Math, asked by nandangoud2015, 1 month ago

In an isosceles trapezium ABCD , the length of AM =

Answers

Answered by imdipanshuraj
0

Complete the question then ask

Answered by bhavikmittal005
0

Answer:

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

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