Let us start with \(A\)

The quadrilateral has equal sides so general perception could be that this is a square however there is one more quadrilateral that has all four sides equal to one another but has an area different from square i.e Rhombus. area of square = side\(^2\) and area of rhombus = \(1/2 D1 * D2\) so a is not true

\(B\)

The quadrilateral contains two right angled triangle hence there are two \(45-45-90\) triangle as such all four sides of the quadrilateral should be equal but this does not necessarily mean the quadrilateral is a square as it could again be a rhombus

\(C\)

In this case we can be certain that the quadrilateral is not a square as all the angles of square are \(90\) degrees. This could be a parallelogram however there could be multiple scenarios for a perimeter of \(16\) for eg. a side of \(6\) and \(2\) or \(7\) and \(1\) or \(5\) and \(3\) with a \(2:1\) ratio of (\(60,120\)) degrees

In all of these variation in sides area will be different also notice that we need a height to get the area of a parellelogram which is not given nor can be derived hence c is also false

\(D\)

For a quadrialteral all angles add to \(360\) degrees hence if all angles are equal \(4x =360\)and as such each angle \((x)\) = \(90\) degrees furthermore given that width is \(0.4\) of length the quadrilateral is not a square or rhombus but a rectangle

we can find the exact length and width of the rectangle \(2(l + 0.4l) = 16\)

or \(1.4 l = 8\)

therefore \(l = 5.70\) approx

similarly we can get the width \(= 2.28\)

and area = \(12.996\)

hence \(D\) is true

\(E\)

Let us take a rectangle with sides \(6\)and\(2\)

Initial perimeter = \(16\)

50% decrease in perimeter so new perimeter = \(8\)

This should bring about \(25\)% reduction in area

original are

a = \(6*2 =12\)

new area = \(12*0.75 = 9\)

However a combination of length and breadth with perimeter \(16\) for a rectangle can be many such as \(7,1\); \(5,3\)

In each case the area is not same and hence when the perimeter and area is reduced by 50% and 25% respectively we get a different area for different set of sides

for \(7,1\) original area = \(7\) after reduction = \(5.25\)

for \(5,3\) original area = \(15\) after reduction = \(11.25\)

We cannot be sure of the area hence E is also false

only D

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This is my response to the question and may be incorrect. Feel free to rectify any mistakes