In a right-angled triangle ABC (angle C = 90 degrees), the height CD is drawn so that the length of the segment BD

In a right-angled triangle ABC (angle C = 90 degrees), the height CD is drawn so that the length of the segment BD is 4 cm longer than the length of the segment CD, AD = 9cm. Find the sides of the triangle ABC.

According to the property of the height of a right-angled triangle drawn from the vertex of the right angle, the height is the average proportional to the projections of the legs on the hypotenuse.

Let the length of the height СD = X cm, then, by condition, the length of the segment is equal to: ВD = (X + 4).

Then: X = √ (9 * (X + 4)) = √ (9 * X + 36).

Let’s square both sides of the equality.

X ^ 2 = 9 * X + 36.

X ^ 2 – 9 * X – 36 = 0.

D = b ^ 2 – 4 * a * c = (-9) ^ 2 – 4 * 1 * (-36) = 81 + 144 = 225.

X1 = (9 – √225) / (2/1) = (9 – 15) / 2 = -6 / 2 = -3. (Doesn’t fit because <0).

X2 = (9 + √225) / (2/1) = (9 + 15) / 2 = 24/2 = 12.

CD = 12 cm, then DB = 12 + 4 = 16 cm.

AB = AD + DB = 9 + 16 = 25 cm.

From the right-angled triangle ACD we define the hypotenuse AC.

AC ^ 2 = SD ^ 2 + AD ^ 2 = 144 + 81 = 225.

AC = 15 cm.

From the right-angled triangle ABC, we define the leg BC according to the Pythagorean theorem.

BC ^ 2 = AB ^ 2 – AC ^ 2 = 625 – 225 = 400.

BC = 20 cm.

Answer: AB = 25 cm, BC = 20 cm, AC = 15 cm.

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