Answer:
[tex]\displaystyle [CQF]=5[/tex]
Step-by-step explanation:
Note that [tex][n][/tex] refers to the area of some polygon [tex]n[/tex].
Diagonal [tex]\overline{AC}[/tex] forms two triangles, [tex]\triangle ABC[/tex] and [tex]\triangle ADC[/tex]. Both of these triangles have an equal area, and since the area of parallelogram [tex]ABCD[/tex] is given as [tex]210[/tex], each triangle must have an area of [tex]105[/tex].
Furthermore, [tex]\triangle ADC[/tex] is broken up into two smaller triangles, [tex]\triangle ADF[/tex] and [tex]\triangle ACF[/tex]. We're given that [tex]\frac{DF}{FC}=2[/tex]. Since [tex]DF[/tex] and [tex]FC[/tex] represent bases of [tex]\triangle ADF[/tex] and [tex]\triangle ACF[/tex] respectively and both triangles extend to point [tex]A[/tex], both triangles must have the same height and hence the ratio of the areas of [tex]\triangle ADF[/tex] and [tex]\triangle ACF[/tex] must be [tex]2:1[/tex] (recall [tex]A=\frac{1}{2}bh[/tex]).
Therefore, the area of each of these triangles is:
[tex][ACF]+[ADF]=105,\\[][ACF]+2[ACF]=105,\\3[ACF]=105,\\[][ACF]=35 \implies [ADF]=70[/tex]
With the same concept, the ratio of the areas of [tex]\triangle AQE[/tex] and [tex]\triangle DQE[/tex] must be [tex]2:1[/tex] respectively, from [tex]\frac{AE}{ED}=2[/tex], and the ratio of the areas of [tex]\triangle DQF[/tex] and [tex]\triangle CQF[/tex] is also [tex]2:1[/tex], from [tex]\frac{DF}{FC}=2[/tex].
Let [tex][DQE]=y[/tex] and [tex][CQF]=x[/tex] (refer to the picture attached). We have the following system of equations:
[tex]\displaystyle \begin{cases}2y+y+2x=70,\\y+2x+x=35\end{cases}[/tex]
Combine like terms:
[tex]\displaystyle \begin{cases}3y+2x=70,\\y+3x=35\end{cases}[/tex]
Multiply the second equation by [tex]-3[/tex], then add both equations:
[tex]\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}[/tex]
