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# From the Information Given in the Figure, Prove that Pm = Pn = $\Sqrt{3}$ × A - Geometry

ConceptSimilarity in Right Angled Triangles

#### Question

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

#### Solution

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,

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

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.

Is there an error in this question or solution?

#### APPEARS IN

Balbharati Solution for Balbharati Class 10 Mathematics 2 Geometry (2018 to Current)
Chapter 2: Pythagoras Theorem
Problem Set 2 | Q: 8 | Page no. 45

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Solution From the Information Given in the Figure, Prove that Pm = Pn = $\Sqrt{3}$ × A Concept: Similarity in Right Angled Triangles.
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