1. **Problem 1A:** Construct a triangle PQR with sides $PQ=7$ cm, $QR=5$ cm, and angle $\angle PQR=120^\circ$. Then construct another triangle SQR with the same area as triangle PQR. 2. **Step 1:** Use the Law of Cosines or geometric construction to draw triangle PQR. 3. **Step 2:** Calculate the area of triangle PQR using the formula for area with two sides and included angle: $$\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR)$$ 4. Substitute values: $$\text{Area} = \frac{1}{2} \times 7 \times 5 \times \sin(120^\circ)$$ 5. Calculate $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$. 6. So, $$\text{Area} = \frac{1}{2} \times 7 \times 5 \times \frac{\sqrt{3}}{2} = \frac{35 \sqrt{3}}{4} \approx 15.13 \text{ cm}^2$$ 7. **Step 3:** To construct triangle SQR with the same area, choose side $QR=5$ cm and find height $h$ such that $$\text{Area} = \frac{1}{2} \times QR \times h = 15.13$$ 8. Solve for $h$: $$h = \frac{2 \times 15.13}{5} = 6.05 \text{ cm}$$ 9. Draw triangle SQR with base $QR=5$ cm and height $6.05$ cm perpendicular to $QR$ to get the same area. --- **Problem 1B:** Given $BC=CD$ and $AE \parallel BD$, find the relationship between triangles $ABC$ and $BDE$. 10. Since $AE \parallel BD$, by the Alternate Interior Angles Theorem, $\angle BAC = \angle BDE$. 11. Also, $BC=CD$ implies triangles share proportional sides. 12. Therefore, triangles $ABC$ and $BDE$ are similar by the AA (Angle-Angle) similarity criterion. **Final answers:** - Area of triangle PQR is $\frac{35 \sqrt{3}}{4}$ cm$^2$. - Triangle SQR constructed with base $5$ cm and height $6.05$ cm has the same area. - Triangles $ABC$ and $BDE$ are similar because $AE \parallel BD$ and $BC=CD$.