Shaded Area Rectangle 923361

1. **State the problem:** We need to find the shaded area inside rectangle $ABDC$ after removing two white triangular regions: triangle $AFD$ and triangle $FBG$. 2. **Given data:** - Rectangle $ABDC$ has width $20$ cm and height $10$ cm. - Point $F$ is on $AB$ such that $FB = 5$ cm, so $AF = AB - FB = 10 - 5 = 5$ cm (since $AB$ is height $10$ cm). - Point $E$ is on $CD$ such that $ED = 8$ cm. - Point $G$ is $5$ cm below $B$. 3. **Calculate the area of rectangle $ABDC$:** $$\text{Area}_{ABDC} = \text{width} \times \text{height} = 20 \times 10 = 200 \text{ cm}^2$$ 4. **Find the area of triangle $AFD$:** - Triangle $AFD$ is formed by points $A$, $F$, and $D$. - $AF$ is vertical segment on $AB$ with length $5$ cm. - $AD$ is horizontal segment on $AD$ with length $20$ cm. - Since $AF$ and $AD$ are perpendicular, triangle $AFD$ is right-angled with legs $AF = 5$ cm and $AD = 20$ cm. - Area formula for right triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ - So, $$\text{Area}_{AFD} = \frac{1}{2} \times 5 \times 20 = 50 \text{ cm}^2$$ 5. **Find the area of triangle $FBG$:** - Triangle $FBG$ is formed by points $F$, $B$, and $G$. - $FB = 5$ cm (given). - $BG = 5$ cm (since $G$ is 5 cm below $B$). - $FB$ and $BG$ are perpendicular (vertical and horizontal segments). - Area: $$\text{Area}_{FBG} = \frac{1}{2} \times 5 \times 5 = \frac{25}{2} = 12.5 \text{ cm}^2$$ 6. **Calculate the shaded area:** $$\text{Shaded area} = \text{Area}_{ABDC} - \text{Area}_{AFD} - \text{Area}_{FBG} = 200 - 50 - 12.5 = 137.5 \text{ cm}^2$$ **Final answer:** The shaded area is $137.5$ cm$^2$.