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ABCD is a square of area 4, which is divided into four non-overlapping triangles as shown in the figure. The sum of the perimeters of the triangles is:

  1. $8(1+\sqrt{2})$
  2. $8(2+\sqrt{2})$
  3. $4(2+\sqrt{2})$
  4. $4(1+\sqrt{2})$
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Let 'a' be the side of the square

ABCD is a square of $a^{2}$ = 4 $\Rightarrow$ a = 2.

AC = BD = $2\sqrt{2}$

perimeters of four triangles = AB + BC + CD + DA + 2(AC + BD) = 8 + 2 ($2\sqrt{2}$ + $2\sqrt{2}$)

   = 8 ( 1 + $2\sqrt{2}$)

Option A is the Correct Answer.

 

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