The radius of the circle centered at point O is 120, the length of chord AB is 144. Find the distance from chord AB to the tangent line k parallel to it.
Let’s use the property of the tangent to the circle: the tangent is perpendicular to the radius of the given circle. Draw the OS radius to the tangent point of the straight line k. Draw the radius AO and BO. The AOB triangle is isosceles. Point E – the point of intersection of the radius OS and the chord AB. Since the triangle AOB is isosceles, then OE is the height and the median drawn to the base AB. AE = BE = 1/2 AB = 1/2 * 144 = 72 (cm) – since OE is the median. In a right-angled triangle, according to the Pythagorean theorem, leg OE ^ 2 = OB ^ 2 – BE ^ 2 = 120 ^ 2 – 72 ^ 2 = 1440 – 5184 = 9216; OE = √9216 = 96 (cm). Hence we have EC = OC – OE = 144 – 96 = 48 (cm).
Answer: 48 cm.
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