1. Statement of the problem: Find the area and centroid of the quadrilateral with vertices $D(5,4)$, $E(8,2)$, $F(7,-2)$, and $G(4,-4)$. 2. Formulas and rules used: Use the shoelace formula for polygon area. $$A=\tfrac12\left|\sum_{i=0}^{n-1}(x_i y_{i+1}-x_{i+1} y_i)\right|$$ Also use the centroid formulas for a polygon given a signed area $A$. $$C_x=\frac{1}{6A}\sum_{i=0}^{n-1}(x_i+x_{i+1})(x_i y_{i+1}-x_{i+1} y_i)$$ $$C_y=\frac{1}{6A}\sum_{i=0}^{n-1}(y_i+y_{i+1})(x_i y_{i+1}-x_{i+1} y_i)$$ 3. Intermediate work: List the vertices in order and repeat the first at the end for the shoelace sum. Compute the cross terms $x_i y_{i+1}-x_{i+1} y_i$ for each edge. For $D\to E$: $5\cdot 2-8\cdot 4=-22$. For $E\to F$: $8\cdot(-2)-7\cdot 2=-30$. For $F\to G$: $7\cdot(-4)-4\cdot(-2)=-20$. For $G\to D$: $4\cdot 4-5\cdot(-4)=36$. Sum of the cross terms $\sum k_i=-36$. Signed area $A_{signed}=\tfrac12(-36)=-18$. Area $A=|A_{signed}|=18$. 4. Centroid calculation: compute numerators then divide by $6A_{signed}$. Compute each term for the $C_x$ numerator. $(5+8)(-22)=-286$. $(8+7)(-30)=-450$. $(7+4)(-20)=-220$. $(4+5)(36)=324$. Sum for $C_x$ numerator $=-632$. Using the centroid formula and $A_{signed}=-18$ gives $C_x=\frac{-632}{6\cdot(-18)}=\frac{632}{108}=\frac{158}{27}\approx 5.85185$. Compute each term for the $C_y$ numerator. $(4+2)(-22)=-132$. $(2+(-2))(-30)=0$. $(-2+ -4)(-20)=120$. $(-4+4)(36)=0$. Sum for $C_y$ numerator $=-12$. Using the centroid formula gives $C_y=\frac{-12}{6\cdot(-18)}=\frac{12}{108}=\frac{1}{9}\approx 0.11111$. Final answer: Area $A=18$ and centroid $C=\left(\frac{158}{27},\frac{1}{9}\right)\approx(5.85185,0.11111)$.