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Question

A 6 cm long chord of a circle is at a distance of 4 cm from the centre of the circle. Find the distance of 8 cm long chord of the same circle from the centre.

The correct answer is
4 cm

Understanding the Circle Chord Problem

This problem involves a circle and two chords. We are given information about one chord (length and distance from the center) and need to use this to find the distance of another chord (given its length) from the center of the same circle. The key is to realize that the radius of the circle remains constant.

Calculating the Circle's Radius

We can use the properties of a circle and the Pythagorean theorem to find the radius. A line segment drawn from the center of a circle perpendicular to a chord bisects the chord.

  • Given: First chord length = 6 cm, Distance from center = 4 cm.
  • The perpendicular from the center bisects the chord, so half the length of the first chord is $\frac{6 \text{ cm}}{2} = 3 \text{ cm}$.
  • This creates a right-angled triangle with the distance from the center (4 cm) as one leg, half the chord length (3 cm) as the other leg, and the radius (r) as the hypotenuse.
  • Using the Pythagorean theorem, $a^2 + b^2 = c^2$: $r^2 = (\text{Distance})^2 + (\text{Half Chord Length})^2$ $r^2 = 4^2 + 3^2$ $r^2 = 16 + 9$ $r^2 = 25$
  • Taking the square root of both sides gives the radius: $r = \sqrt{25}$ $r = 5 \text{ cm}$

The radius of the circle is 5 cm.

Finding the Distance of the Second Chord

Now that we know the radius, we can find the distance of the second chord from the center.

  • Given: Second chord length = 8 cm, Radius (r) = 5 cm.
  • The perpendicular from the center bisects this chord too. Half the length of the second chord is $\frac{8 \text{ cm}}{2} = 4 \text{ cm}$.
  • Again, we form a right-angled triangle. The radius (5 cm) is the hypotenuse, half the chord length (4 cm) is one leg, and the distance (d) from the center to the chord is the other leg.
  • Applying the Pythagorean theorem: $r^2 = d^2 + (\text{Half Chord Length})^2$ $5^2 = d^2 + 4^2$ $25 = d^2 + 16$
  • Rearranging to solve for $d^2$: $d^2 = 25 - 16$ $d^2 = 9$
  • Taking the square root gives the distance: $d = \sqrt{9}$ $d = 3 \text{ cm}$

Therefore, the distance of the 8 cm long chord from the center of the circle is 3 cm.

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Important Questions from Geometry (Notes)

  1. Which of the following is not true for a parallelogram?
  2. The length of major axis and coordinate of vertices for the ellipse $3x^2 + 2y^2 = 6$ respectively are:
  3. If the line through (3, y) and (2, 7) is parallel to the line through (-1, 4) and (0,6), then the value of y is:
  4. The points (K, 2 – 2K), (-K +1,2K) and (-4-K, 6-2K) are collinear if:
    (A) K = $\frac{1}{2}$
    (B) K = $-\frac{1}{2}$
    (C) K = $\frac{3}{2}$
    (D) K = -1
    (E) K = 1
    Choose the correct answer from the options given below:
  5. The asymptotes of the curve $(x^2- a^2)(y^2-b^2) = a^2b^2$ are
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