Question

Prove that in any triangle the length of a side is less than the half of the perimeter. In statement reason please.

Answer 1

Let $a,b,c$ be the sides of a triangle and $p$ its perimeter.

For any triangle, the below is true.

$a,b,c>0 \wedge a<b+c \wedge b<a+c \wedge <a+b$

$p=a+b+c$

So

$\dfrac{p}{2}=\dfrac{a+b+c}{2}$

Now, let's write the inequalities for the thesis and prove they're true.

$a<\dfrac{a+b+c}{2} \wedge b<\dfrac{a+b+c}{2} \wedge c<\dfrac{a+b+c}{2}\\\\2a<a+b+c \wedge 2b<a+b+c \wedge 2c<a+b+c\\\\a<b+c \wedge b<a+c \wedge c<a+b$

As a result, I got the exact same inequalities that are true for any triangle. Q.E.D.

Answer 2

In a triangle, the length of a side is less than the half of the perimeter.

Let's assume that the sides of a triangle are 8cm, 9cm and 10cm. The perimeter of a triangle is gotten by the addition of all its sides. Therefore, the perimeter will be:

= 8cm + 9cm + 13cm

= 30cm

Therefore, half of the perimeter is 30/2 = 15cm

Since the values of the sides are 8cm, 9cm and 10cm, then they're lesser than the half of the perimeter.

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