Two mutually perpendicular chords are drawn in a circle at a distance of 2 from the center, the length of each of them is 8

| June 19, 2021 | education

Two mutually perpendicular chords are drawn in a circle at a distance of 2 from the center, the length of each of them is 8. Into what parts does the intersection point of the chords divide each of them?

Perpendiculars drawn from the center of the circle to the chord divide the chord in half.

Then AM = BM = AB / 2 = 8/2 = 4 cm., DK = CK = CD / 2 = 8/2 = 4 cm.

Chords intersect at an angle of 90, OM is perpendicular to AB, OK is perpendicular to CD, then the quadrangle OMНK is a rectangle, and since OK = OM = 2 cm, then OMНK is a square.

MO = HK = 2 cm.

AH = AM + MH = 4 + 2 = 6 cm. BH = BM – MH = 4 – 2 = 2 cm.

DН = DК + НК = 4 + 2 = 6 cm. CH = СК – НК = 4 – 2 = 2 cm.

Answer: The point of intersection of the chords divides them into 6 cm and 2 cm segments.



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.