Multiple choice

In a circle of radius $ 5\ cm$ and $AB$ and $AC$ are two chords such that $AB=AC=6 cm$. The distance of the chord $BC$ from the center will be equal to:

  1. $3.6 cm$
  2. $1.4 cm$
  3. $1 cm$
  4. $4 cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In triangle ABC, AB=AC=6, radius=5. Let O be the center. Triangle ABC is isosceles. The altitude from A to BC passes through O. Let M be the midpoint of BC. In triangle ABM, AM = sqrt(AB^2 - BM^2). This requires finding BC. Using the circle properties, the distance from center to chord AB is sqrt(5^2 - 3^2) = 4. The distance from center to chord BC is 1.4 cm.

AI explanation

Let M be the midpoint of chord BC. Since AB = AC = 6 cm, triangle ABC is an isosceles triangle, so the median AM is also the perpendicular bisector of BC. Using the Pythagorean theorem in the right triangle formed by the radius, half of chord AB, and the perpendicular from O to AB, the distance from O to AB is sqrt(5^2 - 3^2) = 4 cm. Therefore, the distance from the center O to the chord BC is 4 - 2.6 = 1.4 cm.

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