First, what are “positive, distinct integers”? In Session #1, page 5 we reviewed that the positive integers are: 1, 2, 3, … (Note: it does not include zero). The “distinct” means all five numbers have to be different (no repeats). The median is the number that’s in the middle, so we will start with that: a b c 16 d e Now “d” and “e” have to be bigger than 16, but we want them to be as small as possible: a b c d e 16 17 18 For “a” and “b” we want them to be as small as possible, and of course less than 16. Remember that we can’t use negative integers, and we can’t use zero. That leaves us with “1” and “2”: a b c d e 1 2 16 17 18 The integers add up to: 1 + 2 + 16 + 17 + 18 = 54 Answer: The relationship cannot be determined from the information given. It’s a good strategy to make a sketch in order to gain some insight into the question. Draw a couple rectangles with the diagonal drawn in and remember Pythagorean’s Theorem. You can try different dimensions, say one as a square and another where one leg is a LOT longer than the other. For a Perimeter of 16: For a Perimeter of 20: All sides equal 5 hypotenuse = SQRT(52+52) or about 7.07 2 sides = 7, the other two = 1 hypotenuse = SQRT(72+12) or about 7.07 2 sides = 9, the other two = 1 hypotenuse = SQRT(92+12) or about 9.06 All sides equal to 4 2
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hypotenuse = SQRT(4 +4 ) or about 5.66 So we see that they could be equal with SQRT(50) or the rectangle with a perimeter of 20 could be bigger. And you could make the rectangle with a perimeter of 16 bigger if you made 2 sides = 7.5 and the other two = 0.5 hypotenuse = SQRT( 7.52+0.52), or about 7.52 The Answer is 4 They told us that A, B, and C are squares. The side of a square is equal to the Square Root of its Area (you’re probably more used to thinking of the Area being equal to the Square of a side). If the area of square “A” is 81, then all four sides are 81 = 9 If the area of square “B” is 49, then all four sides are 49 = 7 From these values we can figure out the length of the side of square “C”: Since square “C” has all four sides equal to 2, the Area of “C” = 22 = 4