Question

1. 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 centre, what is the distance of the other chord from Center. ​

Answer 1

Answer:

Step-by-step explanation:

∵ OM 1 AB

∴ is the mid-point of AB.

| The perpendicular from the centre of a circle to a chord bisects the chord

∵ ON ⊥ CD

∴ N is the mid-point of CD.

| The perpendicular from the centre of a circle to a chord bisects the chord

In triangle OMB,

OB2 = OM2 + MB2

| By Pythagoras Theorem

= (4)2 + (3)2

= 16 + 9 = 25

In right triangle OND,

OD2 + ON2 + ND2

| By Pythagoras Theorem

(5)2 = ON2 + ND2

25 = ON2 + 16

ON2 = 25 - 16

ON2 = 9

Hence, the distance of the chord from the centre is 3 cm.

Case II. When the two chords are on the opposite sides of the centre

As in case I

ON = 3 cm.

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