Question

Prove that PQ+QR+RPis grater than 2PS​

Answer 1

Answer:

S is any point on side QR of a ΔPQR. Show that PQ + QR + RP > 2 PS.

Thinking Process

Use the inequality of a triangle i.e., sum of two sides of a triangle is greater than the third side. Further, show the required result

Step-by-step explanation:

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Answer 2

Answer:

Theorem: In a triangle , sum of the length of any two sides is greater than the third side.

Draw the triangle and join the points P and S

In the figure,

In ΔPQS, according to the theorem,

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

In ΔPSR, according to the theorem,

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

Adding (1) and (2),

PQ+QS+SR+PR>PS+PS

PQ+(QS+SR)+PR>2PS

PQ+QR+PR>2PS

Step-by-step explanation:

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