Problem:
$PQRS$ is a trapezium in which base $PQ=2$ units and base $RS = 3$ units. Draw a line from point $P$ to $R$. Triangle $PQR$ is formed. Shade in that triangle.
Now, what fraction of the trapezium is shaded?
I am having some trouble solving this problem as the height of the trapezium is not mentioned. I believe that knowing that height is unnecessary, but I am not too sure.
My Attempt:
Lemma: Given a trapezium with bases $a$ and $b$, and an altitude (height) $h$, the area is $\dfrac{h(a+b)}{2}$.
In this example, $a,b=PQ,RS$ respectively, so the area is $$\dfrac{h(2+3)}{2} = \dfrac{5h}{2}.$$
Now I know that if it were a square, then the two bases would be equal to each other, and the line $PR$ would be splitting it into equal halves.
So, $\Delta PQR = \dfrac 12$.
But, since on of the bases is larger, and the smaller base is a side of $\Delta PQR$, then my assumption is that the area of $\Delta PQR$ is less than $\dfrac 12$.
Nevertheless, the height of $PQRS$ is not stated so I do not know how to solve this. Intuitively, I believe the answer is $\dfrac 13$ but how can it be proven?
Thank you in advance.
Edit:
Doing some research, I found a very similar post over here.
General Rule (edited just now):
After looking at the answer, I came up with a general rule:
You are given a quadrilateral $PQRS$ (i.e. a square; trapezium; etc) in which base $PQ = a$ units; base $RS = b$ units; the two bases are parallel; and height $h\perp PQ\land RS$. By drawing a diagonal line $PR = c$ units to form two congruent triangles $PQR$ and $PSR$ that make up $PQRS$, then by denoting $S_{ABCD}$ the area of $ABCD$ and $\Delta{UVW}$ a triangle $UVW$, $$\frac{S_{\Delta PQR}}{S_{PQRS}} = \frac{a}{a+b}\,\text{ and }\,\frac{S_{\Delta PSR}}{S_{PQRS}} = \frac{b}{b+a},$$ regardless of the value of $h$.
Just clarifying: is this true? If so, does this theorem have a name?
$$\text{Required fraction} = \frac{S_{\triangle PQR}}{S_{PQRS}} = \frac{\frac12 \cdot 2h}{\frac12 \cdot 5h} = \frac25$$