gmat_thingie wrote:Same question for "perimeter"
Two equations worthwhile to know:
Sin 30 = 1/2
Sin 60 = sqrt(3)/2
lets label the point on CD where the two points meet as F
lets label the point on BC where the two points meet as E
Area of reactangle = AD X CD
Area of reactangle = (sqrt 3) x (1 + sqrt 3) = sqrt 3 + 3
Area of trapezium = 1/2 * (AD + CE) * CD
Area of trapezium = 1/2 * (sqrt 3 + 1) * (1 + sqrt 3))
Area of trapezium = 1/2 * (3 + 2 * sqrt 3 + 1) = 1/2 * (4 + 2 * sqrt 3) = 2 + sqrt 3
Area of shaded region = (sqrt 3 + 3) - (2 + sqrt 3) = 1
Perimeter:
If we continue we have the following for the sides of the shaded region:
AB + BE + AE
1+sqrt(3) + (sqrt(3) - 1) + sqrt(2^2 + 2^2)
1+sqrt(3) + (sqrt(3) - 1) + 2sqrt2
2*sqrt(3) + 2sqrt2
