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Page 1 Rebecca Lambert Williams Beyond Pythagoras (Maths Coursework) (Smallest number)2 + (Middle number)2 = (Largest number)2 1. a. Supposedly 52+122=132 52=25 & 122=144 & 132=169 25+144=169 So these numbers satisfy the condition (Smallest number)2 + (Middle number)2 = (Largest number)2 b. Supposedly 72+242=252 72=49 & 242=576 & 252=625 49+576=625 So these numbers satisfy the condition (Smallest number)2 + (Middle number)2 = (Largest number)2 2 a. 5 Area=(5X12) Perimeter=5+12+13=30 2 =30 25 7 Area=(7X24) Perimeter=24+25+7=56 2 =84 24 b. Length of shortest side Length of middle side Length of largest side Perimeter Area 3 4 5 12 6 5 12 13 30 30 7 24 25 56 84 9 40 41 90 180 So 92+402=412 92=81 & 402=1600 & 412=1681 81+1600=1681 So these numbers satisfy the condition (Smallest number)2 + (Middle number)2 = (Largest number) Page 2 Rebecca Lambert Williams Area=(9X40) Perimeter=9+40+41=90 N. Length of shortest side Length of middle side Length of largest side Perimeter Area 1 3 4 5 12 6 2 5 12 13 30 30 3 7 24 25 56 84 4 9 40 41 90 180 5 11 60 61 132 330 6 13 84 85 182 546 2 =180 Investigation I am investigating the relationship between the lengths of the three sides of right-angled triangles, the perimeters and...

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