Question

4. ABCD is a trapezium with AB parallel to
DC. If AB = 10 cm, AD = BC = 4 cm and
ZDAB = ZCBA = 60°, calculate
(1) the length of CD,
(ii) the distance between AB and CD. (ICSE)​

Answer 1

Answer:

let we draw perpendicular from D and C on AB

perpendicular s are DE and CF

Step-by-step explanation:

from triangle ADE,

cos 60=AE/AD

1/2=AE/4

AE=4/2=2

similarly from triangle CBF,

BF=2cm

so,

CD={10-(2+2)}cm=6cm

DE/AD=sin60

DE=√3/2.4=2√3

the distance between AB and CD is 2√3cm