How to find the area of a triangle with two equations?

Question

So I was given the following problem :

ABC is a right angled triangle with the sides $a,b,c$ . Find the area of this triangle, given that $$a+b+c = 22$$ $$a^2+b^2+c^2 = 200$$

I've tried to do a lot of things withose two, and using the pythagore theorem I got :

$$a + b = 12$$ $$a^2 + b^2 = 100$$

Sure, I can turn it to a Polynomial, but I will have two solutions for $a$, and that looks weird . Also the area must be a natural number .

How can I find the area ?

Answer 1

As noted, we have $$a+b=12$$$$a^2+b^2=100$$ As you mentioned, this can be turned into a polynomial. However, notice that $$(a+b)^2-a^2-b^2=2ab=44$$Thus $$ab=22$$ and by taking the half of this we can find the desired area. Sure, there are two solutions. Namely that $$a=6-\sqrt{14}, b=6+\sqrt{14}$$ or $$a=6+\sqrt{14}, b=6-\sqrt{14}$$But notice that in both cases, the triangles are equivalent.

Answer 2

Hint: $ab/2=((a+b)^2-(a^2+b^2))/4$

Answer 3

Yes, you will have two solutions for $a$. That is because if $(a,b,c)$ is a solution then so is $(b,a,c)$. But both lead to the same area. Actually, we do not need to determine $a$, or $b$, all we need is $S=\frac 12ab$: $$12^2=(a+b)^2=a^2+2ab+b^2 =(a^2+b^2)+4S=100 +4S$$ so that $S=11$.

Answer 4

Square the first equation and solve simultaneously for $\frac{ab}{2}$, since that is the area of the triangle by definition. This turns out to be 11 square units.

Answer 5

If you assume that the sides have been labelled so that $a^2 + b^2 = c^2$, then $2c^2 = 200$ becomes $c = 10$ and $a + b = 12$

So $144 = (a+b)^2 = a^2 + b^2 + 2ab = 100 + 2ab$. So $ab = 22$

the area of the triangle is $\dfrac 12 ab = 11$.

Answer 6

HINT:

$$100=a^2+(12-a)^2\iff a^2-12a+22=0$$

Solve for $a,$

Observe that the area $$=\dfrac{(6-\sqrt{14})\cdot(6+\sqrt{14})}2=?$$ which is a natural number