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Yuki888 [10]
3 years ago
5

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

Mathematics
1 answer:
Mama L [17]3 years ago
3 0

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}

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