## Can we Prove that ……..

**The length of any side of a triangle is not more than half of its perimeter**

**Key Concepts**

**Triangle Inequalit**y

**Perimeter**

**Geometry**

## Check the Answer

But try the problem first…

*Answer: Yes we can definitely prove that by Triangle Inequality*

**Mathematical Circles – Chapter 6 – Inequalities Problem 3**

* Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg*

## Try with Hints

First Hint

**We can start this sum by using this picture below**

*The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem *

** b + c > a **

*So can you try to do the rest of the sum ????????*

Second Hint

**According to the question we have to find the perimeter at first**

**Perimeter is the sum of the length of all sides of the triangle = a + b + c **

**And the length of each side is a or b or c.**

**We have to prove : a + b + c > length of any one side**

**This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..**

Final Step

**Here is the rest of the sum** :

**As stated above if we use triangle inequality :**

**b + c > a**

**Lets add a to both the sides **

**a + b + c > a + a**

**a + b + c > 2 a **

**The left hand side of the above inequality is the perimeter of this triangle.**

**perimeter > 2 a **

**So , \(\frac {perimeter}{2} > a \)**

**\(\frac {perimeter}{2} \) = semi perimeter**

*Hence this is proved that the length of one side of a triangle is less than half of its perimeter.*

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