If the area of ABCD is equal to 45, then the perimeter of EFG is equal to x * 1.5 = 45. 45 / 1.5 = 30, so the perimeter of EFG is equal to 30. This means that each side is equal to 10.

The height of the equilateral triangle EFG creates two 30-60-90 triangles, each with a hypotenuse of 10 and a short side equal to 5. We know that the long side of 30-60-90 triangle (here the height of EFG) is equal to √3 times the short side, or 5√3.

We then apply the formula for the area of a triangle, which is 1/2 * b * h. We get 1/2 * 10 * 5√3 = 5 * 5√3 = 25√3.

In general, the height of an equilateral triangle is equal to √3 / 2 times a side of the equilateral triangle. The area of an equilateral triangle is equal to 1/2 * √3s/ 2 * s = √3s²/4.

An equilateral triangle has three congruent sides and results in three congruent angles. This figure results in two special right triangles back to back: 30° – 60° – 90° giving sides of x - x √3 – 2x in general. The height of the triangle is the x √3 side.  So A_triangle = 1/2 bh = 1/2 * 12 * 6√3 = 36√3 cm².

Recall that an equilateral triangle also obeys the rules of isosceles triangles.   That means that our triangle can be represented as having a height that bisects both the opposite side and the angle from which the height is "dropped."  For our triangle, this can be represented as:

Now, although we do not yet know the height, we do know from our 30-60-90 regular triangle that the side opposite the 60° angle is √3 times the length of the side across from the 30° angle. Therefore, we know that the height is 3√3.

Now, the area of a triangle is (1/2)bh. If the height is 3√3 and the base is 6, then the area is (1/2) * 6 * 3√3 = 3 * 3√3 = 9√(3).

The area of a triangle is (1/2)*base*height. We know that the area = 15 cm, and the base is 5 cm, so:

15 = 1/2 * 5 * height

3 = 1/2 * height

6 = height

The sum of the lengths of the shortest sides of a triangle cannot be less than the third side.

3 + 5 = 8 < 9, so 3 can't be a value of x.

There are 2 components to this problem. The first, and easier one, is recognizing how much of the perimeter the equilateral triangles take up—since there are 40 triangles in total, there must be 40 – 4 = 36 of these triangles. By observation, each contributes only 1 side to the overall perimeter, thus we can simply multiply 36(5) = 180" contribution.

The second component is the corner triangles—recognizing that the congruent sides are adjacent to the 5-inch equilateral triangles, and the congruent angles can be found by

180 = 30+2x → x = 75°

We can use ratios to find the unknown side:

75/5 = 30/y → 75y = 150 → y = 2''.

Since there are 4 corners to the square rug, 2(4) = 8'' contribution to the total perimeter. Adding the 2 components, we get 180+8 = 188 inch perimeter.

An altitude drawn in an equilateral triangle will form two 30-60-90 triangles. The height of equilateral triangle is the length of the longer leg of the 30-60-90 triangle. The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90.

The ratio of the length of the hypotenuse to the length of the longer leg of a 30-60-90 triangle is  

The length of the longer leg of the 30-60-90 triangle in this problem is

Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12.

If  and  have the same perimeter, , and , it follows that . The three triangles have the same sidelengths, setting the conditions for the Side-Side-Side Congruence Postulate. 

If , then, since the sum of the degree measures of both triangles is the same (180 degrees), it follows that . Since  and  are congruent included angles of congruent sides, this sets the conditions for the SAS Congruence Postulate. 

In both of the above cases, it follows that .

However, similarly to the previous situation, if , then it follows that , meaning that we have congruent sides and congruent nonincluded angles. However, this is not sufficient to prove congruence.

"Statement III" is the correct response.