Question

The sum of any two sides of a triangle is greater than the third side.

Answer 1

Hint: One of the general properties of the triangle is that the sum of the two sides of a triangle should be greater than the third side of the same triangle. In this question, we need to prove this statement.
Here we will first draw a triangle PQR then extend \[QP\] to A, Such that, \[PA{\text{ }} = {\text{ }}PR\]. Then we will compare the angles using the theorem segment opposite to the larger angle will be larger and smaller will be smaller.

Complete step-by -step solution:
First,let us draw a triangle QPR.

Now, extend QP to A, Such that, \[PA{\text{ }} = {\text{ }}PR\].
\[ \Rightarrow \angle PAR{\text{ }} = \angle PRA\]
Since, By the diagram, \[\angle ARQ{\text{ }} > \angle PRA\]
\[ \Rightarrow \angle ARQ{\text{ }} > \angle PAR\] (As given \[PA{\text{ }} = {\text{ }}PR\])
\[ \Rightarrow QA{\text{ }} > {\text{ }}PQ\]
(Because the side opposite to larger angle is larger and the side opposite to smaller angle is smaller)
\[
   \Rightarrow QP{\text{ }} + {\text{ }}PA{\text{ }} > {\text{ }}QR \\
   \Rightarrow QP{\text{ }} + {\text{ }}PR{\text{ }} > {\text{ }}QR. \\
 \]

Hence, proved that the sum of any two sides of a triangle is greater than the third side.


Note: Similarly, we can prove, \[QP{\text{ }} + {\text{ }}QR{\text{ }} > {\text{ }}PR{\text{ }}or{\text{ }}PR{\text{ }} + {\text{ }}QR{\text{ }} > {\text{ }}QP\]. Also, generalize that, the sum of two sides of a triangle is greater than the third side. Thoroughly, get all the different properties of the triangle.