Question

In the adjoining figure, ABCD is an isosceles trapezium. If ZA= 60°,
DC = 20 cm and AD= 13 cm, find the length of AB if altitude from DC
to AB is 12 cm.
Please give correct answer​

Answer 1

Answer:

In triangle ADM.

cos60

0

=

12

AM

AM=12cos60

0

=6cm

So by symmetry BN=6cm

Hence AB=AM+NM+NB=6+16+6=28cm

Now,

sin60

0

=

12

DM

2

3

=

12

DM

⇒DM=6

3

So are of trapezium =

2

1

×(AB+CD)×DM

=

2

1

(28+16)×6

3

=132

3

cm

2

≈16cm

We get,

length of AB=16cm