Through the center O of the circle inscribed in the triangle ABC, a straight line OK is drawn, perpendicular

| March 23, 2021 | education

Through the center O of the circle inscribed in the triangle ABC, a straight line OK is drawn, perpendicular to the plane of the triangle. Find the distance from point K to the sides of the triangle if AB = BC = 10cm, AC = 12cm, OK = 4cm.

For the solution, we need to determine the radius of the inscribed circle.

R = S / p, where S is the area of ​​the triangle, p is the semiperimeter of the triangle.

p = (AB + AC + CB) / 2 = (10 + 12 + 10) / 2 = 32/2 = 16 cm.

S = √ (p * (p – AB) * (p – AC) * (p – BC)) = √16 * 6 * 4 * 6 = √2304 = 48 cm2.

Then R = 48/16 = 3 cm.

OK = OM = OH = R = 3 cm.

Consider a right-angled triangle KOM, in which the leg KO = 4 cm by condition, the leg OM is equal to the radius of the circle OM = 3 cm.

Let us define by the Pythagorean theorem the hypotenuse KM, which is equal to the distance from point K to the side AC of the triangle.

KM ^ 2 = KO ^ 2 + OM ^ 2 = 4 ^ 2 + 3 ^ 2 = 16 + 9 = 25.

KM = 5 cm.

Since OK = OM = OH = R = 3 cm, the distances from point K to the sides of the triangle AB and BC are equal to KM.

KM = KH = KP = 5 cm.

Answer: The distance from point K to the sides of the triangle is 5 cm.



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