CAT 1995 | Question: 54

Question

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:

• A. $8(1+\sqrt{2})$
• B. $8(2+\sqrt{2})$
• C. $4(2+\sqrt{2})$
• D. $4(1+\sqrt{2})$

Answer 1

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.