**Question: **

In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

\(\sqrt {12} \) | |

\(\sqrt {14} \) | |

\(\sqrt {13} \) | |

\(\sqrt {11} \) |

**Show Answer**

Given that two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.

In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.

We can see that M and N are the mid points of AB and CD respectively.

AM = 3 cm and CD = 2 cm. Let 'OM' be x cm.

In right angle triangle AMO,

$AO^2 = AM^2 + OM^2$

$\Rightarrow$ $AO^2 = 3^2 + x^2$ ... (1)

In right angle triangle CNO,

$CO^2 = CN^2 + ON^2$

$\Rightarrow$ $CO^2 = 2^2 + (OM+MN)^2$

$\Rightarrow$ $CO^2 = 2^2 + (x+1)^2$ ... (2)

We know that both AO and CO are the radius of the circle. Hence $AO = CO$

Therefore, we can equate equation (1) and (2)

$3^2+x^2$=$2^2+(x+1)^2$

$\Rightarrow$ x = 2 cm

Therefore, the radius of the circle

$AO = \sqrt{AM^2 + OM^2}$

$\Rightarrow$ $AO=\sqrt{3^2+2^2}=\sqrt{13}$.

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