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    Question

    The average of the areas of 2 similar triangles is 706.5

    m2 whose perimeters are in the ratio of 6: 11. What is 20% of the difference (in m2) in areas of both triangles?
    A 153 Correct Answer Incorrect Answer
    B 146 Correct Answer Incorrect Answer
    C 124 Correct Answer Incorrect Answer
    D 136 Correct Answer Incorrect Answer

    Solution

    The ratio of the areas of the similar triangles is (6/11)2 =36/121 Given the average area is 706.5 m┬▓, the total area is 1413 m┬▓. Let the areas be AтВБ = 36x and A2 = 121x. ATQ- 36x+121x =1413 157x =1413 x =1413/157 Solving, x = 9 m┬▓. Thus, AтВБ = 324 m┬▓ and A2 = 1089 m┬▓. Difference = 765 m┬▓. 20% of the difference = 153 m┬▓. So, the answer is 153 m┬▓.

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