Trapezoid, Parallelogram, & Triangle

Extend $AE$ and $BG$ to meet at $H$ .

$\triangle AHB$ is similar to $\triangle HDC$ by AA similarity, and since $2AB = CD$ , $\frac{1^2}{2^2} = \frac{A_{\triangle AHB}}{A_{\triangle AHB} + 4}$ , which solves to $A_{\triangle AHB} = \frac{4}{3}$ .

Using $A_\triangle = \frac{1}{2}bh$ , the height of $\triangle AHB$ solves to $h_{\triangle AHB} = \frac{8}{3AB}$ .

$\triangle AHB$ is also similar to $\triangle CFG$ by AA similarity, so $\frac{(\frac{8}{3AB})^2}{h_{\triangle CFG}^2} = \frac{\frac{4}{3}}{3}$ , which solves to $h_{\triangle CFG} = \frac{4}{AB}$ .

The area of the blue parallelogram is therefore $A_{CDEF} = CD \cdot h_{\triangle CFG} = 2AB \cdot \frac{4}{AB} = \boxed{8}$ .

By Cavalieri's principle , we can reshape the big trapezoid $ABCD$ into a right trapezoid while retaining all the same areas and parallel base lengths, as followed:

Now we will create a mirror image of such right trapezoid and join both shapes together to form a bigger isosceles trapezoid, as shown:

Then by drawing a line extending from $FC$ to $A'$ , it is clear that $DA || CA'$ and $AA' = 2AB = CD$ . Thus, $ADCA'$ is also a parallelogram.

Furthermore, by drawing a line $AH$ perpendicular to $DC$ , we can see that $DH = DC - HC = DC - AB = AB$ . Thus, $AH$ bisects $DC$ , and $ADC$ is an isosceles triangle and has an area of $\dfrac{2}{3}$ of the trapezoid $ABCD$ , yielding $\dfrac{2}{3} \times 4 = \dfrac{8}{3}$ . The area of $ADCA'$ is twice that of the isosceles triangle = $\dfrac{16}{3}$ .

On the other hand, with $AA' || FF'$ and $AF'$ intersecting $FA'$ , we can conclude that $\triangle ACA' \sim \triangle FCF'$ . Then the area of $\triangle FCF'$ is twice that of $\triangle FCG$ , resulting in $6$ . Thus, the area ratio of $\triangle ACA' : \triangle FCF' = \dfrac{8}{3} : 6 = 4:9 = 2^2 : 3^2$ .

Hence, it is obvious that the height ratio $BC : CG = 2:3$ .

Finally, since the parallelograms $DCEF$ and $ADCA'$ have the same base, we can calculate the desired area of $DCEF$ by using such ratio: $\dfrac{16}{3}\times \dfrac{3}{2} = \boxed{8}$ .

Let $AB=a$ and the height of trapezoid $ABCD$ be $h$ . Draw a line $BH$ parallel to $AE$ to form parallelogram $ABHD$ . Then $DH = HC = a$ .

The area of trapezoid $ABCD$ , $[ABCD] = \dfrac {a+2a}2 h = \dfrac 32 ah = 4$ . $\implies ah = \dfrac 83$ . Then the area of $\triangle BCH$ , $[BCH] = \dfrac 12 ah = \dfrac 43$ .

We note that $\triangle CFG$ is similar to $\triangle BCH$ , Then the area of the two triangles are directly proportional to the square of the linear dimension. That is $\dfrac {FG^2}{HC^2} = \dfrac {[CFG]}{[BCH]} = \dfrac 3{\frac 43} = \dfrac 94$ , $\implies FG = \dfrac 32 a$ . Then the height of $\triangle CFG$ , $h' = \dfrac 32 h$ , which is also the height of blue parallelogram $CDEF$ .

Therefore, the area of the blue parallelogram $[CDEF] = 2ah' = 2a \times \dfrac 32 h = 3ah = 3 \times \dfrac 83 = \boxed 8$ .

If we take a similar triangle to the Green one, we can fit 3 of them inside the Red trapezium. They each have area $\frac{4}{3}$ and $FG:AB = \sqrt{3} : \sqrt{\frac{4}{3}} = 3:2$ .

Treating Blue as 2 triangles: the ratio of Blue to Green is $2EF : FG = 4 : \frac{3}{2} = 8 : 3$ . Hence the area of Blue is 8.

Extend FC to H, where H lies on the extension of AB. Let K be on BH and L be on Fg so that KL is a straight line perpendicular to BH and FG and passing through C. Let KC = d and CL = k. Then triangle BHC is similar to triangle FCG by 3 angles equal. Then d/x = h/y, and hx = dy. Area of triangle FCG = 3 = yh/2. So hy = 6, h(hx/d) = 6, (h^2)x = 6d, (h^2)(x^2) = 6dx. From trapezoid ABCD, Area = 4 =(1/2)(d)(x + 2x), or 8 = 3dx, and 16 = 6dx. But 6dx = (h^2)(x^2), so hx = 4. Finally, area of parallelogram CDEF = ? = ((1/2)(h)(4x) = (1/2)(4/x)(4x) =16x/2x = 8.