A regular four-pointed star is formed inside a square with area
.
Find the area of the star rounded to the nearest integer.
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( 3 0 0 ) 2 = x 2 + x 2 − 2 ( x ) ( x ) ( cos 1 2 0 )
x 2 = 1 0 0
Area of the star is equal to the area of the square minus the area of the four congruent triangles.
A = 3 0 0 − 4 × 2 1 × x 2 × sin 1 2 0 = 3 0 0 − 2 × 1 0 0 × sin 1 2 0 ≈ 1 2 7