I am delighted to say that there are actually nine ways! 12 x 483 = 5796 42 x 138 = 5796 18 x 297 = 5346 27 x 198 = 5346 28 x 157 = 4396 39 x 186 = 7254 48 x 159 = 7632 4 x 1738 = 6952 4 x 1963 = 7852 I wrote some javascript (don't know any other programming codes!) to come up with the first seven. It was little more than a search through the various permutations, with only a few shortcuts. (It would be interesting if anybody could think of a neater way of finding the solutions.) I overlooked the last two - Henry Ernest Duden takes sole credit for them. The list is exhaustive - there are no other solutions.
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Oh, isn't that just a happy little math problem! If the median of an isosceles trapezoid is 5.5, then the sum of the bases is twice the length of the median. So, the bases could be 6 and 5, or 7 and 4, or 8 and 3, or any other pair of integers that add up to 11. Just remember, there are many happy little solutions to this problem!
the answer.
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There are many limitations that mathematical models have as problem solving tools. There is always a margin of error for example.
in this problem solving in a rectangle width is 3 and the perimeter is 16 what is he length