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

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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.

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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
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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