Each side of a square has length 4 units and its center is at (3,4). If one of the diagonals is parallel to the line `y=x`, then anser the following questions.
The radius of the circle inscribed in the triangle formed by any three vertices is

Options

A. `2(sqrt(2)-1)`
B. `2(sqrt(2)+1)`
C. `sqrt(2)(sqrt(2)-1)`
D. none of these

Answer

1

Explanation

It is given that one of the diagonals of the square is parallel to the line `y=x`
Also, the length of the diagonal of the square is `4 sqrt(2)`
Hence, the equation of one of the diagonals is
`(x-3)/(1//sqrt(2))=(y-4)/(1//sqrt(2))=r = +-2 sqrt(2)`
Hence, `x-3= y-4= +- 2`
or `x=5,1` and `y=6,2`
Hence, two of the vertices are `(1,2)` and `(5,6)`
The other diagonal is parallel to the line `y= -x` , so that its equation is
`(x-3)/(-1//sqrt(2))=(y-4)/(1 //sqrt(2))=r = +- 2 sqrt(2)`
Hence, the two vertices on this diagonal are (1,6) and (5,2)

`AB =4, AC= 4 sqrt(2)`
`:. AE =2 sqrt(2)`
In first figure, `EF+FA = AE`
or `r +sqrt(2) r = 2 sqrt(2)`
or `r= (2sqrt(2))/( sqrt(2)+1)=2sqrt(2)(sqrt(2)-1)`
In second figure, `EG +FG =EF`
or `sqrt(2)r +r=2`
or `r= (2)/(sqrt(2)+1)=2(sqrt(2)-1)`