Question

4. ABCD is a trapezium with AB parallel to
DC. If AB = 10 cm, AD = BC = 4 cm and
DAB = ZCBA = 60°, calculate
(i) the length of CD,
(ii) the distance between AB and CD. (ICSE)
give me the right answer only no promotions ​

Answer 1

Answer:

Distance b/w cd and ab is 14 under root 2

$cd = 4 \sqrt{2 \: cm}$

Step-by-step explanation:

First we construct an triangle inside the trapezium

Than we will find out value of CD with the help of cos45

And than we add both the value Ab and CD