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?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 153.
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides (or perimeters).
The ratio of perimeters = 6:11
The ratio of their areas = (6/11)2 = 36/121
Let the areas of the two triangles be 36x and 121x.
The average of the areas of 2 similar triangles = 706.5 m2
So, (36x + 121x)/2 = 706.5
⇒ 157x / 2 = 706.5
⇒ 157x = 1413
⇒ x = 1413/157 = 9
The two triangles' area is 36 × 9 = 324 m2 and 121 * 9 = 1089 m2.
The difference in areas = 1089 - 324 = 765 m2
So, 20% of the difference = (20/100) × 765 = 153 m2
Therefore, 20% of the difference in the areas of both triangles is 153 m2.