In the figure, \(AB\) is parallel to \(CD\). If \(AB =4 \text{ cm}\) and \(AD=2 \text{ cm}\), angles \(D\) and \(C\) are \(30^\circ\) and \(60^\circ\) respectively. Find the area of the figure (in \(\text{cm}^2\)).
I've been stuck on this.
+128474 Draw AE perpendicuar to DC
Triangle AED will form a 30-60-90 right triangle AE is opposite the 30° angle so it will be 1/2 the length of hypotenuse AD = (1/2)(2) = 1 = height of trapezoid
And DE will be opp the 60° angle so its length = AE * sqrt (3) = 1*sqrt (3) = sqrt (3)
Similarly draw BF perpendicular to DC
This will also form a 30-60-90 right triangle
BF =1
And FC is opposite the 30° angle so its length = 1/sqrt (3) = sqrt (3) / 3
So DC = DE + AB + FC = sqrt (3) + 4 + sqrt (3)/3 = 4 + (4/3)sqrt (3)
The area of the trapezoid= (1/2) height (sum of bases) =
(1/2) (1) ( 4 + 4 + (4/3)sqrt (3) ) =
( 8 + (4/3)sqrt (3) ) / 2 =
[4 + (2/3) sqrt (3)] cm^2
