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?Solution
The ratio of the areas of the similar triangles is (6/11)2 =36/121 Given the average area is 706.5 mĀ², the total area is 1413 mĀ². Let the areas be Aā = 36x and A2 = 121x. ATQ- 36x+121x =1413 157x =1413 x =1413/157 Solving, x = 9 mĀ². Thus, Aā = 324 mĀ² and A2 = 1089 mĀ². Difference = 765 mĀ². 20% of the difference = 153 mĀ². So, the answer is 153 mĀ².
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