1. **State the problem:** We have a parallelogram with sides $a=6$ and $b=8$, and a diagonal of length $12$. We need to find the area $S$ and the height $H$ corresponding to side $a$. 2. **Recall formulas:** - The area of a parallelogram is $S = a \times H$ where $H$ is the height corresponding to side $a$. - Using the law of cosines for the diagonal $d=12$, we have: $$d^2 = a^2 + b^2 - 2ab \cos(\theta)$$ where $\theta$ is the angle between sides $a$ and $b$. 3. **Calculate the angle $\theta$:** $$12^2 = 6^2 + 8^2 - 2 \times 6 \times 8 \cos(\theta)$$ $$144 = 36 + 64 - 96 \cos(\theta)$$ $$144 = 100 - 96 \cos(\theta)$$ $$144 - 100 = -96 \cos(\theta)$$ $$44 = -96 \cos(\theta)$$ $$\cos(\theta) = -\frac{44}{96} = -\frac{11}{24}$$ 4. **Find $\sin(\theta)$:** $$\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(-\frac{11}{24}\right)^2} = \sqrt{1 - \frac{121}{576}} = \sqrt{\frac{455}{576}} = \frac{\sqrt{455}}{24}$$ 5. **Calculate the height $H$:** Height $H$ is the perpendicular from side $a$, so: $$H = b \sin(\theta) = 8 \times \frac{\sqrt{455}}{24} = \frac{8 \sqrt{455}}{24} = \frac{\cancel{8} \sqrt{455}}{\cancel{24} 3} = \frac{\sqrt{455}}{3}$$ 6. **Calculate the area $S$:** $$S = a \times H = 6 \times \frac{\sqrt{455}}{3} = \frac{6}{3} \sqrt{455} = 2 \sqrt{455}$$ **Final answers:** - Height $H = \frac{\sqrt{455}}{3}$ - Area $S = 2 \sqrt{455}$