The chords AB and CD of the circle intersect at point F. The segment DF is 7 cm less than CF

| March 1, 2021 | education

The chords AB and CD of the circle intersect at point F. The segment DF is 7 cm less than CF, AF = 2 cm, AB = 6 cm. Find the length of the segment CF.

Let the length of the segment СF of the chord CD be equal to X cm, then the segment DF = (X – 7) cm.

Length of the segment BF = AB – AF = 6 – 2 = 4 cm.

By the property of intersecting chords, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

AF * BF = CF * DF.

2 * 4 = X * (X – 7).

8 = X2 – 7 * X.

X2 – 7 * X – 8 = 0.

Let’s solve the quadratic equation.

D = b ^ 2 – 4 * a * c = (-7) ^ 2 – 4 * 1 * (-8) = 49 + 32 = 81.

X1 = (7 – √81) / (2 * 1) = (7 – 9) / 2 = -2 / 2 = -1. (Doesn’t fit because <0).

X2 = (7 + √81) / (2 * 1) = (7 + 9) / 2 = 16/2 = 8.

CF = 8 cm.

Answer: The length of the segment CF is 8 cm.



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