In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20

B. 30

C. 40

D. 50

E. 60

(A) The area of the smallest square must be equal to 2² = 4. Since the ratio of the area of the shaded square (it consists of the shaded region and the smallest square) to the area of the smallest square is 5/2, we can solve for the area of the shaded square by solving for

*A* in the following equation:

*A*/4 = 5/2

*A* = 4 × (5/2)

*A* = 10

So, the area of the shaded square is equal to 10, and each side is equal to √10. Now, all we have to do is recognize that the triangles made at each corner of the large square where, two sides are equal to half the side length and the hypotenuse is equal to the edge length of the shaded square, are right triangles and to use the Pythagorean Theorem to solve for half the edge length. Let each side of the triangle be equal to

*x*, then 2

*x*² = 10, so

*x*² = 5, and

*x* = √5. This means that the entire side length of the larger square is twice the length of a side of that triangle, or 2√5. Finally, the area of a square is equal to a side squared, so the answer must be (2√5 )² = 20, or answer choice (A).

----------

**This question is misleading - "area of the shaded square" should not include area of the smallest square.**