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

Question

$12$, $13$, $15$ 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).

Answer 1

Misread the question.

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 $$\frac{12 * 14}{2}= 84$$

Answer 2

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.

**** read the Last comment i fix me answer sorry

Answer 3

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 =$0.5\times 14\times 12=84$

Answer 4

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.

Answer 5

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.