The chords MT and PK intersect at point E so that ME = 12cm TE = 3cm PE = KE find PK

| April 26, 2021 | education

The chords MT and PK intersect at point E so that ME = 12cm TE = 3cm PE = KE find PK and find the smallest value of the radius of this circle

By the property of intersecting chords of a circle, the product of the segments of one chord formed by the intersection point is equal to the product of the segments of the other chord.

Let the length of the segment PE be equal to X cm, then, by condition, KE = PE = X cm.

RE * KE = ME * TE.

X * X = 12 * 3.

X2 = 36.

X = PE = KE = 6 cm.

Let us determine the lengths of the chords.

РK = PE + KE = 6 + 6 = 12 cm.

MT = ME + TE = 12 + 3 = 15 cm.

The smallest radius of a circle with the obtained lengths of chords will be if the diameter of the circle coincides with a chord of a more painful length.

D = MT = 15 cm.

Then Rmin = D / 2 = 15/2 = 7.5 cm.

Answer: The length of the chord RK is 12 cm, the smallest value of the radius of the circle is 7.5 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.