As #triangleABD# is a right triangle, we have, using the Pythagorean theorem, #bar(BD)=sqrt(25^2+25^2)=25sqrt(2)#. Because #triangleBDE# is equilateral, this implies that #bar(ED)=25sqrt(2)#.

Next, we can note that #triangleABD# is a #45-45-90# right triangle, meaning the angle #angleADB = 45^@#. Additionally, as #triangleBDE# is equilateral, we have #angleBDE=60^@#. Using those facts, along with #angleADF=180^@#, we have:

#angleEDF = angleADF - angleBDE-angleABD#

#=180^@-60^@-45^@#

#=75^@#

As #triangleEDF# is a right triangle, we can use the sine and cosine functions to say

#bar(EF)/bar(ED)=sin(75^@)#

#=>bar(EF)=bar(ED)sin(75^@)=25sqrt(2)sin(75^@)#

and

#bar(DF)/bar(ED)=cos(75^@)#

#=>bar(DF)=bar(ED)cos(75^@)=25sqrt(2)cos(75^@)#

As #bar(DF)# and #bar(EF)# are the base and height, respectively, of the hatched triangle, we can use the formula #A=1/2bh# to obtain the area of the hatched triangle as

#A =1/2bar(DF)*bar(EF)#

#=1/2(25sqrt(2)cos(75^@))(25sqrt(2)sin(75^@))#

#=625cos(75^@)sin(75^@)#

#=156.25#