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The perimeter of an equilateral triangle is 11 inches more than the perimeter of a square, and the side of the triangle is 6 inches longer than the side of the square. Find the side of the triangle

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u can put that to two equations as (a = side of triangle, b = side square) 3a =4b+11 a=b+6
Pathagorean Theorem: \[A^{2}+B^[2]=C^[2]\]
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not yet i'm having trouble with it
(a = side of triangle, b = side square). 3a is the perimeter of triangle, 4b perimeter of square. from question , a perimeter relationship can be found as 3a =4b+11 (perimeter of triangle = perimeter of square+ 11)
similarly the side relationship can be found out as a= b+6 nw u hav 2 equations , can't u solve it.........

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