From the information given in the figure 2.31, prove that \(PM = PN = \sqrt{3} \times a\)
Figure: Points M, Q, S, R, N lie on a line in that order. MQ = QR = RN = a, PQ = PR = a, and PS ⊥ QR with S on seg QR.
Answer: \(\triangle PQR\) is equilateral with side a, so PS = \(\frac{\sqrt{3}}{2}a\) and S is the midpoint of QR; then MS = NS = \(\frac{3a}{2}\) and Pythagoras theorem gives PM = PN = \(\sqrt{3}a\).
PQ = PR = QR = a, so \(\triangle PQR\) is equilateral. The perpendicular PS bisects QR, so QS = SR = \(\frac{a}{2}\), and
\[PS = \frac{\sqrt{3}}{2} a\]MS = MQ + QS = \(a + \frac{a}{2} = \frac{3a}{2}\). In right-angled \(\triangle PSM\),
\[PM^2 = PS^2 + MS^2\] \[= \frac{3a^2}{4} + \frac{9a^2}{4} = 3a^2\] \[PM = \sqrt{3}\,a\]Similarly SN = SR + RN = \(\frac{3a}{2}\), so in right-angled \(\triangle PSN\), \(PN^2 = 3a^2\) and \(PN = \sqrt{3}\,a\).
\(\therefore PM = PN = \sqrt{3} \times a\). Hence proved.
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