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

Question

Competition Mode Question https://gmatclub.com/forum/download/file.php?id=55487 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 Are You Up For the Challenge: 700 Level Questions 2020-06-22_1206.png Untitled.png

yashikaaggarwal · Answer

Let The side of small square = 4
side of bigger square = 2 root 5
x and y are the value of degrees of triangle with side 4 and 2
using similarity, triangles have x and y as their degree measure.
Using Pythagoras theorem, the side of unshade portion of bigger square = 8/root 5

saurabh851 · Answer

Is Ans C (1/5)

Need to check the OA.

Another way to do so....

Area of small square is 4 (suppose)
Side of big square is 2 root5 (applying Pythagoras theorem)
Area of big square = 20

If we reposition A over the small square, total area would be 20
Area of portion left would be = 20-16 = 4

Fraction of area shaded in Red = 4/ 20 = 1/5

Hit Kudos if you like it....

unraveled · Answer

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

Attachment:

Squares.png [ 44.29 KiB | Viewed 3512 times ]

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

nick1816 · Answer

Untitled (1).png 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/ = 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}

indugopal1991 · Answer

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

indugopal1991 · Answer

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

GMATGuruNY · Answer

https://i.ibb.co/YbDCQ6H/overlapping-squares-angles-ABC.png 
Let angle BCA = x and angle CAB = y, as shown in the figure above.

Since x+y=90, the result is the following figure:
 https://i.ibb.co/brk7KL1/overlapping-squares-angles.png 
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:
 https://i.ibb.co/spfnddc/overlapping-squares-sides-ABC.png

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:
 https://i.ibb.co/yPv12sr/overlapping-squares.png

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}