Question

fast answer no useless answer

Answer 1

look at attachment for procedure

Answer 2

$\underline{\bold{Given:-}}$

∆PQR is a triangle and S is any point in the interior of a triangle PQR.

$\underline{\bold{To\:prove:-}}$

SQ + SR < PQ + PR

$\underline{\bold{Construction:-}}$

Join PS, QS and RS respectively and extend QS on PR and mark it as A.

So,

QA = QS + SA

PR = PA + AR

$\underline{\bold{Proof:-}}$

In ∆PQA,

[⭐The sum of any two sides of a triangle is always greater than the third side⭐]

PQ + PA > QA

PQ + PA > QS + SA. .........(1)

Similarly, In ∆ SAR

AR + SA > SR. ..........(2)

On adding eq (1) and (2).

PQ + PA + AR + SA > QS + SA + SR

On subtracting SA from both sides

PQ + PA + AR > QS + SR

PQ + PR > SQ + SR

SQ + SR < PQ + PR

Hence proved.