuse in the ratio of 5: 12. Find the area of the triangle if the center of the circle is removed from the vertex of the right angle at a distance of √18.

ОМ = ОЕ and is equal to the radius of the inscribed circle. Consider a right-angled triangle CEO, in which the hypotenuse OC= √18, and the legs OE and CE are equal to the radius R of the circle.

By the Pythagorean theorem OC ^ 2 = R ^ 2 + R ^ 2 = 2 * R ^ 2.

(√18) ^ 2 = 2 * R ^ 2.

R ^ 2 = 18/2 = 9.

R = 3 cm.

Then OM = OE = CM = CE = 3 cm.

According to the condition VK / AK = 5/12.

Let one part of the ratio be X cm, then AK = 12 * X, BK = 5 * X, and the hypotenuse AB = AK + BK = 17 * X.

According to the properties of tangents drawn from one point, ВM = ВK = 5 * X, AE = AK = 12 * X.

Then the leg BC = 5 * X + 3, and the leg AC = 12 * X + 3.

By the Pythagorean theorem AB ^ 2 = AC ^ 2 + BC ^ 2.

(17 * X) ^ 2 = (12 * X + 3) ^ 2 + (5 * X + 3) ^ 2.

289 * X ^ 2 = 144 * X ^ 2 + 72 * X + 9 + 25 * X ^ 2 + 30 * X + 9.

60 * X ^ 2 – 51 * X – 9 = 0.

D = b ^ 2 – 4 * a * c = (-17) ^ 2 – 4 * 20 * (-3) = 289 + 240 = 529.

X1 = (17 – √529) / (2 * 20) = (17 – 23) / 40 = -6/40 = -0.15 (Not suitable, since X cannot be <0).

X2 = (17 + √529) / (2 * 20) = (17 + 23) / 40 = 40/40 = 1.

Then AB = 5 + 12 = 17 cm.

AC = 12 + 3 = 15 cm.

BC = 5 + 3 = 8 cm.

Determine the area of ​​the triangle.

S = AC * BC / 2 = 15 * 8/2 = 60 cm2.



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