Two parallel chords $A B$ and $C D$ are $3.9$ cm apart and lie on opposite sides of the centre of a circle. If $A B = 1.4$ cm and $C D = 4$ cm, find the radius of the circle.

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Solution

The correct option is C $3.2$ cm
Given,
$A B = 1.4 c m$
$C D = 4 c m$
$P Q = 3.9 c m$
We know that the perpendicular dropped from the centre of the circle on the chord bisects the chord.
Therefore,
$C Q = Q D = 2 c m$
$A P = A B = O .7 c m$
Let the radius of the circle be $r$ and $O Q = x$
Thus,
$O P = 3.9 - x$
Now,
$C O^{2} = O Q^{2} + C Q^{2}$
$=> r^{2} = x^{2} + 2^{2}$
$=> r^{2} = x^{2} + 4$ (i)
Again,
$O A^{2} = O P^{2} + A P^{2}$
$=> r^{2} = (3.9 - x)^{2} + (0.7)^{2}$
$=> r^{2} = (3.9)^{2} - 2 (3.9) x + x^{2} + 0.49$
$=> x^{2} + 4 = (3.9)^{2} - 7.8 x + x^{2} + 0.49$ (from (i)
$=> 7.8 x = 15.21 + 4 + 0.49$
$=> 7.8 x = 19.7$
$=> x = 2.5$
Therefore,
$r^{2} = (2.5)^{2} + 4$
$r = 3.2 c m$
Radius of the circle is $3.2 c m$

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Two parallel chords AB and CD are 3.9 cm apart and lie on opposite sides of the centre of a circle. If AB=1.4 cm and CD=4 cm, find the radius of the circle.

Two parallel chords $A B$ and $C D$ are $3.9$ cm apart and lie on opposite sides of the centre of a circle. If $A B = 1.4$ cm and $C D = 4$ cm, find the radius of the circle.