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ABCD and EFGH are squares as shown in the diagram.
Assume side of larger square, ABCD, is equal to 2x. Therefore BF = FC = CG = GD = DH = HA = AE = EB = x.
Using pythagoras theorem, we can also say that EH = HG = GF = FE = \(\sqrt{2}x\)
i.e side of smaller square,EFGH, is equal to \(\sqrt{2}x\)
Cory is moving anticlockwise. Annie and bob are moving towards each other but we don't know if they are moving anticlockwise or clockwise. However, we do know that:
1) Annie and Bob meet for the second time at H
2) Speed (Annie) : Speed (Bob) = 1:3 i.e bob travels three times as fast as annie
Distance between Annie and Bob = 4a. Thus if Annie and Bob start from point B and D, they will cover distance in the ratio of their speed i.e annie will cover distance = a while bob will cover distance = 3a. So they will meet at either point E or F, if they move anticlockwise or clockwise respectively.
Now that they are at the same point, to meet for the second time, they will have to cover the entire length of the square i.e 8a and distance covered will again be a ratio of their speed i.e Annie will cover 2a while Bob will cover 6a. Consequently, they will meet at either H or G, if they continue moving anticlockwise or clockwise respectively.
Since we know that they meet for the second time at H, we can deduce that they moved anticlockwise
Overall, Annie has travelled 3a while Bob has travelled 9a. Ratio of Speed (Annie) : Speed (Cory) = 1:\(5\sqrt{2}\). In the same time, distance covered will be proportional to their speed.
Therefore, distance covered by Cory = \(5\sqrt{2}\) * 3a = 15\(\sqrt{2}\)a
Since answers are given in terms of side of the square ABCD (2a) & side of the square EFGH (\(\sqrt{2}x\)) , lets convert distance covered by cory to similar terms
15\(\sqrt{2}\)a = 7.5 * \(\sqrt{2}\) * side of the square ABCD = 15 * side of the square EFGH
Ans = A