Question

Prove that perimeter of the triangle is greater than the sum of it's three altitudes.Please provide the diagram with the answer....

Answer 1

The sides are AB, BC and CA. The altitudes are AD, BE and CF are the altitudes  

Being the sides of triangles as altitudes then sides of triangles have to be the hypotenuse.  

It is so clear that hypotenuse is always greater than any one of the other sides  

So AB>AD and BC>BE and CA>CF  

Now adding AB+BC+CA>AD+BE+CF  

Hence proved

Answer 2

Answer:

Let there be a triangle ABC with its altitudes D, E and F from vertices A, B and C respectively.

The altitudes form a right angled triangle at their corresponding bases. Also in a right angled triangle the hypotenuse is the longest side. Taking the right angles formed by the altitudes and the sides as the hypotenuse, we observe in each triangle, the side forms the longest side, i.e.

In triangle ABD, AB is the longest side.

In triangle ACF, AC is the longest side.

In triangle CBE, BC is the longest side.

So, adding all these three sides, we find that the perimeter of a triangle is greater than the sum of its three altitudes.

Hence proved.

Step-by-step explanation: