1. **Problem statement:** We are given a parallelogram with diagonals of lengths 28 and 30, and one side length of 13. We need to find the other side length and the area of the parallelogram. 2. **Formula and important rules:** - Let the sides be $a$ and $b$, with $a=13$ given. - The diagonals $d_1$ and $d_2$ satisfy the relation $$d_1^2 + d_2^2 = 2(a^2 + b^2)$$ - The area $A$ of a parallelogram is given by $$A = ab \sin \theta$$ where $\theta$ is the angle between sides $a$ and $b$. - Using the law of cosines on the diagonals, we can find $\cos \theta$ and then $\sin \theta$. 3. **Find the other side $b$:** Given $d_1=28$, $d_2=30$, and $a=13$, substitute into the diagonal formula: $$28^2 + 30^2 = 2(13^2 + b^2)$$ Calculate squares: $$784 + 900 = 2(169 + b^2)$$ $$1684 = 2(169 + b^2)$$ Divide both sides by 2: $$842 = 169 + b^2$$ Subtract 169: $$b^2 = 842 - 169 = 673$$ Take square root: $$b = \sqrt{673} \approx 25.93$$ 4. **Find the angle $\theta$ between sides:** Using the law of cosines for one diagonal: $$d_1^2 = a^2 + b^2 - 2ab \cos \theta$$ Substitute known values: $$28^2 = 13^2 + (\sqrt{673})^2 - 2 \times 13 \times \sqrt{673} \cos \theta$$ Calculate: $$784 = 169 + 673 - 2 \times 13 \times \sqrt{673} \cos \theta$$ Simplify: $$784 = 842 - 26 \sqrt{673} \cos \theta$$ Rearranged: $$26 \sqrt{673} \cos \theta = 842 - 784 = 58$$ Solve for $\cos \theta$: $$\cos \theta = \frac{58}{26 \sqrt{673}} \approx \frac{58}{26 \times 25.93} = \frac{58}{674.18} \approx 0.086$$ 5. **Calculate $\sin \theta$:** $$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - 0.086^2} = \sqrt{1 - 0.0074} = \sqrt{0.9926} \approx 0.996$$ 6. **Calculate the area:** $$A = ab \sin \theta = 13 \times 25.93 \times 0.996 \approx 335.7$$ **Final answers:** - Other side $b \approx 25.93$ - Area $A \approx 335.7$