Question

The value of $\alpha$ for which side of nth square equals the diagonals of $(n+1)$ th square is-

• A. 1/3
• B. 1/4
• C. 1/2
• D. $1/\sqrt{2}$

Answer 1

Answer: (C)

1/2

Given that $a_n$ denotes the length of side of nth square.
$\Rightarrow$Diagonal of $(n+1)th$ square $=\sqrt{2}{a}_{n+1}$
Side of nth square is equal to diagonals of (n+1)th square.
$\Rightarrow a_n=\sqrt{2}{a}_{n+1}$
$\Rightarrow a_n^2=2a_{n+1}^2$.......(1)
Side of nth square $a_n$ is divided in $\alpha :1-\alpha$ ratio. Hence the sides of nth square and (n+1)th square are related such that $a_{n+1}^2=a_n^2({\alpha }^2+(1-\alpha {)}^2)$......(2)
From (1) and (2) we get,
$a_n^2=2(2{\alpha }^2-2\alpha +1)a_n^2$
$\Rightarrow \alpha =1/2$.