Question

only for genius........

Answer 1

In  triangle STR,

ST + TR > SR  …….(i)  (Sum of two sides of a triangle is greater than the third side)

In triangle PQT,

PQ + PT > QT  …..(ii)  (Sum of two sides of a triangle is greater than the third side)

Adding (i) and (ii), we get

ST + TR  + PQ + PT > SR + QT

PQ + PR + ST > SR + QS + ST  (PT + TR = PR  and QT = QS + ST)

PQ + PR > SQ + SR

Hence proved.

Answer 2

Hello here is your answer by Sujeet yaduvanshi ☝☝☝☝☝☝☝☝☝☝


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Question:-)
GIVEN:-)

S is any point in it's interior of ∆PQR


To prove:-)..SQ+SR<PQ+PR

Construction:- To produce QS length to meeting PR Length In Angle' T'.

NOW.....


PROOF:-)
In Given ∆PQT

so...

PQ+PT>QT

so..

The sum of length of any two sides of a ∆ is greater then the Given two sides.....



Now....

PA+PT>QS+ST

so..

QT=QS+ST


Now...

In ∆SRT..

TR+ST>SR


The sum of length of any two sides of a ∆ is greater then the Given two sides.....


From Equation...
1 And 2

PQ+PT+TR+ST>QS+ST+SR

PQ+PR+ST>QS+ST+SR


PQ+PR>SQ+SR

SQ+SR<PQ+PR


that's

proved
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Sujeet Kumar Yaduvanshi