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, $$PA=PR$$. 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, $$PA=PR$$.
$$\Rightarrow \mathrm{\angle }PAR=\mathrm{\angle }PRA$$
Since, By the diagram, $$\mathrm{\angle }ARQ>\mathrm{\angle }PRA$$
$$\Rightarrow \mathrm{\angle }ARQ>\mathrm{\angle }PAR$$ (As given $$PA=PR$$)
$$\Rightarrow QA>PQ$$
(Because the side opposite to larger angle is larger and the side opposite to smaller angle is smaller)
$$\begin{gathered} \Rightarrow QP+PA>QR \\ \Rightarrow QP+PR>QR. \end{gathered}$$

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


Note: Similarly, we can prove, $$QP+QR>PRorPR+QR>QP$$. 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.