Maharashtra State BoardSSC (English Medium) 9th Standard
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In the Given Figure, Point S is Any Point on Side Qr of δ Pqr Prove that : Pq + Qr + Rp > 2ps - Geometry

Sum

In the given figure, point S is any point on side QR of Δ PQR Prove that : PQ + QR + RP > 2PS

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Solution

In △PQS
PQ + QS > PS                     ...(1)   (Sum of two sides of a traingle is greater than the third side)
In △PRS
RP + RS > PS                     ... (2)   (Sum of two sides of a traingle is greater than the third side)
Adding (1) and (2), we get
PQ + QS + RP + RS  > PS + PS
⇒ PQ + QR + RP > 2PS 

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APPEARS IN

Balbharati Mathematics 2 Geometry 9th Standard Maharashtra State Board
Chapter 3 Triangles
Problem Set 3 | Q 5 | Page 49
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