The radii of two concentric circles are x cm and 26 cm. P and S are the points on larger circle and Q and R are points on smaller circle. If PQRS is a straight line and QR = 40 cm and PS = 48 cm. then what is the value of x? (x < 26 cm)

$10\sqrt{5}$ cm

Construct a perpendicular OT form the center to the straight lines PQRS.

In a bigger circle, PS is the chord with a length of 48 cm.

OT is perpendicular to PS

So, PT = ST = 24 cm

In \(\Delta \)OTP,

OP = 26 cm   (Radius)

PT = 24 cm

Using Pythagoras' theorem

\( {OT }^{2 } \) = \( {OP }^{2 } \) + \( {PT }^{2 } \)

=  \( {OT }^{2 } \) = \( {26}^{2 } \) + \( {24 }^{2 } \)

=  \( {OT }^{2 } \) = 676 - 576

=  \( {OT }^{2 } \) = 100

= OT = 10 cm

In a smaller circle, QR is the chord with a length of 40 cm.

OT is perpendicular to QR

So, QT = RT = 20 cm

In \(\Delta \)OTQ,

OQ = x cm  (Radius)

QT = 20 cm

OT = 10 cm

Using Pythagoras' theorem

\( {OQ }^{2 } \) = \( {OT }^{2 } \) + \( {QT }^{2 } \)

= \( {OT }^{2 } \) = \( {20 }^{2 } \) + \( {10 }^{2 } \)

= \( {OT }^{2 } \) = 400 + 100

= \( {OT }^{2 } \) = 500

= OQ = 10\(\sqrt {5 }\) cm

Therefore, OQ is 10\(\sqrt {5 }\) cm