Vertex A of the large square is at the midpoint of the small square’s

Vertex A of the large square is at the midpoint of the small square’s side, as shown above. What fraction of the total area is shaded in red?

A. 1/7
B. 1/6
C. 1/5
D. 1/4
E. 1/3

Let the sides of Sq. EFBG be 'a'.
\(AF = \frac{a}{2}\)
Now, △AFB and △JEA are similar triangle(AA). \(\frac{FB}{AF} = \frac{EA}{EJ}\)
\(\frac{a}{\frac{a}{2}} = \frac{\frac{a}{2}}{EJ}\)
\(EJ = \frac{a}{4}\)

Also, △AFB and △GIB are similar triangle(AA). FB/AB = IB/GB
\(\frac{a}{\frac{\sqrt{5}}{2}a} = \frac{IB}{a}\)
\(IB = \frac{2a}{\sqrt{5}}\)
Therefore \(IC = \frac{\sqrt{5}}{2}a - \frac{2}{\sqrt{5}}a = \frac{a}{2\sqrt{5}}\)

Area of shaded region = Area of △JEA + Area of Rec. ICDH
= \(\frac{1}{2} * \frac{a}{2} * \frac{a}{4} + \frac{a}{2\sqrt{5}} * \frac{\sqrt{5}}{2}a\)
= \(\frac{5a^2}{16}\)

Total area = Area of Sq.ABCD + Area of △AFB + Area of △JEA
= \((\frac{\sqrt{5}}{2}a)^2 + \frac{1}{2} * a * \frac{a}{2} + \frac{1}{2} * \frac{a}{2} * \frac{a}{4}\)
= \(\frac{25}{16} a^2\)

Required ratio = \(\frac{\frac{5}{16}a^2}{\frac{25}{16} a^2} = \frac{1}{5}\)

Suppose length of the side of small square = 1

triangle CBA is similar to triangle AJI

Hence, Area of CBA/Area of AJI = \(\frac{(BC)^2}{(AJ)^2} = \frac{4}{1}\)

Area of AJI/[Area of AJI +Area of CBA] = 1/5 .........(1)

Now, extend CD to F. Since, triangle CEJ is similar to triangle CBA, \(JE = \frac{√5}{2}\).

CJ = \(\sqrt{CE^2 + JE^2 }\) = 2.5

Hence, DJ = 2.5-2 = 0.5

Area of FEGH / Area of ACEG \(=\frac{DJ}{JC} = \frac{0.5}{2.5} = \frac{1}{5}\)........(2)

From (1) and (2)

Area of shaded region/ total area = \(\frac{1}{5}\)

If we consider smaller square have side, 10, area = 100.

By ball parking we are considering the bigger one having side, 11, area = 121.

So, square shaded area = 11*2 and triangle area - 1/2*5*2.5=6.25

Total shaded area = 22+6.25 = 22.25.

Small swuare comes into bigger one, so shaded zone = 28.25/121 = approximately 25% = 1/4

If we consider smaller square have side, 10, area = 100.

By ball parking we are considering the bigger one having side, 11, area = 121.

So, square shaded area = 11*2 and triangle area - 1/2*5*2.5=6.25

Total shaded area = 22+6.25 = 22.25.

Small square comes into bigger one, so shaded zone = 28.25/121 = approximately 25% = 1/4

Let angle BCA = x and angle CAB = y, as shown in the figure above.

Since x+y=90, the result is the following figure:

Every triangle in the figure above has the same combination of angles: x-y-90
Implication:
The triangles are all SIMILAR.

Let BC=4, yielding the following figure:


In triangle ABC, the sides are in the following ratio: 2-4-2√5.
Since the triangles are all similar, each must be composed of three sides in this ratio.
The result is the following figure:


Thus:
Red triangle \(= \frac{1}{2}*1*2 = 1\)
Red rectangle = ACKJ - ACDH \(= (2√5)^2 - (2√5 * \frac{8}{√5}) = 20-16 = 4\)
Total area = ACKJ+ AFG + ABC \(= (2√5)^2 + (\frac{1}{2}*1*2) + (\frac{1}{2}*2*4) = 20+1+4 = 25\)
\(\frac{red}{total} = \frac{1+4}{25} = \frac{5}{25} = \frac{1}{5}\)

­­­­Hi! Please find attached pictures for an answer using coordinate geometry..equation of lines and perpednicular distance of a point from the line..

PS. I understand the solution goes beyond GMAT topics but its usefull for engineers!­