Question

ABCD is a trapezium with AB parallel to DC.
If AB = 10 cm, AD = BC = 4 cm and
∆DAB = ∆CBA = 60°, calculate
(i) the length of CD;
(ii) the distance between AB and CD.​

Answer 1

$\Huge\mathfrak{answer \: \: \: ! }$

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