It's been a while since I've done any geometry so I'm a bit confused by this question.
We have a triangle $\triangle PQR$ whose total area is $90 \mathrm{cm}^2$. Another triangle $\triangle PTU$ is formed inside $\triangle PQR$ where the point $T$ is such that $PT=2TQ$ and $U$ is such that $QU = 2UR$. The points $T,U$ are on the edges $PQ$ and $QR$, respectively.
So we can say that $TQ= x, PT = 2x, PQ =3x$ and $UR =y, UQ = 2y, QR= 3y$.
I'm trying to find the area of the triangle $\triangle PTU$ but I'm having some difficulties. If we let $\angle PQR=\gamma, \angle PRQ = \beta, \angle QPR = \alpha$, then $$y=kx$$ where $k=\frac{\sin(\alpha)}{\sin(\beta)}$ and $$x^2 \frac{\sin(\gamma)\sin(\alpha)}{\sin(\beta)}=20$$ via the Sine Rule and the Cosine Rule.
I seem to have a few too many variables to solve this though and was wondering if I could get some pointers. Thanks!
PS: This is supposed to be solvable using elementary methods (middle-high school), so nothing crazy in your solutions (even Routh's theorem would probably be a bit much). Sorry for the lack of a diagram too!
$PT=2QT$, $QU=2UR$. Let $PH$ be the altitude from $P$ to edge $QR$. Thus $Area(\triangle PQR)=3Area(\triangle PUR)=90$, since they have the same altitude $PH$, and their bases satisfy $QR=3UR$.
Let $UG$ be the altitude from $U$ to edge $PQ$: you get $Area(\triangle PUT)=2Area(\triangle QUT)$, since they have the same altitude $UG$, and their bases satisfy $PT=2QT$. Thus $Area(\triangle PUT)=40$.