Question

S is any point on side QR of triangle PQR . show that PQ + QR + PR > 2PS ​

Answer 1

$\Large{ \underline{\underline{ \bf{To \:Prove:}}}}$

S is any point on side QR of triangle PQR . show that PQ + QR + PR > 2PS.

$\Large{ \underline{\underline{ \bf{Proof:}}}}$

In ∆PQS,

By using inequality of Triangles, The sum of any two sides of the ∆ is greater than the third side:

➛ PQ + QS > PS ----------(1)

And In ∆PRS,

Again by the same inequality,

➛ PR + SR > PS ----------(2)

Adding (1) and (2),

➛ PQ + QS + PR + SR > PS + PS

➛ PQ + (QS + SR) + PR > 2PS

➛ PQ + QR + PR > 2PS

Hence, Proved !!

Note that!!

• We have used the triangular inequality which states that sum of any two sides of a ∆ is greater than the 3rd side.

• There is one more similar inequality which says that the Difference of any two sides is always lesser than the 3rd side in a ∆.

Answer 2

$\underline{\underline{\sf{\clubsuit \:\:Question}}}$

• S is any point on side QR of triangle PQR .Show that PQ + QR + PR > 2PS ​

$\underline{\underline{\sf{\clubsuit \:\:Given}}}$

• S is any point on side QR of triangle PQR

$\underline{\underline{\sf{\clubsuit \:\:To \:Show}}}$

• PQ + QR + PR > 2PS

$\underline{\underline{\sf{\clubsuit \:\:Proof}}}$

In △ PQR :

PQ + QS > PS  $\sf{.......(1)}$

[Sum of any two sides of a triangle is greater than the third Side]

In △ PQR :

SR + RP > PS  $\sf{.......(2)}$

[Sum of any two sides of a triangle is greater than the third Side]

On adding (1) and (2) :

PQ + QS + SR + PR > PS + PS

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

⇒ PQ + QR + PR > 2PS

$\bf{Hence\:\:Proved\:!!!}$