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)
Answer & explanation
Correct answer: option 2
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