S is any point on side QR of a triangle PQR. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(N/A) Given: $A$ point $S$ on side $QR$ of $	riangle PQR$.To prove: $PQ + QR + RP > 2 \, PS$.Proof: In $	riangle PQS$,we have:$PQ + QS > PS$ ..... $

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Given: $A$ point $S$ on side $QR$ of $\triangle PQR$.
To prove: $PQ + QR + RP > 2 \, PS$.
Proof: In $\triangle PQS$,we have:
$PQ + QS > PS$ ..... $(1)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Now,in $\triangle PSR$,we have:
$RS + RP > PS$ ..... $(2)$
[Since the sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding $(1)$ and $(2)$,we get:
$PQ + QS + RS + RP > 2 \, PS$
Since $QS + RS = QR$,we have:
$PQ + QR + RP > 2 \, PS$
Hence,proved.

$S$ is any point on side $QR$ of a $\triangle PQR$. Show that: $PQ + QR + RP > 2 \, PS$.

Similar Questions

"If two sides and an angle of one triangle are equal to two sides and an angle of another triangle,then the two triangles must be congruent." Is the statement true? Why?

Two lines $l$ and $m$ intersect at the point $O$ and $P$ is a point on a line $n$ passing through the point $O$ such that $P$ is equidistant from $l$ and $m$. Prove that $n$ is the bisector of the angle formed by $l$ and $m$.

$AD$ is a median of the triangle $ABC$. Is it true that $AB + BC + CA > 2AD$? Give reason for your answer.

In the given figure,$XP = XS$,$XQ = XR$ and $\angle PXR = \angle SXQ$. Prove that $PQ = SR$.

$ABCD$ is a quadrilateral such that $AB = AD$ and $CB = CD$. Prove that $AC$ is the perpendicular bisector of $BD$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module