# Finding the area of a triangle with sides $25$, $39$, $56$, using right triangles

I am stuck on this problem for the past few days and I can't completely figure it out.....

I have to figure out the area of this triangle with side lengths $$25,39,56$$ using right angled triangles.

I previously did a question which had a triangle with sides $$13,14,15$$ like this, using right triangles $$9,12,15$$ and $$5,12,13$$

However I can't make any such combination from $$25,39,56$$.. I computed the area using Heron's to find any combination... As the area is $$420$$ I thought to find a side which would give an integer height which turns out to be $$15$$ on base $$56$$...

But I can't find any right triangle with side $$15$$ which would yield the solution..

Can somebody provide me any insight on how should I proceed?

• Hint: drop the altitude $h$. then use Pythagorus to get two simultaneous quadratic equations (in two variables). Eliminate $h$ and solve.
– lulu
Jun 19, 2017 at 9:43
• Your height of 15 is actually correct. One triangle is 15, 20, 25. Jun 19, 2017 at 9:47

If you draw the height onto the side of length $56$, it will have a length of $15$ (as you already found) and split this side into $20+36$, and the whole triangle into the two pythagorean triangles:

• $15, 20, 25$, which is a scaled version of the $3, 4, 5$ triangle.
• $15, 36, 39$, which is a scaled version of the $5, 12, 13$ triangle.

The previous question also used scaled versions of the $3,4,5$ and $5, 12, 13$ triangles.

If the triangle's sides u are looking for must be Natural numbers then the problem has no solution.Else What Mr.Rozenberg said is right https://math.stackexchange.com/a/2328281/445265

Draw $\Delta ABC$, where $AB=39$, $AC=25$ and $BC=56$.

Let $BD$ is an altitude of $\Delta ABC$.

Since $\measuredangle BAC>90^{\circ}$, we see that $A$ placed between $D$ and $C$.

Thus, $AD=19.8$ and $BD=33.6$,

which gives two right-angled triangles: $\Delta BDC$ and $\Delta BDA$.