$ABCD$ is a rectangle of area $210$ cm$^2$. $L$ is a mid-point of $CD$ . $P,Q$ trisect $AB$ . $AC$ cuts $LP,LQ$ at $M,N$ respectively.

Question

$ABCD$ is a rectangle of area $210$ cm$^2$. $L$ is a mid-point of $CD$ . $P,Q$ trisect $AB$ . $AC$ cuts $LP,LQ$ at $M,N$ respectively. Find $[\Delta LMN]$

What I Tried: For a geometry Problem, it's always better to start with a figure :-

I have no good idea for this, it's also hard to do angle-chasing in a rectangle. And neither I find similar triangles in this picture.

It will be good if someone posts a hint to start with this.

Answer 1

I'm taking cosmo5's advice by using a $7\times3$ rectangle then just scaling by 10:

I made my own drawing including the lines and marking the points A, B, C, D, P, Q, and L on the Desmos graphing calculator. (The rest will be algebra and geometry, so I didn't just use the graph to cheat.)

The first thing to notice is that, if A is at the origin of a graph, and B is at the point $(0, 3)$, then we know that P rests at $(0, 2)$ and Q at $(0, 1)$, thus trisecting $\overline{AB}$. This means that $\overline{CD}$ is bisected by point L at $(0, 1.5)$. Thus we know that the following lines are true: $\overline{PL}=\frac{-x}{14}+2$; $\overline{QL}=\frac{x}{14}+1$; and $\overline{AC}=\frac{3x}{7}$.

Setting the formulæ for $\overline{QL}$ and $\overline{PL}$ equal to the formula $\overline{AC}$, we can find where they intersect (thus N and M, respectively). Doing so reveals that M sits at $(4,\frac{12}{7})$ and N sits at $(\frac{14}{5},\frac{42}{35})$.

From here, we have our triangle LMN (the one in question). To figure out its area, we utilise Heron's formula which is $A=\sqrt{(s(s-a)(s-b)(s-c)}$ such that $s=\frac{a+b+c}{2}$ such that $a, b$, and $c$ are the lengths of the sides of the triangle. So, to find the lengths, we use the distance formula ($d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$) between the points. We'll define $a\equiv\overline{LM}$, $b\equiv\overline{MN}$, and $c\equiv\overline{LN}$. Doing this reveals that reveals that: $a=\sqrt{9+(1.5-\frac{12}{7})^2}$, $b=\sqrt{(4-\frac{14}{5})^2+(\frac{12}{7}-\frac{42}{35})^2}$, and $c=\sqrt{(7-\frac{14}{5})^2+(1.5-\frac{42}{35})^2}$. Plugging these into the formula gives an answer of $0.9$. Because we scaled down by 10, we know that the area of $\triangle LMN=9$.