1. **Problem statement:** We have a rectangle ABCD formed by two adjacent squares ABFE and CDEF, each with side length 6 cm. Point G is the midpoint of DE, and BG intersects EF at H. We need to find the area of the shaded pentagon CDGHF. 2. **Setup and coordinates:** - Since ABFE and CDEF are squares of side 6 cm, place point A at origin $(0,0)$. - Then $B=(6,0)$, $F=(0,6)$, $E=(6,6)$ for square ABFE. - For square CDEF adjacent to ABFE, $C=(12,0)$, $D=(12,6)$. - $G$ is midpoint of $DE$, so $D=(12,6)$ and $E=(6,6)$, thus $G=\left(\frac{12+6}{2},6\right)=(9,6)$. 3. **Find point H:** - $B=(6,0)$ and $G=(9,6)$, line BG has slope $m=\frac{6-0}{9-6}=\frac{6}{3}=2$. - Equation of BG: $y=2(x-6)$ or $y=2x-12$. - EF is horizontal line at $y=6$ from $E=(6,6)$ to $F=(0,6)$. - Find intersection H by setting $y=6$ in BG: $6=2x-12 \Rightarrow 2x=18 \Rightarrow x=9$. - So $H=(9,6)$, which coincides with $G$. 4. **Vertices of pentagon CDGHF:** - $C=(12,0)$ - $D=(12,6)$ - $G=(9,6)$ - $H=(9,6)$ (same as G) - $F=(0,6)$ Since $G$ and $H$ are the same point, the pentagon reduces to quadrilateral $CDGF$. 5. **Calculate area of quadrilateral CDGF:** - Use shoelace formula for points in order $C(12,0)$, $D(12,6)$, $G(9,6)$, $F(0,6)$. - Sum1 = $12*6 + 12*6 + 9*6 + 0*0 = 72 + 72 + 54 + 0 = 198$ - Sum2 = $0*12 + 6*9 + 6*0 + 6*12 = 0 + 54 + 0 + 72 = 126$ - Area = $\frac{|198 - 126|}{2} = \frac{72}{2} = 36$ cm$^2$. 6. **Final answer:** The shaded area of pentagon CDGHF is $36$ cm$^2$.