Answer:
Explanation:
1) Check the first picture, to visualize what the question has told. An equilateral triangle DLM inscribed in a square ABCD. A triangle like that has three 60º angles.
2) Looking to DCL and DBE we have two right triangles. The next natural step, is doing Pythagorean theorem.
Since this is an equilateral triangle, all three sides have the same size, we can set them as equal:
3} This allows to replace x for y. We also have, since there is symmetry. We can check that's on the left, a right isosceles triangle (ALE) whose hypotenuse is
. Hence, the hypotenuse would be
. Again starting from the fact this is an equilateral triangle
4) One and only one value must fit. So let's try plugging the minor one.
If one vertex of the triangle is on the square, it can't be an outlying value as the absolute value of 38.64. Furthermore the greatest value for a line segment of this square would be the value of the diagonal, in this case
So, it is 10.35