A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.
Let AB be the chord of the given circle with centre O. The perpendicular distance from the centre of the circle to the chord is 8 cm.
Join OB.
Then OM = 8 cm and AB = 30 cm
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore M B=\left(\frac{A B}{2}\right)=\left(\frac{30}{2}\right) \mathrm{cm}=15 \mathrm{~cm}$
From the right ΔOMB, we have:
OB2 = OM2 + MB2
⇒ OB2 = 82 + 152
⇒ OB2 = 64 + 225
⇒ OB2 = 289
$\Rightarrow O B=\sqrt{289} \mathrm{~cm}=17 \mathrm{~cm}$
Hence, the required length of the radius is 17 cm.
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