AB and CD are two parallel chords of a circle such that
AB = 10 cm and CD = 24cm. If the chords are on the opposite
sides of the centre and the distance between them is 17 cm,
the radius of the circle is

14 cm

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10 cm

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13 cm

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15 cm

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Solution

The correct option is C 13 cm
$r^{2} = x^{2} + 5^{2}$
[perpendicular from centre bisects the chord]

$r^{2} = (17 - x)^{2} + 12^{2}$

Now $x^{2} + 25 = (17 - x)^{2} + 144$
[perpendicular from centre bisects the chord]
$\Rightarrow x = 12$

$r^{2} = 144 + 25$
$r = 13 c m$

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Similar questions

$P Q$ and $R S$ are two parallel chords of a circle such that $P Q = 10 c m$ & $R S = 24 c m$ . If the chords are on the opposite sides of the centre and the distance between them is $17 c m$ , find the radius of the circle.

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are two parallel chords of a circle, which are on opposite sides of the centre, such that

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and the distance between them is

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A B = 10 c m

and

C D = 24 c m

and

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. The distance between AB and CD is

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. Calculate the radius of circle.

A B

and

C D

are two parallel chords of a circle such that

A B = 10

cm and

C D = 24

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17

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AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24cm. If the chords are on the opposite sides of the centre and the distance between them is 17 cm, the radius of the circle is

The correct option is C 13 cmr2=x2+52 [perpendicular from centre bisects the chord] r2=(17−x)2+122 Now x2+25=(17−x)2+144 [perpendicular from centre bisects the chord] ⇒x=12 r2=144+25 r=13 cm

AB and CD are two parallel chords of a circle such that
AB = 10 cm and CD = 24cm. If the chords are on the opposite
sides of the centre and the distance between them is 17 cm,
the radius of the circle is

The correct option is C 13 cm
$r^{2} = x^{2} + 5^{2}$
[perpendicular from centre bisects the chord]

$r^{2} = (17 - x)^{2} + 12^{2}$

Now $x^{2} + 25 = (17 - x)^{2} + 144$
[perpendicular from centre bisects the chord]
$\Rightarrow x = 12$

$r^{2} = 144 + 25$
$r = 13 c m$