Question

1.
In a circle of radius 13 cm, PQ and RS are two parallel
chords of length 24 cm and 10 cm respectively.
calculate the distance between the chords if they are
(ii) on the same side of the centre
(iii) on the opposite sides of the
centre​

Answer 1

Given:

• Radius of circle $= 13cm$

• Chord $PQ = 24cm$

• Chord $RS = 10 cm$

To Find:

• Distance between the chords if they are

            (i) on the same side of the center

            (ii) on the opposite sides of the  center​

Solution:

(i) on the same side of the center

As perpendicular from center bisects the chord

⇒                    $AQ=12cm\\ BC=5cm$

By Pythagoras theorem,

          ⇒      $AO^2 = OQ^ 2 - AQ ^2$

                       $=169-144 =25$

              ⇒ $AO=5 cm$

         ⇒  $BO ^2 = OS ^2 -BS^2$

                  $=169-25 =144$

         ⇒$BO=12 cm$

          $AB = BO - AO$

               $=12 - 5 =7$

Thus, $AB=7cm.$

 (ii) on the opposite sides of the  center​

Perpendicular from center bisects the chord

      ⇒              $AQ=12cm, BS=5cm$

By Pythagoras theorem,

                  ⇒    $OA^2 = OQ^2- AQ ^2$

                             $= 169-144 = 25cm$

                        $OA = 5cm$

               ⇒  $OB^2 = OS^2 -BS ^2$

​                  $= 169-25 =144cm$

               ⇒ $AB = OA + OB$

                     $AB = 12+ 5 = 17$

⇒ Thus, $AB=17cm$