+817 Inside a square with side length $10$, two congruent equilateral triangles are drawn such that they share one side and each has one vertex on a vertex of the square. What is the total area of the two equilateral triangles?
+129771 The height of one of these triangles = (1/2) diagonal of the square = (1/2) 10 *sqrt (2) = 5sqrt 2
Half the base length = 5 sqrt (2) / sqrt (3) ) ...so the whole base length = 10sqrt (2) / sqrt (3)
Total area of both triangles = 2 (1/2) ( 5 sqrt (2)) (10 sqrt (2) / sqrt (3) =
50 * 2 / sqrt (3) =
100 /sqrt (3) ≈ 57.735
