Question

ABCD is a trapezium with parallel to DC if AB =10cm, AD=BC=4cm and angle DAB=Angle CBA=60 degree calculate the length of CD , the distance between AB and CD​

Answer 1

the distance between AB and CD​ = 2√3 cm  & Difference between AB & CD =4 cm

Step-by-step explanation:

Let say DM ⊥ AB

=> Sin60  = DM/AD

=> √3 / 2 =  DM/4

=> DM = 4√3 /2

=> DM = 2√3

the distance between AB and CD​ = 2√3

Cos 60 = AM/AD

=> 1/2 = AM/4

=> AM = 2

Similarly  CN ⊥ AB

=> BN = 2

CD = AB - (AM + BN)

=> CD = 10 - (2 + 2)

=> CD = 6

Difference between AB & CD = 10 - 6 = 4 cm

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Answer 2

60=AM/AP

1/2=AM/4

AM=2

BN=2

CD=AB-(AM+BW)

CD=10-(2+2)

CD=6

Difference between AB & CD = 10-6 = 4Cm