the base of an isosceles triangle is 2 root 2 if the medians to the equal sides intersect at right angles then the height of the triangle in

Look at the following :

I think you want to find the height of the triangle containing the isosceles triangle

AB = 2sqrt (2)   

Let  AEB  be isosceles......with  angles ABE, BAE  equal....so  AE  = BE

Bisect  AEB.......by SAS......BH = AH = EH

And ABE  = 90......so  ABE, BAE  = 45

And BH =  sqrt (2).......so EH = sqrt (2)

Let E be the intersection of medians BG and AF

Draw GF through D.....drop  perpendiculars from F, G to interesect AB at I and J  and by AAS,  triangle FIB is congruent to  triangle GJA.  ...then FI  = GJ ...so.. GF  is parallel to AB

And triangle AGF  is similar to triangle ABC.....so  DF  = 1/2  of HB  =  sqrt(2)/2

So  CD  =   1/2 CH   

And EDF  is right  and  angle DFE  = BAE  =  45

So DEF  = 45...so triangle EDF  is 45 - 45 - 90  with DF = DE  = sqrt (2) / 2

So.....HD  = EH + DE  =  sqrt (2) +  sqrt (2) / 2

And..........By  similar triangles......  2 CD  = 2 HD  =  HC

So  .....HC  =    2 [  sqrt (2) +  sqrt (2) / 2 ]   =  3sqrt (2)  = height of ABC