ExplanationAn inscribed figure is one that is drawn inside another with only the sides touching, so square T will be placed inside of square S.
Hence the target diagram the question, both with square T as large as possible and with T as small as possible:
As we can see, the closer the vertex of square T to a vertex of square A, the larger the area of square T. Square T has its least possible area
when its vertex bisects each side of square S.
Now use square S to find the length of one side of square T:
The triangle that constitutes the space between square T and square S is a 45:45:90 triangle. This helps us determine that the length of one
side of square T is \(5 \sqrt{2}\) .
Finally, find the area of square T:
\(Area = side^2\)
\(Area = (5 \sqrt{2} )^2 = 50\).
Hence option D is correct. Thank you for your efforts. But why it's strange that it must be the bisector! I understand that it it's at circumscribed square corners, then the area would be 100. On the opposite, if it's at the middle of the circumscribed square side, it will be 50. As it approaches the sides of circumscribed vertex at the corners, the area is getting larger and larger. But still the whole thing is strange! Is not the inscribed figure should have a fixed area or it's made of Rubber!