1. **State the problem:** We need to find the area of a Y-shaped polygon with given side lengths 4.5 m, 12 m, 16 m, 21 m, and 5 m. 2. **Approach:** To find the area of a complex polygon, we can divide it into simpler shapes such as triangles and rectangles, calculate their areas, and then sum them. 3. **Divide the polygon:** Based on the shape description, split the Y-shaped polygon into three triangles meeting at the central notch. 4. **Use Heron's formula for each triangle:** Heron's formula for the area of a triangle with sides $a$, $b$, and $c$ is: $$ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $$ where $s = \frac{a+b+c}{2}$ is the semi-perimeter. 5. **Calculate areas of each triangle:** - Triangle 1 with sides 4.5 m, 12 m, and 16 m: $$s_1 = \frac{4.5 + 12 + 16}{2} = 16.25$$ $$\text{Area}_1 = \sqrt{16.25(16.25-4.5)(16.25-12)(16.25-16)} = \sqrt{16.25 \times 11.75 \times 4.25 \times 0.25}$$ - Triangle 2 with sides 16 m, 21 m, and 5 m: $$s_2 = \frac{16 + 21 + 5}{2} = 21$$ $$\text{Area}_2 = \sqrt{21(21-16)(21-21)(21-5)} = \sqrt{21 \times 5 \times 0 \times 16} = 0$$ (This triangle is degenerate because one side equals the sum of the other two, so area is zero.) - Triangle 3 with sides 12 m, 21 m, and 5 m (assuming this is the third triangle): $$s_3 = \frac{12 + 21 + 5}{2} = 19$$ $$\text{Area}_3 = \sqrt{19(19-12)(19-21)(19-5)} = \sqrt{19 \times 7 \times (-2) \times 14}$$ (Negative value under the root means invalid triangle, so this triangle does not exist.) 6. **Conclusion:** Only Triangle 1 is valid. Calculate its area: $$\text{Area}_1 = \sqrt{16.25 \times 11.75 \times 4.25 \times 0.25} = \sqrt{202.246} \approx 14.22$$ 7. **Final answer:** The area of the polygon is approximately **14.22 square meters**.