Question

In the given isosceles trapezium ABCD, AB~n DC and AD = BC, If D = 60°, AB = 18 cm and AD
= 12 cm, find the length of side DC.
​

Answer 1

Step-by-step explanation:

Through B, draw a straight line parallel to AD which meets CD at E.

Now, since AB∥DE and AD∥BE,

∴ABED is a parallelogram.

Thus, ED=AB=18cm

As, BE∥AD and CD is a transversal,

∴∠BED=∠D=60

o

(∵∠ABE=∠D).

Since, in an isosceles trapezium, the base angles are equal, ∠C=∠D=60

o

.

In ΔBEC,∠BEC+∠ECD+∠CBE=180

o

(Angle sum property of a triangle)

⇒60

o

+60

o

+∠CBE=180

o

⇒120

o

+∠CBE=180

o

⇒∠CBE=180

o

−120

o

⇒∠CBE=60

o

As the measure of ∠BEC=60

o

,∠ECB=60

o

and ∠CBE=60

o

,ΔCBE is an equilateral triangle, Thus

∴CE=BC=12cm(BC=AD=12cm)

Now, CE+ED=12+18

⇒DC=30cm