Question

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Answer 1

Answer:

Consider the longer of the two sides PQ and PR. Let's say that it is PQ (though it doesn't matter, and you can pick either if they are equal). 

PS must be shorter than or equal to PQ. (It's only equal if S and Q are the same point.) 

It is a rule that the sum of any two sides of a triangle must be greater than the third side. So we also know that PR + QR > PQ 

Since PR + QR > PQ 

We know PQ + PR + QR > 2PQ 

(just added PQ to both sides) 

And since PQ >= PS, 

We know 2PQ >= 2PS 

(multipled both sides by 2) 

From those two (PQ + PR + QR > 2PQ, 2PQ >= 2PS), we know: 

PQ + PR + QR > 2PS

Step-by-step explanation:

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Answer 2

SOLUTION:-

Given:

In ∆PQR, S is any point on side QR.

To prove:

PQ + QR + RP>2PS

Proof:

In ∆PQS,

PQ + QS >PS..........(1)

[sum of two sides of a ∆ is greater than the third side]

&

In ∆PRS,

SR + RP >PS............(2)

[Sum of two sides of a ∆ is greater than the third side]

Adding both equation (1) & (2), we get;

=) PQ + QS +SR +RP >2PS

=) PQ+(QS+SR)+ RP> 2PS

=) PQ + QR + RP > 2PS [proved]

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