Some amount of billiard balls were arranged in an equilateral triangle. And three balls were extra. When the same set of billiard balls were arranged into triangle in which each side has one more ball than in the first arrangement there were four billiard balls shortage. How many balls were in the set?

25 The number of balls to make an equilateral triangle with the sides of length n is expressed by the formula n(n+1)/2 So to express the number of balls you have is b = n(n+1)/2 + 4 With the larger arrangement you have b = (n+1)(n+2)/2 - 3 Both of those qualities are equal to each other, so set an equation where they're equal. Then solve for n n(n+1)/2 + 4 = (n+1)(n+2)/2 - 3 Distribute the n term on the left (n^2 + n)/2 + 4 = (n+1)(n+2)/2 - 3 Distribute the /2 on the left. 0.5n^2 + 0.5n + 4 = (n+1)(n+2)/2 - 3 Multiply (n+1)(n+2) on right, then distribute the /2 0.5n^2 + 0.5n + 4 = 0.5n^2 + 1.5n + 1 - 3 Subtract 0.5n^2 + 0.5n from both sides 4 = n + 1 - 3 Add 2 to each side 6 = n So the original triangle had sides with a length of 6, for a total number of balls of 6(6+1)/2 = 21 with 4 extra giving 25 balls. Let's check with the next larger triangle 7(7+1)/2 = 28 with 3 balls shortage. Which 25 balls would make happen. So the number of balls in the set is 25

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d. it is a convex pentagon because it has five sides and none of the sides would extend into the inside of the polygon.

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answer: 1 + 1 = 11

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