1. **State the problem:** We need to find the total area enclosed by two triangles joined at a vertex, sharing a side. The left triangle has a side length of 5 cm and an included angle of 105°. The right triangle has a side length of 13 cm and an included angle of 40°. 2. **Formula used:** The area of a triangle given two sides and the included angle is $$\text{Area} = \frac{1}{2}ab\sin(C)$$ where $a$ and $b$ are the sides and $C$ is the included angle. 3. **Calculate the area of the left triangle:** Given side $a=5$ cm, side $b$ is the shared side (unknown), and angle $C=105^\circ$. 4. **Calculate the area of the right triangle:** Given side $a=13$ cm, side $b$ is the shared side (same as above), and angle $C=40^\circ$. 5. **Find the length of the shared side:** Since the two triangles share the vertex and side, the sum of the angles at the vertex is $105^\circ + 40^\circ = 145^\circ$. The remaining angle between the two triangles is $180^\circ - 145^\circ = 35^\circ$. 6. **Use the Law of Cosines to find the shared side $x$:** $$x^2 = 5^2 + 13^2 - 2 \times 5 \times 13 \times \cos(35^\circ)$$ Calculate: $$x^2 = 25 + 169 - 130 \times \cos(35^\circ)$$ $$x^2 = 194 - 130 \times 0.8192 = 194 - 106.5 = 87.5$$ $$x = \sqrt{87.5} = 9.35 \text{ cm (approx)}$$ 7. **Calculate the area of the left triangle:** $$\text{Area}_1 = \frac{1}{2} \times 5 \times 9.35 \times \sin(105^\circ)$$ $$= 2.5 \times 9.35 \times 0.9659 = 22.6 \text{ cm}^2$$ 8. **Calculate the area of the right triangle:** $$\text{Area}_2 = \frac{1}{2} \times 13 \times 9.35 \times \sin(40^\circ)$$ $$= 6.5 \times 9.35 \times 0.6428 = 39.1 \text{ cm}^2$$ 9. **Total area enclosed:** $$\text{Total Area} = 22.6 + 39.1 = 61.7 \text{ cm}^2$$ **Final answer:** The total area enclosed is approximately **61.7 cm\textsuperscript{2}** to three significant figures.