If the side of the smallest square is 1 and the side of the largest sq

It's a WRONG question as the information is not sufficient.

The information about the quadrilateral which is Neither smallest nor the biggest is missing. Calling that square would have solved the question this way

base of teh triangle is Diagonal of Square with side 1 i.e. \(√2\)

Height of the traingle is the distance between the diagonal of small square and diagonal of leftmost square = Diagonal of the square with side 2 (because both diagonals are parallel) = \(2√2\)

Area of the triangle = (!/2)*Base*Height \(= (1/2)*(√2)*√(2√2) = 2\)

Answer: Option C

If the image is rotated a bit, the triangle will be in this position. Now the triangle base is diagonal of smaller square and height is diagonal of the larger square.

Base = \(\sqrt{2}\)

Height = \(2\sqrt{2}\)

Area of triangle = \(\frac{1}{2} * \sqrt{2} * 2\sqrt{2}\)

= 2

OPTION : C

Hi GMATinsight

I have solution for this. Will post it in a while. However, you have got the solution for this. then how can the question be wrong?

SOlution is already mentioned in my post above.

I was only mentioning that the question should have specified that every quadrilateral is a square here. ALso, it would have been better to name the vertices and quadrilaterals while giving their references.

can you please elaborate the reasoning behind your statement . thank you

Note that the question mentions largest and smallest squares which makes us assume that the leftmost quadrilateral should be a square too but it should be explicitly mentioned.

Assume that the figure lies on the co-ordinate axis as shown.
The base of the required triangle lies on a line which is 45 degrees to the y axis (the red line)
The third vertex of the triangle lies on the dotted line which is also 45 degrees to the y axis. Hence these two lines are parallel and vertical distance between them will always be the same (the green lines)
No matter what the actual size of the leftmost square (as shown by three different squares), the altitude of the triangle will be the green line only.

Base of the triangle is the diagonal of the square of side 1. So the length of the diagonal is \(\sqrt{2}\)
Altitude of the triangle is the diagonal of teh square of side 2. So the length of the diagonal is \(2\sqrt{2}\)

Area of triangle = \((1/2)*\sqrt{2}*2*\sqrt{2} = 2\)
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