AB is the diameter of a circle centered at point O, the chord EF intersects the diameter at point K, EK = 4

| July 3, 2021 | education

AB is the diameter of a circle centered at point O, the chord EF intersects the diameter at point K, EK = 4, KF = 6, OK = 5. 1) Find the radius of the circle 2) Find the distance from point O to the chord BF 3) Find an acute angle between AB and the chord EF 4) What is the chord FM if the chord EM is parallel to AB

Let the radius of the circle be X cm, then the length of the segment BK = (X + 5), the length of the segment AK = (X – 5) cm.

By the property of intersecting chords: AK * BK = KF * EK.

(X – 5) * (X + 5) = 24.

X ^ 2 – 25 = 24.

X ^ 2 = 49.

X = R = 7 cm.

Let us draw the radius ОF and in the triangle OKF, by the cosine theorem, we determine the cosine of the angle OKF.

OF ^ 2 = OK ^ 2 + KF ^ 2 – 2 * OK * KF * CosOKF.

49 = 25 + 36 – 2 * 5 * 6 * CosOKF.

60 * CosOKF = 61 – 49 = 12.

CosOKF = 12/60 = 0.2.

Angle OKF = arcos (0.2) = 78.460.

Since EM is parallel to AB, the angle FEM = FRM = 78.460. The central angle FОМ rests on the arc FM as well as the inscribed angle FЕМ, then FОМ = 2 * FЕМ = 156.920.

In triangle FOM, apply the cosine theorem. FM ^ 2 = FO ^ 2 + OM ^ 2 – 2 * FO * OM * CosFOM.

FM ^ 2 = 49 + 49 – 2 * 7 * 7 * (-0.92) = 188.16.

FM =

Answer: The radius is 7 cm, the OKF angle is 78.460, the FM chord is 13.71 cm.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.