Question

EFGH is an isosceles trapezium. If /G is 60 , HE = 20cm, and GH = 15cm, find the length of FG.​

Answer 1

Answer:

In triangle ADM.

cos600=12AM 

AM=12cos600=6cm

So by symmetry BN=6cm

Hence AB=AM+NM+NB=6+16+6=28cm

Now,

sin600=12DM

23=12DM

⇒DM=63

So are of trapezium =21×(AB+CD)×DM

=21(28+16)×63=1323cm2≈16cm

We get,

length of AB=16cm