Question

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

Answer 1

`\displaystyle [CQF]=5`

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
`(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=(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
`(AE)/(ED)=2`, and the ratio of the areas of
`\triangle DQF` and
`\triangle CQF` is also
`2:1`, from
`(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]=(-35)/(-7)=\boxed{5}`