Question

prove that any two sides of a triangle are together greater than the the third side​

Answer 1

Answer:

hi mate

Here we will prove that the sum of any two sides of a triangle is greater than the third side.

Given: XYZ is a triangle.

Inequalities in Triangles

To Prove: (XY + XZ) > YZ, (YZ + XZ) > XY and (XY + YZ) > XZ

Construction: Produce YX to P such that XP = XZ. Join P and Z.

The Sum of any Two Sides of a Triangle is Greater than the Third Side

0Save

Statement

1. ∠XZP = ∠XPZ.

2. ∠YZP > ∠XZP.

3. Therefore, ∠YZP > ∠XPZ.

4. ∠YZP > ∠YPZ.

5. In ∆YZP, YP > YZ.

6. (YX + XP) > YZ.

7. (YX + XZ) > YZ. (Proved)

Reason

1. XP = XZ.

2. ∠YZP = ∠YZX + ∠XZP.

3. From 1 and 2.

4. From 3.

5. Greater angle has greater side opposite to it.

6. YP = YX + XP

7. XP = XZ

Similarly, it can be shown that (YZ + XZ) >XY and (XY + YZ) > XZ.

Answer 2

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\huge\bold{\mathtt{QUESTION}}$

➡ Prove that any two sides of a triangle are together greater than the the third side .

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\huge\bold{\mathtt{ANSWER}}$

☯ Theorem states that -

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

↪Let PQR is a triangle.

☯ To prove : QP + PR > QR

☯ Proof :

Extend QP to A,

Such that, PA = PR

⇒ ∠ PAR = ∠ PRA

↪Now , from 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.

$\large\bold{\mathrm{\red{Hence\:Proved}}}$

★ 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 .

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬