Question

Decide !!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Answer:

$\displaystyle [CQF]=5$

Step-by-step explanation:

Note that $[n]$ refers to the area of some polygon $n$.

Diagonal $\overline{AC}$ forms two triangles, $\triangle ABC$ and $\triangle ADC$. Both of these triangles have an equal area, and since the area of parallelogram $ABCD$ is given as $210$, each triangle must have an area of $105$.

Furthermore, $\triangle ADC$ is broken up into two smaller triangles, $\triangle ADF$ and $\triangle ACF$. We're given that $\frac{DF}{FC}=2$. Since $DF$ and $FC$ represent bases of $\triangle ADF$ and $\triangle ACF$ respectively and both triangles extend to point $A$, both triangles must have the same height and hence the ratio of the areas of $\triangle ADF$ and $\triangle ACF$ must be $2:1$ (recall $A=\frac{1}{2}bh$).

Therefore, the area of each of these triangles is:

$[ACF]+[ADF]=105,\\[][ACF]+2[ACF]=105,\\3[ACF]=105,\\[][ACF]=35 \implies [ADF]=70$

With the same concept, the ratio of the areas of $\triangle AQE$ and $\triangle DQE$ must be $2:1$ respectively, from $\frac{AE}{ED}=2$, and the ratio of the areas of $\triangle DQF$ and $\triangle CQF$ is also $2:1$, from $\frac{DF}{FC}=2$.

Let $[DQE]=y$ and $[CQF]=x$ (refer to the picture attached). We have the following system of equations:

$\displaystyle \begin{cases}2y+y+2x=70,\\y+2x+x=35\end{cases}$

Combine like terms:

$\displaystyle \begin{cases}3y+2x=70,\\y+3x=35\end{cases}$

Multiply the second equation by $-3$, then add both equations:

$\displaystyle \begin{cases}3y+2x=70,\\-3y-9x=-105\end{cases}\\\\\rightarrow 3y-3y+2x-9x=70-105,\\-7x=-35,\\x=[CQF]=\frac{-35}{-7}=\boxed{5}$