If triangle A has a smaller perimeter than triangle B, is the area of A also always smaller than that of B?

Question

If triangle A has a smaller perimeter than triangle B, is the area of A also always smaller than that of B?

The specific problem that got me interested in this question is that you are given a triangle whose sides are respectively less than 2, 3 and 4. You are asked to find the maximum area possible with these constraints.

Answer 1

Hint : Use Heron's Formula :

$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \space\space\space\space\space\space\space \text{where } s = \frac{a+b+c}{2}$$

Answer 2

To answer the question in your title: no. You don't need to calculate it, but imagine a triangle whose side-lengths are $1,1,1$. A triangle with side-lengths $5,5,10$ has a much larger perimeter, but an area of $0$.

(If you don't like degenerate triangles for some reason, a triangle with side-lengths $5,5+\epsilon,10$ for a really small value of $\epsilon$ has an area as small as you want to make it.)

Answer 3

Not necessarily, consider triangles with sides $2, 2, 2$ and $3, 3, 1$. The former triangle has area of $\sqrt 3$, the latter has $\sqrt 2$

Answer 4

The smallest pair of positive Pythagorean triples which provide a counter example are:

• $12,16,20$ with perimeter $48$ and area $96$
• $7,24,25$ with perimeter $56$ and area $84$