Parallelogram Riddle II

Since $ABCD$ is a parallelogram, we can conclude that $\angle ABG = \angle GEF$ and $\angle BAG = \angle GFE$ due to the angles of parallel lines. Thus, $\triangle ABG$ is similar to $\triangle EFG$ because all three interior angles in both triangles are equal.

Hence, the area ratio between these similar triangles $ABG : EFG$ equals $27 : 12$ or $9 : 4$ . Given the base length ratio of $AB : EF$ be $k$ , the height ratio of the bigger to smaller triangles will also be $k$ . In other words, the area ratio then is in terms of base $b$ and height $h$ $\dfrac{1}{2}\times kb \times kh : \dfrac{1}{2}\times b \times h$ or simply $k^2$ .

Hence, $k^2 = \dfrac{9}{4}$ . Thus, $k = \dfrac{3}{2}$ .

Then by creating a new parallel line $HI$ passing the point $G$ , the length ratio $AI : ID = 3:2$ , expressed as $k$ earlier. Furthermore, with $HI // AB$ , $ABHI$ is also a parallelogram and so has twice the area as $\triangle ABG$ : the area of $ABHI = 2\times 27 = 54$ .

Also, we also obtain that $AI : AD = 3:5$ from the previous finding, and then the area ratio, with height $s$ relative to $AD$ , of $ABHI:ABCD = s\times AI : s\times AD = 3:5$ . Therefore, the area of $ABCD = \dfrac{5}{3} \times 54 = 90$ .

Finally, the yellow region has area of $90 - 27 - 12 = \boxed{51}$ .

My solution is similar to Worranat's , but not identical.

We can shear this parallelogram horizontally into a rectangle without changing areas (by Cavalieri's principle). Thus we may assume that $ABCD$ is indeed a rectangle, which makes everything a bit easier to think through. Since the area of the similar triangles $ABG$ and $EFH$ are in a proportion of $9:4$ , the linear dimensions are in a proportion of $3:2$ . In particular, if $h$ is the height of triangle $ABG$ over the side $AB$ , then $h+\frac{2h}{3}=AD$ so $h=\frac{3(AD)}{5}$ . The area of triangle $ABC$ is $27=\frac{h(AB)}{2}=\frac{3(AD)(AB)}{10}$ so that the area of the whole rectangle is $(AD)(AB)=90$ . The area of the yellow region is $90-27-12=\boxed{51}$ .

We have 2 similar triangles the red and blue ones, with ratio of sides = √27/√12 = 3/2. So AB=3x,EF=2x,BH=3y,CH=2y. And the area of ΔABG=27 = (3x)(3y)/2,then xy=6. The area of yellow region = 15xy – 39 = 51.

If you are in a rush cheat by using a rectangle and have BA=3 EF=2 and find out AD=30 and do 30*3-39.

As alternate interior angles of parallel lines, $\angle ABG \cong \angle FEG$ and $\angle BAG \cong \angle EFG$ , so $\triangle ABG \sim \triangle FEG$ by AA similarity. Therefore, we can extend $GE$ to the same length of $GB$ to $H$ and extend $GF$ to the same length of $GA$ to $I$ to create parallelogram $ABIH$ with the same height of parallelogram $ABCD$ . Let $JK$ and $JL$ be the midsegments of parallelograms $ABCD$ and $ABIH$ , respectively.

The ratio of the sides of the $\triangle FEG$ to $\triangle ABG$ is the square root of the ratio of their areas, which is $\sqrt{\frac{12}{27}} = \frac{2}{3}$ , so $GK = \frac{2}{3}JG$ .

Since the ratio of the base of parallelogram $ABCD$ to the base of parallelogram $ABIH$ is $\frac{AD}{AH} = \frac{JK}{JL} = \frac{JG + GK}{JG + GL} = \frac{JG + \frac{2}{3}JG}{2JG} = \frac{5}{6}$ , and since the area of parallelogram $ABIH = 4 \cdot 27 = 108$ , the area of parallelogram $ABIH$ is $\frac{5}{6} \cdot 108 = 90$ .

Therefore, the area of the yellow region of parallelogram $ABCD$ is $90 - 27 - 12 = \boxed{51}$ .