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,

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

In ∆PMS,

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

In ∆PNS,

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

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 = ${\displaystyle \sqrt{3}\times a}$   ....(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 = ${\displaystyle \sqrt{3}\times a}$   ....(ii)

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

∴ PM = PN =$$\sqrt{3}$$× a.