Two chords AB and CD are drawn in a circle centered at point O. Lines AB and CD are perpendicular and intersect at point M

| September 15, 2021 | education

Two chords AB and CD are drawn in a circle centered at point O. Lines AB and CD are perpendicular and intersect at point M, which lies outside the circle. In this case, AM = 36, BM = 6, CD = 4√46. Find OM.

Let us draw perpendiculars OK and OP to the chords AB and CD, then the quadrangle OKMP is a rectangle. OK = PM = (AM – BM) / 2 + BM = (36 – 6) / 2 + 6 = 15 + 5 = 21 cm.

CD = CD / 2 = 4 * √46 / 2 = 2 * √46, since OK divides CD in half.

From a right-angled triangle OKD, OD ^ 2 = R ^ 2 = OK ^ 2 + KD ^ 2 = 441 + 184 = 625.

ОD = R = ОВ = 25 cm.

From the right-angled triangle OPB, we determine the length of the leg OP.

OR ^ 2 = KM ^ 2 = OB ^ 2 – BP ^ 2 = 625 – 225 = 400.

OR = KM = 20 cm.

From a right-angled triangle OKM, OM ^ 2 = OK ^ 2 + KM ^ 2 = 441 + 400 = 841.

OM = 29 cm.

Answer: The length of the segment OM is 29 cm.



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