Assuming that O is at the center of the square, and that #AO_|_OD#, we can deduce that triangle #AOD# is a '45-45-90' triangle. This means that angles #/_ODA# and #/_OAD# are 45 degrees.
The triangular relationship for this is that #AO=DO# and we can now solve for #AD# using two methods:
Method 1: Pythagorean Theorem
The Pythagorean Theorem is as follows:
#a^2+b^2=c^2#
This equation is the relationship between the two sides (#a# and #b#) and the hypotenuse, #c#. Writing it with the triangle segments:
#AO^2+DO^2=AD^2#
We know that #AO=DO#, so lets use that relationship:
#AO^2+AO^2=AD^2 rArr 2AO^2=AD^2#
#AD=sqrt(2AO^2) rArr AD=AOsqrt(2)#
#AD=35sqrt(2)# cm
Method 2: Sine/Cosine Relationship
If #/_ODA# is 45 degrees, and #AO# is 35 cm, then the Sine relationship, #"Opposite"/"Hypotenuse"# gives us:
#sin(45)=35/(AD)#
#sin(45)=sqrt(2)/2 = 1/sqrt(2) rArr 1/sqrt(2)= 35/(AD)#
#AD=35sqrt(2)# cm
