Question

If a square paper is
cut into four equal
parts such that the
are not
parts
quadrilaterals, which
of the following can
be the perimeter of
one of the parts?
O 2 + 3v2 units
O 3
3 + 3V2 units
I 2 + 2v3 units
3 + 2/2 units​

Answer 1

Answer:

5.4/3

Explanation:

$02 + 3v2 = 4v \\ 3 + 3.2 = 4.0 \\ = 5.4 3$

Answer 2

Answer:

Option b is correct,  $3+3\sqrt2$

Explanation:

If the parts are not quadrilateral, it will be a triangle. Assuming the square paper of side length 'a'.

Length of diagonal of the square, D =  $\sqrt {a^2+a^2}$

                                                      =  $a\sqrt2$

Half of D  =   $\frac{a\sqrt2}{2}$

                = $\frac{a}{\sqrt2}$

The part which is in the shape of the triangle will have the following side measurements:-

x =  Length of a side of the square   =  $a$

y =  Half of the length of half of the diagonal of the given square =  $\frac{a}{\sqrt2}$

z =  Half of the length of half of the diagonal of the given square  =  $\frac{a}{\sqrt2}$

Therefore, the perimeter of the one part will be

$x+y+z$    =    $a+\frac{a}{\sqrt2}+\frac{a}{\sqrt2}$

                  =    $a+a\sqrt2$

According to the above perimeter, Option b is correct,

$a+a\sqrt2$   $=$  $3+3\sqrt2$.

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