Question

(14. ABCD is a square of area 4, which is divided into four non overlapping triangles as shown in the fig. Then
the sum of the perimeters of the triangles is
с
(a)8(2+V2)
(b)8(1+V2)
(0)4(1+V2)
(d)4(2 +12)
(e) None of these
​

Answer 1

Answer:

Ans. 4(1+√2)

Step-by-step explanation:

if area of square is 4, then each side is 1.

solve for diagonal by pythagorean theorem.

1²+1²= (diagonal)²

diagonal = √2.

for the side of triangle, diagonal divided by 2.

one side of triangle = √2/2

solve for perimeter of one triangle.

Ptriangle = 1 + (√2/2)+ (√2/2)

               = 1 + √2

Since there are 4 triangles so,

Sum of the perimeters of triangles = 4(1 + √2).

Hope this helps.

Answer 2

Answer:

answer is 4(2+2√2).

i hope its correct...