Question

Prove that the sum of two sides of triangle is always greater than third side. ​

Answer 1

Answer:

${hypotenuse}^{2} = {perpendicular}^{2} +{base}^{2}$

Pythagoras Theorem

Answer 2

Answer:

The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side.

Theorem : The sum of two sides of a triangle is greater than the third side

Let PQR is a triangle.

We have to prove that : QP + PR > QR

Proof:

Extend QP to A,

Such that, PA = PR

⇒ ∠ PAR = ∠ PRA

Since, By the diagram,

∠ ARQ > ∠ PRA

⇒ ∠ ARQ > ∠ PAR

⇒ QA > PQ ( Because, the sides opposite to larger angle is larger and the sides opposite to smaller angle is smaller )

⇒ QP + PA > QR

⇒ QP + PR > QR.

Hence, prove,

Note: similarly we can proved, QP + QR > PR or PR + QR > QP

Thus, The sum of two sides of a triangle is greater than the third side

hope this will help you