An Area Problem

$m_{AC} = \dfrac{a}{4} \implies y = \dfrac{a}{4}x$ and $\overleftrightarrow{BD}$ is $y = 4 \implies \dfrac{ax}{4} = 4 \implies x = \dfrac{16}{a} \implies$

$P:(\dfrac{16}{a},4)$

$m_{AE} = \dfrac{a}{4 + a} \implies y = \dfrac{a}{4 + a}x$ and $\overleftrightarrow{CF}$ is $x = 4 \implies y = \dfrac{4a}{4 + a} \implies$

$Q:(4,\dfrac{4a}{4 + a})$

$A_{\triangle{ABP}} = \dfrac{1}{2}(AB)(BP) = 2BP = 2(\dfrac{16}{a}) = \dfrac{32}{a}$

$A_{\triangle{AQF}} = \dfrac{1}{2}(AF)(QF) = 2QF = 2(\dfrac{4a}{4 + a}) = \dfrac{8a}{4 + a}$

$A_{ABDF} = 16$

$\implies$ The total area $A_{T} = \dfrac{88}{15} = 16 - (\dfrac{32}{a} + \dfrac{8a}{4 + a}) =$ $8(\dfrac{a^2 + 4a - 16}{a(a + 4)}) \implies$

$\dfrac{a^2 + 4a - 16}{a(a + 4)} = \dfrac{11}{15} \implies 15a^2 + 60a - 240 = 44a + 11a^2 \implies$

$4a^2 + 16a - 240 = 0 \implies a^2 + 4a - 60 = 0 \implies (a - 6)(a + 10) = 0$

and $a \neq -10 \implies a = \boxed{6}$

Label the figure as shown. Then we note that the area of the red region is given by

$[ABC] - [AFG] - [BCH] = \frac {88}{15}$

$[ABC] = \dfrac {4a}2 = 2a$ . Since $\triangle AFG$ and $\triangle ABC$ are similar, we have $[AFG] = \dfrac {AG^2}{AC^2} \times [ABC] = \dfrac {(a-4)^2}{a^2} \times 2a = \dfrac {2(a-4)^2}a$ . Again $\triangle BCH$ and $\triangle BDE$ are similar, $[BCH] = \dfrac {BC^2}{BD^2} \times [BDE] = \dfrac {4^2}{(a+4)^2} \times \dfrac {a(a+4)}2 = \dfrac {8a}{a+4}$ . Then we have:

$\begin{aligned} 2a - \frac {2(a-4)^2}a - \frac {8a}{a+4} & = \frac {88}{15} \\ 2a - 2a + 16 - \frac {32}a - \frac {8a}{a+4} & = \frac {88}{15} \\ 2 - \frac 4a - \frac a{a+4} & = \frac {11}{15} \\ \frac {4a + 16 + a^2}{a^2+4a} & = 2 - \frac {11}{15} \\ 1 + \frac {16}{a^2 + 4a} & = 2 - \frac {11}{15} \\ \frac {16}{a^2 + 4a} & = \frac 4{15} \\ \frac 4{a^2 + 4a} & = \frac 1{15} \\ a^2 + 4a - 60 & = 0 \\ (a-6)(a + 10) & = 0 & \small \blue{\text{Since }a > 0} \\ \implies a & = \boxed 6 \end{aligned}$

With the labelling shown on the figure, we have

$\ \$

$\triangle ABI\sim \triangle AEF\Rightarrow \dfrac{IB}{FE}=\dfrac{AB}{AE}\Rightarrow \dfrac{x}{a}=\dfrac{4}{4+a}\Rightarrow x=\dfrac{4a}{4+a}$

$\triangle GHC\sim \triangle GAB\Rightarrow \dfrac{HC}{AB}=\dfrac{GC}{GB}\Rightarrow \dfrac{y}{4}=\dfrac{a-4}{a}\Rightarrow y=\dfrac{4(a-4)}{a}$

$\ \$

$\begin{aligned} \left[ AHCI \right] & =\left[ AHC \right]+\left[ ACI \right] \\ & =\dfrac{1}{2}HC\cdot AD+\dfrac{1}{2}CI\cdot AB \\ & =\dfrac{1}{2}\left[ 4y+4\left( 4-x \right) \right] \\ & =2y-2x+8 \\ \end{aligned}$

Combining we get

$2\dfrac{4\left( a-4 \right)}{a}-2\dfrac{4a}{4+a}+8=\dfrac{88}{15}\Leftrightarrow \ldots \Leftrightarrow {{a}^{2}}+4a-60=0\overset{a>0}{\mathop{\Leftrightarrow }}\,\boxed{a=6}.$