Explanation

An 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.

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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!

The perimeter of square S is 40 implies each side of S is 10, which also means that the diagonal of square S is 10. In the picture, the diagonals of square S, split square T into 4 isosceles right (45-45-90) triangles, which, as you know, have length ratios of \(x:x:x\sqrt{2}\). As you can see, the sides of square T represent the hypotenuse of each of the smaller 4 triangles, thus each side of triangle T has a length of \(5\sqrt{2}.\)

Since the area of the triangle is \(x^2\), \((5\sqrt{2})^2\) = 25*2 = 50, thus choice (D).

The explanation provided by Sandy is correct.

Regards
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You have pointed out absolutely correctly that area would approach 100 inscribed squares vertices approaches the corners of the outer square. Now visualize the inner square (rotating) with vertices near the outer square vertices and start slowly moving away to the center, 'My bisector assumption', and then keep on moving till the corners of the squares again reach corners of the larger square (after crossing the midpint and ).

As you may have notices as you move the inscribed squares vertices closer to the mid point of the larger square the area keeps on decreasing and once it crosses the mid point the area starts increasing again! This is because of the symmetry of the square in question.

If no information is given how exactly the inner square is you can place it in any orientation as you like such that the area of the inner square is minimum. Hence the inner square is a variable whose exact orientation needs to be defined.

So in order to solve this problem the inner square is assumed to be variable. Just like a rubber! (a variable whose values are not fixed!)
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Just out of curiosity, why can't a square of side 7 be inscribed in a square of side 10? This way the least area would be 49.

Please see my explanation above why you have to follow certain parameters.

Regards
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A square of side 7 cannot be inscribed as a square of side 7 would not touch all the sides of the larger square.

The mininimum size of the square inside is \(5\sqrt{2}\).
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Yes, by definition, inscribed square in a square act like a rubber. For that reason, the question asks "least possible area " not greater than the least.

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