1. **Problem statement:** Construct a triangle $\triangle ABC$ with sides $a=6.2$ cm, $b=5.5$ cm, and $c=6.5$ cm, then construct a parallelogram with one side $7$ cm whose area equals that of $\triangle ABC$. 2. **Formula for area of triangle using Heron's formula:** $$s = \frac{a+b+c}{2}$$ $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$ where $s$ is the semi-perimeter. 3. **Calculate semi-perimeter:** $$s = \frac{6.2 + 5.5 + 6.5}{2} = \frac{18.2}{2} = 9.1$$ 4. **Calculate area of $\triangle ABC$:** $$\text{Area} = \sqrt{9.1(9.1-6.2)(9.1-5.5)(9.1-6.5)} = \sqrt{9.1 \times 2.9 \times 3.6 \times 2.6}$$ Calculate inside the root: $$9.1 \times 2.9 = 26.39$$ $$3.6 \times 2.6 = 9.36$$ $$26.39 \times 9.36 = 246.98$$ So, $$\text{Area} = \sqrt{246.98} \approx 15.72 \text{ cm}^2$$ 5. **Construct parallelogram with side 7 cm and area equal to $\triangle ABC$:** Area of parallelogram = base $\times$ height Let height be $h$, base = 7 cm $$7 \times h = 15.72$$ Divide both sides by 7: $$\cancel{7} \times h = \frac{15.72}{\cancel{7}}$$ $$h = 2.2457 \text{ cm}$$ 6. **Conclusion:** Construct a parallelogram with base 7 cm and height approximately 2.25 cm to have the same area as $\triangle ABC$. --- **Note:** The second question about proving equal areas of parallelograms ABCD and PQRS is not solved here as per instructions to solve only the first problem.