In a circle with centre $O, A B$ and $C D$ are two parallel chords on the same side of the diameter. If $A B=12 \mathrm{~cm}, C D=18$ $\mathrm{cm}$ and distance between the chords $\mathrm{AB}$ and $\mathrm{CD}$ is $3 \mathrm{~cm}$, then find the radius of the circle (in $\mathrm{cm}$ ).
Answer & explanation
Correct answer: option 2
Perpendicular drawn from center of circle to chord which bisects the chord.
For chord AB,
OP is perpendicular drawn from center and OB is the radius.
AP = PB = \(\frac{12}{2}\) = 6 cm
In right angled triangle POB,
Using Pythagoras theorem,
= \( {OB }^{ 2} \) = \( {PB }^{2 } \) + \( {OP }^{2 } \)
= \( {OB }^{ 2} \) = \( {6 }^{2 } \) + \( {OP }^{2 } \)
Similarly, for chord CD
CO = OD = \(\frac{18}{2}\) = 9 cm
In right angled triangle QDC,
= \( {OD }^{ 2} \) = \( {QD }^{2 } \) + \( {OQ }^{2 } \)
= \( {OD }^{ 2} \) = \( {9 }^{2 } \) + \( {OQ }^{2 } \)
Then,
= OP = OQ + QP
= OP = OQ + 3
But,
= OD = OB ..(Radii of circle)
= \( {OD }^{ 2} \) = \( {9 }^{2 } \) + \( {6 }^{2 } \) = 117
= OD = 3\(\sqrt {13 }\) cm.
Therefore, radius of circle is 3\(\sqrt {13 }\) cm.