Question

what is the maximum number of squares of size 2x2 that can be fit in a right angled isosceles triangle of base b.

Answer 1

Since the triangle is isosceles, the given base would also be equal to the height. Now in the diagonal part, we would always need an extra length of 2 units in both height and base of the triangle to accommodate a triangle. (The CF and AM segment of the triangle in the image. The part that does not contribute to any square). In the remaining length of base, we can construct length / 2 squares. Since each square is of 2 units, same would be the case of height, there is no need to calculate that again.

So, for each level of given length we can construct “(length-2)/2” squares. This gives us a base of “(length-2)” above it. Continuing this process to get the no of squares for all available “length-2” height, we can calculate the squares.

while length > 2

answer += (length - 2 )/2

length = length - 2

For more effective way, we can use the formula of sum of AP n * ( n + 1 ) / 2, where n = length – 2

Hope it helps

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Have a great day ahead

Answer 2

Right Angled Isosceles Triangle

Explanation:

• A side of the square should be parallel to the base of the triangle  
• Since the triangle is isosceles, the given base would likewise be equivalent to the tallness. Presently in the corner to corner part, we would consistently require an additional length of 2 units in both tallness and base of the triangle to oblige a triangle. (The CF and AM section of the triangle in the picture. The part that doesn't add to any square). In the rest of the length of base, we can build length/2 squares. Since each square is of 2 units, same would be the situation of stature, there is no compelling reason to figure that once more  
• In this way, for each degree of given length we can develop "(length-2)/2" squares
• This gives us a base of "(length-2)" above it. Proceeding with this procedure to get the no of squares for all accessible "length-2" stature  while length > 2  answer += (length - 2 )/2  length = length - 2