You are given a 2D 0-indexed integer array dimensions.
For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i.
Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area.
\n\n\n
Example 1:
\n\n\nInput: dimensions = [[9,3],[8,6]]\nOutput: 48\nExplanation: \nFor index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.\nFor index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.\nSo, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48.\n\n\n
Example 2:
\n\n\nInput: dimensions = [[3,4],[4,3]]\nOutput: 12\nExplanation: Length of diagonal is the same for both which is 5, so maximum area = 12.\n\n\n
\n
Constraints:
\n\n1 <= dimensions.length <= 100dimensions[i].length == 21 <= dimensions[i][0], dimensions[i][1] <= 100给你一个下标从 0 开始的二维整数数组 dimensions。
对于所有下标 i（0 <= i < dimensions.length），dimensions[i][0] 表示矩形 i 的长度，而 dimensions[i][1] 表示矩形 i 的宽度。
返回对角线最 长 的矩形的 面积 。如果存在多个对角线长度相同的矩形，返回面积最 大 的矩形的面积。
\n\n\n\n
示例 1：
\n\n\n输入：dimensions = [[9,3],[8,6]]\n输出：48\n解释：\n下标 = 0，长度 = 9，宽度 = 3。对角线长度 = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487。\n下标 = 1，长度 = 8，宽度 = 6。对角线长度 = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10。\n因此，下标为 1 的矩形对角线更长，所以返回面积 = 8 * 6 = 48。\n\n\n
示例 2：
\n\n\n输入：dimensions = [[3,4],[4,3]]\n输出：12\n解释：两个矩形的对角线长度相同，为 5，所以最大面积 = 12。\n\n\n
\n\n
提示：
\n\n1 <= dimensions.length <= 100dimensions[i].length == 21 <= dimensions[i][0], dimensions[i][1] <= 100sqrt(length2 + width2)."
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