Question 250362
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If the width is fixed at 15, then two times 15 is 30.  Subtract that from 100 to get 70.  That leaves 70 to split between two instances of the length.  Half of 70 is 35.  35 would be fine except that lengths of 35 would give a perimeter of exactly 100.  The problem requires that the perimeter be greater than 100.  So the length has to be greater than 35.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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