$ABC$ is an equilateral triangle with side length 4. $M$ is the midpoint of $\overline{BC}$, and $\overline{AM}$ is a diagonal of square $ALMN$. Find the area of the region common to both $ABC$ and $ALMN$.

The area of the shaded region common to both ABD and ALMN is 4.39 m²

Step-by-step explanation:

Given that ABC is an equilateral triangle, we have;

The shaded area = The area formed of the two triangles formed by the cross section of the square and the equilateral triangle

The base length of the two triangles = the height of the equilateral triangle = 4×sin(60°) = 2·√3

The angle formed by the square and the base of the equilateral triangle = 45°

Therefore, the base triangle formed by the square and the equilateral triangle outside the shaded region has angles 60, 45 and 75

Which gives 2/sin(75) = x/sin(60)

x = 2×sin(60)/sin(75) = 1.79 m

The height of the shaded triangle is given as follows

h² + h² = x²

2·h² = 3.22

h² = 1.61

h = √1.61= 1.27

The area of the shaded region = h × Height of the equilateral triangle

The area of the shaded region = 1.27×2·√3 = 4.39 m²

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