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In the Given Figure, ∆Pqr is an Equilateral Triangle. Point S is on Seg Qr Such Thatn Qs =N 1 3 Qr. Prove that : 9 Ps2 = 7 Pq2 - Geometry

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In the given figure, ∆PQR is an equilateral triangle. Point S is on seg QR such thatn QS =n\[\frac{1}{3}\] QR.

Prove that : 9 PS= 7 PQ2

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Solution

Let the side of equilateral triangle ∆PQR be x.
PT be the altitude of the ∆PQR.

We know that, in equilateral triangle, altitude divides the base in two equal parts.
∴ QT = TR = \[\frac{1}{2}QR = \frac{x}{2}\]

Given: QS =

\[\frac{1}{3}\] QR = \[\frac{x}{3}\]
\[\therefore ST = QT - QS = \frac{x}{2} - \frac{x}{3} = \frac{x}{6}\]
According to Pythagoras theorem,
In ∆PQT
\[{PQ}^2 = {QT}^2 + {PT}^2 \]
\[ \Rightarrow \left( x \right)^2 = \left( \frac{x}{2} \right)^2 + {PT}^2 \]
\[ \Rightarrow x^2 = \frac{x^2}{4} + {PT}^2 \]
\[ \Rightarrow {PT}^2 = x^2 - \frac{x^2}{4}\]
\[ \Rightarrow {PT}^2 = \frac{3 x^2}{4}\]
\[ \Rightarrow PT = \frac{\sqrt{3}x}{2}\]
In ∆PST
\[{PS}^2 = {ST}^2 + {PT}^2 \]
\[ \Rightarrow {PS}^2 = \left( \frac{x}{6} \right)^2 + \left( \frac{\sqrt{3}x}{2} \right)^2 \]
\[ \Rightarrow {PS}^2 = \frac{x^2}{36} + \frac{3 x^2}{4}\]
\[ \Rightarrow {PS}^2 = \frac{x^2 + 27 x^2}{36}\]
\[ \Rightarrow {PS}^2 = \frac{28 x^2}{36}\]
\[ \Rightarrow {PS}^2 = \frac{7 x^2}{9}\]
\[ \Rightarrow 9 {PS}^2 = 7 {PQ}^2\]

Hence, 9 PS= 7 PQ2.

Concept: Property of 30°- 60°- 90° Triangle Theorem
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APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Problem Set 2 | Q 16 | Page 44
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