The area of a triangle with sides $11$, $12$, and $x$ is $28.9$. Find $x$.

Question

I’ve been stuck on this question for days now, and honestly can’t figure it out. It’s not isosceles or right angled, and there’s no angles so I can’t use the sine or cosine rule. Any ideas?

The area of a triangle with sides $11$, $12$, and $x$ is $28.9$. Find $x$.

Answer 1

By Heron's formula $$16\cdot28.9^2=(23+x)(1+x)(x-1)(23-x)$$ $$13363.36=(x^2-1)(529-x^2)$$ Substitute $y=x^2$: $$13363.36=(y-1)(529-y)=-529+530y-y^2$$ $$y^2-530y+13892.36=0$$ Solving the quadratic formula gives $y=502.345$ or $y=27.6550$, yielding $x=22.413$ or $x=5.2588$. Both possibilities yield valid triangles.

Answer 2

I would say that maybe

$A=\frac12 ab\sin \varphi$, A = 28.9, a = 11, b = 12 $\Rightarrow \varphi$,

$x=\sqrt{a^2+b^2-2ab\cos \varphi}$