A square has sides of length 2 cm - Leaving Cert Mathematics - Question b - 2017

The perimeter of the first square is:

$ext{Perimeter}_1 = 4 imes 2 = 8 ext{ cm}$

The perimeter of the second square is:

$ext{Perimeter}_2 = 4 imes ext{√2} = 4 ext{√2} ext{ cm}$

In general, the perimeter of the n-th square can be expressed as:

ext{Perimeter}_n = 4 imes 2 imes rac{1}{ ext{√2}^{n-1}} = rac{8}{ ext{√2}^{n-1}}

To find the sum of the perimeters as the number of squares approaches infinity, we use the formula for the sum of a geometric series:

S_∞ = rac{a}{1 - r}

where:

• a is the first term (8 cm),
• r is the common ratio

For our case:

• For square 2, the common ratio is: r = rac{1}{ ext{√2}}

Thus, substituting into the geometric series formula gives:

S_∞ = rac{8}{1 - rac{1}{ ext{√2}}} = rac{8}{rac{ ext{√2}-1}{ ext{√2}}} = rac{8 ext{√2}}{ ext{√2}-1}

To rationalize the denominator:

S_∞ = rac{8 ext{√2}( ext{√2}+1)}{( ext{√2}-1)( ext{√2}+1)} = rac{8 ext{√2}( ext{√2}+1)}{2-1} = 8 ext{√2}( ext{√2}+1) = 8 + 8 ext{√2}

So, the final result in the form a + b√c is: $8 + 8 ext{√2} ext{ cm}$