Easy solution with dictionary tricks for python3
Complexity\n- Time complexity:\nO(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:\nO(n)\n Add your space complexity here, e.g. O(n) \n\n# 
# Complexity\n- Time complexity:\n$$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nfrom math import sqrt\nclass Solution:\n    def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int:\n        dim_length = dict()\n        index = 0\n        for dim in dimensions:\n            length = int(sqrt((dim[0] ** 2) + (dim[1] ** 2)))\n            area = (dim[0] ** 2) + (dim[1] ** 2)\n            dim_length[index] = length, area, dim[0]*dim[1]\n            index += 1\n        max_length = max(dim_length, key=dim_length.get)\n        return dimensions[max_length][0] * dimensions[max_length][1]\n```
['Python3']