Find the area in a right-angled triangle ABC (angle C = 90 degrees) AB = 10 cm R of the drawn circle = 2 cm.

| September 24, 2021 | education

From the point O, the center of the circle, we will construct the radii OK, OH and OM to the points of tangency.

The quadrangle CKOH is a square, then CH = CK = R = 2 cm.

By the property of tangents drawn from one point, AH = AM, BK = BM.

Let the length AH = AM = X cm, then AC = X + 2 cm, BM = BK = 10 – X cm.

BC = 10 – X + 2 = 12 – X cm.

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

100 = (12 – X) ^ 2 + (X + 2) ^ 2.

100 = 144 – 24 * X + X ^ 2 + X ^ 2 + 4 * X + 4.

2 * X ^ 2 – 20 * X + 48 = 0.

X ^ 2 – 10 * X + 24 = 0.

Let’s solve the quadratic equation.

X1 = 4, X2 = 6.

If AH = 4, then AC = 4 + 2 = 6 cm, BC = 12 – 4 = 8 cm.

The area of ​​the triangle is: S = AC * BC / 2 = 6 * 8/2 = 24 cm2.

If AH = 6, then AC = 6 + 2 = 8 cm, BC = 12 – 6 = 6 cm.

The area of ​​the triangle is: S = AC * BC / 2 = 8 * 6/2 = 24 cm2.

Answer: The area of ​​the triangle is 24 cm2.



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