In the isosceles trapezoid ABCD, draw two perpendiculars AM and BL from A and B to the side CD.
Now, LM = BA = 5 m
CD = CL + LM + MD
= CL + 5 + CL (CL = MD)
= 2CL + 5
But, CD = 11
Therefore, 2CL + 5 = 11
2CL = 11 - 5
= 6
CL = 6/2 = 3m
MD = 3m
Now, consider the right triangle AMD.
We have,
[tex]AM^{2} = AD^{2} - MD^{2}[/tex]
= [tex]4^{2} - 3^{2}[/tex]
= 16 - 9
= 7
Hence, height of the isosceles trapezoid = AM = [tex]\sqrt{7} m[/tex]
