S is any point on triangle pqr show that pq+pr>2ps
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Solution of this question :-
• GIVEN:- In Δ PQR, S is any point on the side QR.
• TO SHOW:- PQ + PR + QR > 2ps
• PROVE:- In ΔPQS
Sum of any two sides is greater than the third side.
So, PQ + QS > PS ......(1)
In ΔPSR
Sum of any two sides is greater than the third side.
So, PR + SR > PS ...(2)
Adding equations (1) and (2)
PQ + PR + QS + SR > PS + PS
PQ + PR + QR > 2PS [ QS + SR = QR]
I hope this will help you please mark my answer as brainliest answer.
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