Question

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

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

Hint:

When summing perimeters of shapes within a figure, always check for shared edges—these should be counted only once if the problem asks for the combined external perimeters.

Answer 1

The Correct Option is B

To solve the problem, we begin by understanding the properties of the square ABCD. Given that its area is 4, the side length of the square is \(s\), where \(s^2=4\). Therefore, \(s=2\).

Next, examine the structure of the square divided into four triangles. Each triangle's vertices lie on the vertices and midpoints of the square. We denote the midpoints of sides AB, BC, CD, and DA as points E, F, G, and H, respectively, to form the triangles AEB, BFC, CGD, and DHA.

The side length of each triangle from midpoint to vertex (e.g., AD or DH) is half the side length of the square, so each enclosing triangle has a leg length of 1 unit. The hypotenuse for triangles such as AEB follows the diagonal split from vertex A to midpoint E or B, categorizing them as right-angled isosceles triangles with leg lengths of 1. 

Utilize the Pythagorean Theorem to find the hypotenuse of one such triangle:

c=12+12=2

This calculation yields \(\sqrt{2}\).

Now calculate the perimeter for one such triangle:

P=1+1+2=2+2

Therefore, the total perimeter of all four triangles is:

T=4×(2+2)=8(1+2)

Thus, the sum of the perimeters of the triangles is indeed \(8(1+\sqrt{2})\).