# Maximum Area of a Triangle when 1 Side, Perimeter Known

This is an example of a "quantitative comparison" question the GRE would test.

Suppose the following information is known:

• one side of a triangle has length 12

• the perimeter of the triangle is 40

What is greater, the area of the triangle or 72?

Either (A) the area of the triangle is greater, (B) 72 is greater, (C) the two quantities are equal, or (D) the relationship cannot be determined from the information given.

Here is my strategy, which may be inefficient given time constraints of the GRE.

First, I eliminate (A) because I know I can make the area of the triangle as close to $0$ as desired, simply by making one of the angles of the triangle as close $0$ as desired. For the same reason, I can eliminate option (C).

Now I wish to show that the area of the triangle could possibly be greater than $72$. If I cannot, then choice (B) is correct. This is where I ask your help -- how can one easily find the maximum area of a triangle when only the length of one side and the perimeter is known?

Thank you!

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## 3 Answers

In your case, the area of the triangle can be made to be greater than $72$ if the two remaining sides both have length $14.$ That is, if the triangle is $12-14-14,$ then it is isosceles, so we can find its area pretty easily: Drop an altitude to the $12$-length side, dividing that side into two segments of length $6.$ Then the altitude has length $\sqrt{14^2-6^2} = \sqrt{196-36} = \sqrt{160} = 4\sqrt{10},$ by the Pythagorean theorem, so the area of this triangle is $\dfrac12 (12)(4\sqrt{10}) = 24\sqrt{10},$ which is about $75.89.$

In general, I believe that the area of a triangle given one side and the perimeter is maximized when the two remaining sides have the same length. This should be provable algebraically using Heron's formula.

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Thank you so much! I'm not familiar with Heron's formula. However a quick Google search for "when is area of triangle maximized" returns "right triangle" in the very first result, which is strange. However, another page I found (mathopenref.com/triangleareaperim.html) has a neat interactive demo which leads me to think isosceles maximizes the area. – EthanAlvaree Aug 10 '14 at 22:17

if $AC=CB=14$ then $S_{ABC}=4\sqrt{10}>72.$ But if $Ac=8.1$ then $S<14<72.$

(D) is correct

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You mean $AC=CB=14$, right? What does $S$ denote? Also, is it always true that isosceles triangles maximize the area? – EthanAlvaree Aug 10 '14 at 22:27
@Mathemanic - Oops I fixed it. Yes it is. I will prove it. – nadia-liza Aug 10 '14 at 22:33
@Mathemanic Prove that isosceles triangles maximize the area. Let $AC=x.$ Using Heron's formula we have $S(x)=S_{ABC}^2=p(p-AB)(p-CB)(p-AC)=20*8*(20-x)(8-x).$ since $S'(12)=0,$ function S(x) has maximum point at $x=12.$ – nadia-liza Aug 10 '14 at 22:46
You mean $x=14$, correct? – EthanAlvaree Aug 10 '14 at 22:54
yes, You are right. $x=14.$ – nadia-liza Aug 10 '14 at 23:01

With a fixed base, the area is maximum for the largest height. Also, with fixed base and fixed perimeter, the locus of $C$ is the ellipse of foci $A$ and $B$, hence the apex is on the perpendicular bisector of $AB$.

Then by Pythagoras, $h=\sqrt{\left(\frac{40-12}2\right)^2-\left(\frac{12}2\right)^2}$.

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