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non-overlapping triangles

by sanju09 » Mon Mar 12, 2012 2:32 am
The diagonals of square ABCD of area 4 divide the square into four non-overlapping triangles. What is the sum of the perimeters of those triangles?
(A) 4 (1 +√ 2)
(B) 4 (2 +√ 2)
(C) 8 (1 +√ 2)
(D) 8 (2 +√2)
(E) 4 (4 +3√ 2)
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by killer1387 » Mon Mar 12, 2012 2:48 am
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by Anurag@Gurome » Mon Mar 12, 2012 5:42 am
sanju09 wrote:The diagonals of square ABCD of area 4 divide the square into four non-overlapping triangles. What is the sum of the perimeters of those triangles?
(A) 4 (1 +√ 2)
(B) 4 (2 +√ 2)
(C) 8 (1 +√ 2)
(D) 8 (2 +√2)
(E) 4 (4 +3√ 2)
Area of square = 4
If each side of square = s, then s² = 4 implies side of square, s = 2

Image

In triangle ABC, AC² = 2² + 2² = 8 implies AC = 2√2
So, AE = 2√2/2 = √2
Now let us find the perimeter of 1 triangle.
Perimeter of 1 triangle = 2 + √2 + √2 = 2 + 2√2 = 2(1 + √2)
Therefore, perimeter of 4 triangles = 4 * 2(1 + √2) = [spoiler]8(1 + √2)[/spoiler]

The correct answer is C.
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