In the circle radius $13 c m, P Q$ and $R S$ are two parallel chords of length $24 c m$ and $10 c m$ respectively calculate the distance between the chords if they are
One the same side of the centre.

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Solution

As perpendicular from centre bisects the chord
$\Rightarrow A Q = 12$ cm, $B C = 5$ cm
By Pythagoras theorem,
$A O^{2} = O Q^{2} - A Q^{2} = 169 - 144 = 25$
$\Rightarrow A O = 5 c m$
$B O^{2} = O S^{2} - B S^{2} = 169 - 25 = 144 \Rightarrow B O = 12 c m$
$\Rightarrow A B = 7 c m$ .

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In the circle 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 areOne the same side of the centre.

As perpendicular from centre bisects the chord⇒AQ=12cm, BC=5cmBy Pythagoras theorem,AO2=OQ2−AQ2=169−144=25⇒AO=5cmBO2=OS2−BS2=169−25=144⇒BO=12cm⇒AB=7cm.

In the circle radius $13 c m, P Q$ and $R S$ are two parallel chords of length $24 c m$ and $10 c m$ respectively calculate the distance between the chords if they are
One the same side of the centre.

As perpendicular from centre bisects the chord
$\Rightarrow A Q = 12$ cm, $B C = 5$ cm
By Pythagoras theorem,
$A O^{2} = O Q^{2} - A Q^{2} = 169 - 144 = 25$
$\Rightarrow A O = 5 c m$
$B O^{2} = O S^{2} - B S^{2} = 169 - 25 = 144 \Rightarrow B O = 12 c m$
$\Rightarrow A B = 7 c m$ .