In the figure, STQ = SRP, PT = TQ = 6cm and QS = 5cm. Find SR
 

  • A \(\frac{47}{5}\)
  • B 5
  • C \(\frac{32}{5}\)
  • D \(\frac{22}{5}\)
jamb 1988mathematics

The correct answer is A. \(\frac{47}{5}\)

From similar triangle, \(\frac{QS}{QP} = \frac{TQ}{QR} = \frac{5}{12} = \frac{6}{QR}\)

QT = \(\frac{6 \times 12}{5} = \frac{72}{5} = SR = QR - QS\)

= \(\frac{72}{5} - 5 = \frac{72 - 25}{5}\)

= \(\frac{47}{5}\)

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