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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.

 

 Jan 14, 2021
 #1
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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

 

 

cool cool cool

 Jan 14, 2021

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