Question

Choose the correct answer from the alternative given.
PQ and RS are two parallel chords of a circle. If PQ = 30 cm, RS = 16 cm and distance between PQ and RS is 23 cm, then the radius of the circle.

• A. $12$
• B. $17$
• C. $34$
• D. $19$

Answer 1

Answer: (B)

$17$

See the figure. from centre O of the circle perpendicular OL is drawn to chord PQ and perpendicular OM is drawn to chord RS.

PL = $\frac{PQ}{2}=\frac{30}{2}$= 15cm

and RM = $\frac{RS}{2}=\frac{16}{2}$= 8cm

Let OL = x cm then, OM = (23 x) cm

In $\Delta OLP,O{P}^2=P{L}^2+L{O}^2$

or, $r^{=}{15}^2$ + $x^2$ (1) Again in $\Delta ,O{R}^2=O{M}^2+R{M}^2$

or $r^2$=$(23x{)}^2+8^2$ (2) Equating equation (1) and (2). ${15}^2+x^2(23-x{)}^2+8^2$

225 + $x^2={23}^2$ $46x+x^2+64$

$225=52946x+64=225=59346x$

$x=\frac{368}{46}$ = 8

Thus from(1).$r^2={15}^2+8^2$= 225+64 = 289

r = $\sqrt{289}$ =17cm

Radius = 17 cm

Hence, radius of the circle is 17 cm.