Question

What is the height of an isosceles trapezoid, if the lengths of its bases are 5m and 11m, and the length of a leg 4m

Answer 1

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,  

$AM^{2} = AD^{2} - MD^{2}$

                  = $4^{2} - 3^{2}$

                  = 16 - 9

                 = 7

Hence, height of the isosceles trapezoid = AM = $\sqrt{7} m$