Area in a Triangle

Let $AA' = h$ and $BD'=D'E'=D'A'=a$ . Then we have:

$\begin{cases} AB: & h^2 + (3a)^2 = 9^2 & ...(1) \\ CA: & h^2+(3a+5)^2 = 13^2 & ...(2) \end{cases} \implies (2) - (1): \ 30a + 25 = 88 \implies a = 2.1$

The area of $\triangle ABC$ , $[ABC] = \dfrac {BC\cdot h}2 = 2.5 h$ . The area of the red region:

$\begin{aligned} [DEE''D''] & = [CEE']-[CDD']-[D'D''E''E'] \\ & = [CEE']-[CDD']- \big([BE'E''] - [BD'D'']\big) & \small \blue{\text{By similar triangles}} \\ & = \left(\frac {2a+5}{3a+5} \right)^2 [AA'C] - \left(\frac {a+5}{3a+5} \right)^2 [AA'C] - \left(\frac {2a}{3a} \right)^2 [AA'B] + \left(\frac a{3a} \right)^2 [AA'B] \\ & = \frac {(2a+5)^2-(a+5)^2}{(3a+5)^2} [AA'C] - \frac {2^2-1^2}{3^2} [AA'B] \\ & = \frac {3a^2+10a}{(3a+5)^2} \cdot \frac {(3a+5)h}2 - \frac 13 \cdot \frac {3ah}2 \\ & = \frac {(3a^2+10a)h}{2(3a+5)} - \frac {ah}2 \\ & = \frac {34.23h}{22.6} - \frac {2.1h}2 \\ & \approx 0.4646h \end{aligned}$

Therefore $x = \dfrac {[DEE''D'']}{ABC} \approx \dfrac {0.4646h}{2.5h} \approx 0.1858 = 18.58 \%$ . $\implies \boxed{18.5\%<x<19\%}$ .

The area of $\triangle ABC$ is $\frac{9 \sqrt{51}}{4}$ . Since line segments $BD'$ , $D'E'$ , and $E'A'$ each has the length one-third of the line segment $BA'$ , line segments $AG$ , $GF$ , and $FB$ must have lengths one-third of $AB$ , which would make them equal to $3$ .

We notice that triangles $ABH$ , $AFD$ , and $AGE$ are all similar. Since $\frac{\text{area of} \space \triangle AFD}{\text{area of} \space \triangle ABH} = \frac{AF^{2}}{AB^{2}}$ $= \frac49$ and $\frac{\text{area of} \space \triangle AGE}{\text{area of} \space \triangle ABH} = \frac{AG^{2}}{AB^{2}} = \frac19$ because of a theorem of similar triangles , the area of the red shape is only one-third of the area of $\triangle ABH$ .

Through the use of Law of Cosines , we find that $BH = \frac{45\sqrt{51}}{113}$ and $HA = \frac{819}{113}$ . Then the area of triangle $ABH$ is $\frac{567\sqrt{51}}{452}$ and one-third of that is $\frac{189 \sqrt{51}}{452}$ which is the area of the red shape.

$\small \dfrac{\text{area of} \space DEGF}{\text{area of}\space \triangle ABC} = \dfrac{\frac{189 \sqrt{51}}{452}}{\frac{9 \sqrt{51}}{4}} = \dfrac{756\sqrt{51}}{4068\sqrt{51}} = \dfrac{21}{113} ≈ 0.18584$

Thus, $\boxed{18.5\% < x < 19\%}$ .