PQ and RS are two parallel chords of a circle such that PQ is 48 cm and RS is 40 cm. If the chords are on the opposite sides of the centre and the distance between them is 22 cm, what is the radius (in cm) of the circle?

25

According to the question,

PM = \(\frac{PQ}{2}\)

RN = \(\frac{RS}{2}\)

Now, in \(\Delta \)PMO, we have,

\( {PM }^{2 } \) + \( {OM }^{2 } \) = \( { r}^{2 } \)

= \( {24 }^{2 } \) + \( {x }^{2 } \) = \( { r}^{2 } \)    ..(1)

And in \(\Delta \)RNO, we have

\( {20 }^{2 } \) + \( {22\; - \; x }^{2 } \) = \( { r}^{2 } \)   ..(2)

From equation 1 and 2,

\( {24 }^{2 } \) + \( {x }^{2 } \) = \( {20 }^{2 } \) + \( {22\; - \; x }^{2 } \)

= \( {24 }^{2 } \) - \( {20 }^{2 } \) = \( {22\; - \; x }^{2 } \) - \( {x }^{2 } \)

= (24 - 20)(24 + 20) = (22 - x + x)(22 - x - x)

= 4 x 44 = 22(22 - 2x)

= 8 = 2(11 - x)

= 4 = 11 -x

= x = 7

Now putting x = 7 in equation 1,

r = √(\( {24 }^{2 } \) + \( {7 }^{2 } \)) = 25

Therefore, the length of radius is 25 cm.