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

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

Learn more:

Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 ...

brainly.in/question/4741126

A trapezium PBCQ with its parallel sides QC and PB in the ratio of 7

hope it helps u

please mark as brainliat answer

Step-by-step explanation:

Answer 2

Answer:

ABCD is trapezium with AB parallel to DC. If AB=10 cm, AD = BC = 4 cm and ∠DAB=∠CBA=60∘

, then length of CD is equal to

A) 3 cm

B) 6 cm

C) 7 cm

D) 7.5 cm

Correct Answer:

B) 6 cm

Description for Correct answer:

From △ADP,

APAD=cos60∘

=> x4=12

=> x = 2

Therefore, CD = AB - 2x = 10 - 4 = 6 cm