MathFun4Kids

Solution: Let $AC$ and $BD$ intersect at $E$, $x=BD$, $y=AE$, $z=CE$, $AC=y+z$.

Because $BC=CD$, $AC$ is angle bisector of $\angle{BAD}$. By Angle Bisector Theorem, we have $\dfrac{BE}{DE}=\dfrac{AB}{AD}=\dfrac{5}{8}$. Because $BE+DE=BD=x$, we have $BE=\dfrac{5}{13}x$, $DE=\dfrac{8}{13}x$.

By Ptolemy’s Theorem, we have $BD\cdot AC=AB\cdot CD+AD\cdot BC$. Hence, $$x(y+z)=5\cdot 3+3\cdot 8=39\tag{1}$$

Because $BC=CD$, we have $\angle{BAC}=\angle{DAC}=\angle{BDC}=\angle{CAD}$.

Because $\angle{BCE}=\angle{ADE}$, $\triangle{BCE}\sim\triangle{ADE}$. Therefore $$\dfrac{CE}{DE}=\dfrac{BE}{AE}=\dfrac{BC}{AD}\ \ \ \ i.e.\ \ \ \ \dfrac{z}{\dfrac{8}{13}}=\dfrac{\dfrac{5}{13}x}{y}=\dfrac{3}{8}$$ We have $$y=\dfrac{40}{39}x\tag{2}$$ $$z=\dfrac{3}{13}x\tag{3}$$

Solving equation (1), (2) and (3), and ignoring negative solutions, we have $$x=\dfrac{39}{7}\ \ \ \ \ \ \ y=\dfrac{40}{7}\ \ \ \ \ \ z=\dfrac{9}{7}$$

Therefore, $BD=x=\dfrac{39}{7}$, $AC=y+z=7$. Therefore, the length of the shorter diagonal is $\boxed{\dfrac{39}{7}}$.

Note: since $\triangle{BCE}\sim\triangle{ACB}$, we have $\dfrac{BC}{CE}=\dfrac{AC}{BC}$, i.e. $\dfrac{3}{z}=\dfrac{y+z}{3}$. We have $$z(y+z)=9\tag{4}$$ We obtain the same result by solving equation (2), (3) and (4).

Solution: Draw line AB, BC, CD, DE, EA, and label the area of $5$ newly formed triangles as $a$, $b$, $c$, $d$, and $e$ respectively. Additionally, label the area of the pentagon $FGHIJ$ as $f$, we have the following diagram:

Based on the common ratios of areas for triangles sharing the same bases, we have the following:

$$\dfrac{Area\ \triangle{CDG}}{Area\ \triangle{DEG}}=\dfrac{Area\ \triangle{CBG}}{Area\ \triangle{BEG}}$$

i.e. $$\dfrac{a}{b+15}=\dfrac{e+7}{f+18}\tag{1}$$

Similarly, we have the following additional equations:

$$\dfrac{b}{a+15}=\dfrac{c+8}{f+10}\tag{2}$$

$$\dfrac{c}{b+8}=\dfrac{d+3}{f+25}\tag{3}$$

$$\dfrac{d}{c+3}=\dfrac{e+10}{f+15}\tag{4}$$

$$\dfrac{e}{d+10}=\dfrac{a+7}{f+18}\tag{5}$$

Since there are $6$ variables in the above 5 equations, we can add co-linear constraint $C$, $F$, and $G$.

By using Barycentric Coordinate System for triangles, we have the following non-normalized coordinates for $C$, $F$, and $G$, relative to $\triangle{ABD}$:

$$C=(d+e+10,-(a+e+7),a+f+25)\ \ \ \ \ \ F=(c+8,b,0)\ \ \ \ \ \ G=(e+7,0,a)$$

Because $C$, $F$, and $G$ are co-linear, with Barycentric Coordinate System, we have:

$$\begin{vmatrix} d+e+10 & -(a+e+7) & a+f+25\\ c+8 & b & 0 \\ e+7 & 0 & a \end{vmatrix}=0\tag{6}$$

Expanding the above 6 equations, we have:

$$\begin{cases} \ \ a(f+18)=(e+7)(b+15)\\ \ \ b(f+10)=(a+15)(c+8)\\ \ \ c(f+25)=(b+8)(d+3)\\ \ \ d(f+15)=(c+3)(e+10)\\ \ \ e(f+18)=(d+10)(a+7)\\ \ \ ab(d+e+10)+a(c+8)(a+e+7)-b(e+7)(a+f+25)=0 \end{cases}$$

By using a symbolic Computer Algebra System, such as Wolfram Alpha or Wolfree Alpha, with an input such as:

a>0,b>0,c>0,d>0,e>0,f>0,a(f+18)=(b+15)(e+7), b(f+10)=(c+8)(a+15), 
c(f+25)=(d+3)(b+8), d(f+15)=(e+10)(c+3), e(f+18)=(a+7)(d+10),
ab(d+e+10)+a(c+8)(a+e+7)-b(e+7)(a+f+25)=0

We have the following unique solution:

$$a=21,\ \ \ \ \ b=20,\ \ \ \ \ c=7,\ \ \ \ \ d=\dfrac{15}{2},\ \ \ \ \ e=14,\ \ \ \ \ f=17$$

Therefore the area of the pentagon is $\boxed{17}$ square unit.

Note: Please refer to Barycentric Coordinates in Olympiad Geometry or Barycentric Coordinates for the Impatient for additional information related to Barycentric Coordinates.