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Question
In ΔPQR, PS ⊥ QR ; prove that: PQ + PR > QR and PQ + QR >2PS.
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Solution
In ΔPQR,
PQ + PR > QR (∵ Sum of two sides of a triangle is always greater than third aside.)
In ΔPQS,
PQ + QS > PS (∵ Sum of two sides of a triangle is always greater than third aside.) .....(i)
In ΔPRS,
PR + SR > PS (∵ Sum of two sides of a triangle is always greater than third aside.) .....(ii)
Adding (i) and (ii),
PQ + QS + PR + SR > 2PS
PQ + (QS + SR) + PR > 2PS
PQ + QR + PR > 2PS
Since PQ + PR > QR
⇒ PQ + QR > 2PS.
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