The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
Solution:
The perpendicular drawn from the center of the circle to the chords bisects it.
Draw two parallel chords AB and CD of lengths 6 cm and 8 cm. Let the circle's center be O. Join one end of each chord to the center. Draw 2 perpendiculars OM and ON to AB and CD, respectively, which bisects the chords.
AB = 6 cm CD = 8 cm MB = 3 cm ND = 4 cm
Given OM = 4 cm and let ON = x cm Consider ΔOMB
By Pythagoras theorem,
OM² + MB² = OB²
4² + 3² = OB²
OB² = 25
OB = 5 cm
OB and OD are the radii of the circle.
Therefore OD = OB = 5 cm.
Consider ΔOND
ON² + ND² = OD²
x² + 4² = 5²
x² = 25 - 16
x² = 9
x = 3
The distance of the chord CD from the center is 3 cm.
☛ Check: NCERT Solutions for Class 9 Maths Chapter 10
Video Solution:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the centre?
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.6 Question 3
Summary:
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the center, we have found that the distance of the other chord CD from the center is 3 cm.
☛ Related Questions:
- Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center.
- Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
- ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
- AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.