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True...well, it doesn't HAVE to have a larger perimeter (obviously it could).

The best you can do with regard to maximizing area versus perimeter in a triangle is an equilateral triangle.

Area = 1/2*b*h for all triangles.

Now, imagine that the base of the equilateral triangle is stationary and that you just want to move the top vertex over a bit (KEEPING THE HEIGHT THE SAME).

Well, clearly the area will remain constant, since the base and height are the same, but what will happen to the perimeter? It'll increase!

Hmm...not convinced yet I know...let's compare the perimeter for a given area by pitting an equilateral triangle versus a right triangle.

Base = 2
Height = sqrt(3)

Area = sqrt(3)

The base is the same as the length of every side, so we know the total perimeter is 6.

OK, so onto the right triangle...

Now the top vertex has been moved all the way to the right or left, so that it's vertically parallel with either the bottom left or bottom right vertex. It doesn't really matter, but what does matter is that we have a right triangle.

One leg is the height and the other is the base, so we have the following:

c^2 = a^2 + b^2
c^2 = 3+4 = 7
c = sqrt(7)

Perimeter, therefore, is 2+sqrt(3)+sqrt(7) ~= 6.3

So, as we can see, despite having equal areas, one triangle has a greater perimeter than another.

Now that we've shown it mathematically, keep the concept in mind: polygons with equal sides will always give you the most area for a given perimeter. This means that you can have multiple perimeter values for a given area and vice versa. Knowing this alone will save you a lot of math in the future.