Question

In a circle of radius 10 centimetres, two parallel chords of lengths 16 and 12 centimetres are drawn on either side of a diameter. What is the distance between them​

Answer 1

Answer:

The distance between the chords is 14 cm.

Step-by-step explanation:

In the figure,

⇒  'O' is the center of the circle.

⇒  'AB' and 'CD' are parallel chords with lengths 12 cm and 16 cm respectively.

⇒ 'OB' and 'OD' are the radius of the circle which is 10 cm.

The question is about the distance between the chords. That is the length of 'BQ'.

We know that the perpendicular drawn from the center of the circle to the chord divides the chord into equal parts.

OQ and OP are the perpendiculars drawn. Then,

PB = 6 cm

QD = 8 cm

Now we have two right-angle triangles OQD and OPB. Therefore, by the Pythagorean theorem,

⇒ OP² = OB² - PB²

   OP² = 10² + 6²

         = 100 - 36

         = 64

    OP = $\sqrt{64}$

         = 8 cm

⇒ OQ² = OD² - QD²

         = 10² + 8²

         = 100 - 64

         = 36

   OQ = $\sqrt{36}$

         = 6 cm

Hence the distance between the chords,

    BQ = OP + OQ

         = 6 + 8

     BQ = 14 cm