# Area of an acute triangle, given two sides and an altitude

$12$, $13$, $15$ are the lengths (perhaps not in order) of two sides of an acute-angled triangle and of the height over the third side of triangle. Find the area of the triangle (no calculator allowed).

• What formulas do you know/have for the area of a triangle? Mar 12, 2017 at 23:55

No need to use Heron's formula, the key here is to figure out which lengths are for sides, and which for height.

So 15, 13 are sides, and 12 is the height. This is because from one point, the perpendicular distance (height length) to a line is the shortest.

Let's calculate the length of the third side.

Length of the third side = $\sqrt{13^2-12^2} + \sqrt{15^2-12^2}=14$

(Note that for above calculation for third side length, we used the fact that the triangle is acute-angled. If not, there could be another solution as $\sqrt{15^2-12^2} - \sqrt{13^2-12^2}=4$)

Thus the area =$0.5\times 14\times 12=84$

• To answer a retracted comment: Why 12 is the height? - Because height perpendicular to the third side. From one point, the perpendicular distant to a line (third side) is the shortest. Thus it has to be that 12 is the height. Mar 13, 2017 at 0:13
• Absolutely right. Good solution. (+1)
– lulu
Mar 13, 2017 at 0:15

Not an answer, but interesting stuff too long for a comment.

The 13-14-15 triangle was famous when I was on the math team in high school in New York in the early fifties. It has an integral altitude (12) to side 14.

To see why, build it by gluing together the length 12 sides of the 9-12-15 and 5-12-13 right triangles.

This construction generalizes in the obvious way.

High school competitions then were a lot easier than they are now.

• That is an interesting result.
– Mick
Mar 13, 2017 at 10:51

In other words you have two sides and the altitude from the corner where they meet. In that situation the altitude must be shorter than either of the sides, so $12$ is the altitude. (The altitude lies in the interior of the triangle because the triangle is acute).

The altitude now splits your triangle in two right triangles. You can compute the third side of each of them by Pythagoras (both of them turn out to be integers), and then it is easy to find their area and add together.

I get an area of $84$ this way.

• None of the answers I see are very explicit about the hypothesis that the triangle is acute. If we assume instead an obtuse triangle, $12$ is still the altitude, but now the last side becomes $9-5$ instead of $9+5$. So without knowing about acute or obtuse, the area is one of $\frac12 \cdot 12(9 \pm 5)$. Mar 13, 2017 at 10:11

Draw the triangle and the height. Denote by $x,y$ the two segments determined by the height on the triangle.

First, since the perpendicular to a line is smaller than any of the secants, the altitude must be 12.

You have $$12^2+x^2=13^2 \Rightarrow x^2=13^2-12^2= (13-12)(13+12)=25 \\ 12^2+y^2=15^2 \Rightarrow y^2=15^2-12^2= (15-12)(15+12)=3*=81$$

Therefore the last side is $\sqrt{25}+\sqrt{81}=14$

Then the area is $$\frac{12 * 14}{2}= 84$$

• The answer is 6 x √65. I do get Area= 74.8 square units, which is similar to 20√14. Not sure how it arrived at this solution
– Vyas
Mar 12, 2017 at 23:59
• @Vyas Fixed it. The answer $6 \sqrt{65}$ is wrong, what you have is a $5, 12, 13$ and a $9, 12, 15$ right triangle, which are easy to remember, glued along the $12$ side. Mar 13, 2017 at 0:09

I know a "fórmula" to calculate it called heron theorem or formula

Sides are a,b, c, and S is equal to (a+b +c)/2 Thus The área of any triangle is: A=sqrt(S*(S-a)(S-b)(S-c)) Then now just replace the figures.

• The problem, as stated, does not give the three sides but two sides and the height over the third side. That altitude divides the triangle into two right triangles. If the two sides have length a and b and the height is c, we have one right triangle with hypotenuse of length a and one leg c and another with hypotenuse b and leg c. The two parts of the third side are $\sqrt{a^2- c^2}$ and $\sqrt{b^2- c^2}$. Add those to get the length of the base and use "one half height times base". Mar 13, 2017 at 0:11