Question

From the information given in the figure, prove that PM = PN =  \[\sqrt{3}\]  × a

Answer 1

Since ∆PQR is an equilateral triangle, PS is the perpendicular bisector of QR.
∴ QS = SR = \[\frac{a}{2}\]       ...(1)

Now, According to Pythagoras theorem,
In ∆PQS,

\[{PQ}^2 = {QS}^2 + {PS}^2 \]
\[ \Rightarrow a^2 = \left( \frac{a}{2} \right)^2 + {PS}^2 \]
\[ \Rightarrow {PS}^2 = a^2 - \frac{a^2}{4}\]
\[ \Rightarrow {PS}^2 = \frac{4 a^2 - a^2}{4}\]
\[ \Rightarrow {PS}^2 = \frac{3 a^2}{4}\]
\[ \Rightarrow PS = \frac{\sqrt{3}a}{2} . . . \left( 2 \right)\]

In ∆PMS,

\[{PM}^2 = {MS}^2 + {PS}^2 \]
\[ \Rightarrow {PM}^2 = \left( a + \frac{a}{2} \right)^2 + \left( \frac{\sqrt{3}}{2}a \right)^2 \]
\[ \Rightarrow {PM}^2 = \left( \frac{3a}{2} \right)^2 + \left( \frac{\sqrt{3}}{2}a \right)^2 \]
\[ \Rightarrow {PM}^2 = \frac{9 a^2}{4} + \frac{3 a^2}{4}\]
\[ \Rightarrow {PM}^2 = \frac{12 a^2}{4}\]
\[ \Rightarrow {PM}^2 = 3 a^2 \]   ...Taking square root
\[ \Rightarrow PM = \sqrt{3}a . . . \left( 3 \right)\]

In ∆PNS,

\[{PN}^2 = {NS}^2 + {PS}^2 \]
\[ \Rightarrow {PN}^2 = \left( a + \frac{a}{2} \right)^2 + \left( \frac{\sqrt{3}}{2}a \right)^2 \]
\[ \Rightarrow {PN}^2 = \left( \frac{3a}{2} \right)^2 + \left( \frac{\sqrt{3}}{2}a \right)^2 \]
\[ \Rightarrow {PN}^2 = \frac{9 a^2}{4} + \frac{3 a^2}{4}\]
\[ \Rightarrow {PN}^2 = \frac{12 a^2}{4}\]
\[ \Rightarrow {PN}^2 = 3 a^2 \]
\[ \Rightarrow PN = \sqrt{3}a . . . \left( 4 \right)\]

From (3) and (4), we get
PM = PN =\[\sqrt{3}\] × a

Hence, PM = PN =\[\sqrt{3}\]× a.

Answer 2

From figure,

In  ∆PMR

MQ = QR = a   ...(given)

∴ Q is a midpoint of MR.

∴ seg PQ is the median.

∴ According to Apollonius's theorem,

PM² + PR² = 2PQ² + 2MQ²

∴ PM² + a² = 2a² + 2a²

∴ PM² + a² = 4a²

∴ PM² = 4a² − a²

∴ PM² = 3a²    ...Taking square root

PM = `sqrt3 xx a`   ....(i)

From figure,

In  ∆PNQ

NR = RQ = a   ...(given)

∴ R is a midpoint of NQ.

∴ seg PR is the median.

∴ According to Apollonius's theorem,

PN² + PQ² = 2PR² + 2NR²

∴ PN² + a² = 2a² + 2a²

∴ PN² + a² = 4a²

∴ PN² = 4a² − a²

∴ PN = 3a²    ...Taking square root

PN = `sqrt3 xx a`   ....(ii)

From (3) and (4), we get
PM = PN =\[\sqrt{3}\] × a

∴ PM = PN =\[\sqrt{3}\]× a.