PQ and RS are two parallel chords of a circle of length 14 cm and 48 cm, respectively, and lie on the same side of the centre O. If the distance between the chords is 17 cm, what is the radius (in cm) of the circle ?

25

Let OX be perpendicular to RS and OY be perpendicular to PQ

Let OX be d

and XY = 17

So,

OY = (d + 17)

Applying pythagoras theorem in \(\Delta \)OXS

\( {OS }^{2 } \) = \( {OX }^{2 } \) + \( {XS }^{2 } \)

⇒ \( {OS }^{2 } \) = \( {d }^{2 } \) + \( {24}^{2 } \)

Again,

In \(\Delta \)OYQ

\( {OQ }^{2 } \) = \( {OY }^{2 } \) + \( {YQ }^{2 } \)

⇒ \( {OQ }^{2 } \) = \( {(d + 17) }^{2 } \) + \( {7 }^{2 } \)

Now,

\( {d }^{2 } \) + \( {24 }^{2 } \) = \( {(d + 17) }^{2 } \) + \( {7 }^{2 } \)  [OS = OQ = radius of the circle]

⇒  \( {d }^{2 } \) + 576 = \( {d }^{2 } \) + \( {17 }^{2 } \) + d x 17 + 49

⇒ 576 =  289 + 34d + 49

⇒ 34d = 576 - 289 - 49

⇒ 34d = 238

⇒ d = \(\frac{238}{34}\)

⇒ d = 7,

So,

\( {OS }^{2 } \) = \( {7 }^{2 } \) + \( {24 }^{2 } \)

⇒ \( {OS }^{2 } \) = 49 + 576

⇒ \( {OS }^{2 } \) = 625

⇒ OS = 25

Therefore, radius of the circle is 25 cm.