Question

In triangle PQR, PS is the median of the triangle, prove that PQ + QR + RS > 2 PS

Answer 1

Concept to be used :

• Inequality in a triangle : The sum of any two sides of a triangle is always greater than the third side .
• Median of a triangle : It is the line segment that joins a vertex of a triangle to the mid point of the opposite side .
• A triangle has exactly three medians .
• The point of intersection of the medians of a triangle is called its centroid .

Given :

• PS is a median of ∆PQR

To prove :

• PQ + QR + RP > 2PS

Proof :

[Please refer to the attachment for figure]

Clearly ,

The median PS has divided the ∆PQR into two triangles , namely ∆PQS and ∆PRS .

Now ,

In ∆PQS ,

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

Also ,

In ∆PRS ,

PR + RS > PS --------(2)

Now ,

Adding inequations (1) and (2) , we get ;

=> PQ + QS + PR + RS > PS + PS

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

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

=> PQ + QR + RP > 2PS

Hence proved .

Answer 2

Answer:

Step-by-step explanation: