Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10th
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The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2 - Mathematics

Sum

The perpendicular PS on the base QR of a ∆PQR intersects QR at S, such that QS = 3 SR. Prove that 2PQ2 = 2PR2 + QR2 

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Solution

Given QS = 3SR

QR = QS + SR

= 3SR + SR = 4SR

SR = `1/4` QR  ...(1)

QS = 3SR

SR = `"QS"/3`  ...(2)

From (1) and (2) we get

`1/4  "QR" = "QS"/3`

∴ QS = `3/4` QR  ...(3)

In the right ∆PQS,

PQ2 = PS2 + QS2   ...(4)

Similarly in ∆PSR

PR2 = PS2 + SR2   ...(5)

Subtract (4) and (5)

PQ2 – PR2 = PS2 + QS2 – PS2 – SR2

= QS2 – SR2

PQ2 – PR2 = `[3/4 "QR"]^2 - ["QR"/4]^2`

From (3) and (1)

= `(9"QR"^2)/16 - "QR"^2/16`

= `(8"QR"^2)/16`

PQ2 – PR2 = `1/2 "QR"^2`

2PQ2 – 2PR2 = QR2

2PQ2 = 2PR2 + QR2

Hence the proved.

Concept: Pythagoras Theorem
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APPEARS IN

Samacheer Kalvi Mathematics Class 10 SSLC Tamil Nadu State Board
Chapter 4 Geometry
Exercise 4.3 | Q 7 | Page 188
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