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(A) Let $n$ be the number of balls in each side of the equilateral triangle.The total number of balls in the triangle is given by the sum of the first

Vedclass · Accepted Answer

$190$

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(A) Let $n$ be the number of balls in each side of the equilateral triangle.The total number of balls in the triangle is given by the sum of the first $n$ natural numbers: $S = \frac{n(n+1)}{2}$.According to the problem,adding $99$ balls allows them to form a square with side length $(n-2)$.Thus,the equation is: $\frac{n(n+1)}{2} + 99 = (n-2)^2$.Multiplying by $2$: $n^2 + n + 198 = 2(n^2 - 4n + 4)$.$n^2 + n + 198 = 2n^2 - 8n + 8$.Rearranging the terms: $n^2 - 9n - 190 = 0$.Factoring the quadratic equation: $(n - 19)(n + 10) = 0$.Since $n$ must be positive,$n = 19$.The number of balls used to form the equilateral triangle is $\frac{19(19+1)}{2} = \frac{19 	imes 20}{2} = 190$.

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