I am having a question that is it possible to find what will be the maximum area of a triangle if suppose we know any two of the sides and none of the angles? We don't know whether one of these sides is the largest or not.

Can anyone answer considering the two given sides to be m and n?


4 Answers 4


It's easy to answer the question just remembering that the area of a triangle is base $\times$ height divided by $2$. Fix one side as the base, and rotate the other side... when is the height (and thus area) of the triangle maximal? When the other side is at $90$ degrees from the first, in which case you have a right triangle with an area equal to half the product of the two starting sides.

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    $\begingroup$ Correct intuition >>>>> an arbitrary formula (especially since OP stated in a comment that they didn't understand the formula in Emre's answer). $\endgroup$ Jan 14, 2017 at 0:36
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    $\begingroup$ You can actually easily and rigorously prove the intuition (that the height of the triangle is maximal when the angle is 90 degrees) through e.g. Pythagoras' theorem! $\endgroup$
    – Anonymous
    Jan 14, 2017 at 0:52

Well, let $\theta$ be the angle between the sides of length $m$ and $n$. Then, by the are formula $S = \frac12\sin\theta mn$ gives that area is less than or equal to $\frac12mn$

  • $\begingroup$ It is known as SAS formula for triangle's area. You can check out the following link. mathsisfun.com/algebra/… $\endgroup$
    – Emre
    Jan 13, 2017 at 11:37
  • $\begingroup$ @Resorcinol Let $h$ be the height of the triangle. Then, if $n$ is the length of the base side, and $m$ the other known side-length, $\frac{1}{2}hn$ is, of course, the area of the triangle. At this point, we have the setup in Anonymous's answer, at which point intuition (specifically, the idea that diagonally slanted lines are less "tall" than vertical lines of the same length) can get us the rest of the way to our answer. But using the definition of $\sin$, we can also derive the above formula for the height of an arbitrary triangle with the given constraints, since $h=m\sin\theta$. $\endgroup$ Jan 14, 2017 at 0:42

Here is an image to complement the classic answer given in Anonymous's response:

enter image description here

The horizontal black line segment is the initial fixed side.

The length of the second side is the radius of the red circle. All possible triangles with sides of these respective lengths can be generated by these radii; three examples are given in the picture: one in blue, one in green, and one in black. In each case, the base $\times$ height formula has the same base, and one sees that the height is greatest when the two line segments are perpendicular to one another.


If you draw a triangle and label sides a b and c, then you can apply the Law of Cosines. Given any 2 sides a and b, angle C (or gamma) is necessarily between a and b. By applying this, you can solve for side c.

Once you have sides a b and c, you can apply Heron's Formula to find the area of the triangle. You may also be able to use geometry to find the base and height of the triangle.

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    $\begingroup$ OK, so now we know that the area of the triangle is $$\sqrt{\tfrac12(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(\tfrac12(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})-a)(\tfrac12(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})-b)(\tfrac12(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})-\sqrt{a^2+b^2-2ab\cos\gamma})}$$ and all we need to do is maximize that expression with respect to $\gamma$. Good luck! $\endgroup$ Jan 13, 2017 at 13:35
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    $\begingroup$ @DavidRicherby huge applause for diligent formula parsing in that comment. $\endgroup$
    – Joffan
    Jan 13, 2017 at 22:04
  • $\begingroup$ Heron's Formula is nice, but rarely is it convenient or useful to do in a given amount of time. Usually, you can set up a fairly simple expression with some simple geometry/trig. $\endgroup$ Jan 14, 2017 at 4:23
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    $\begingroup$ @DavidRicherby actually it simplifies to $\frac12ab\sin\gamma$ with simple algebra and Pythagorean trigonometric identity. $\endgroup$
    – Ruslan
    Jan 14, 2017 at 9:02
  • $\begingroup$ @Ruslan Sure, but that's not terribly easy to see by looking at it, and you might make a stupid mistake while attempting to simplify the formula. Per my comment on Emre's answer, deriving that formula directly from the basic formula for the area of a triangle and the definition of $\sin$ is essentially trivial. $\endgroup$ Jan 17, 2017 at 1:39

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