GMAT CLUB OLYMPICS: Seven identical square are drawn in the rectangle

Question

https://gmatclub.com/forum/download/file.php?id=62921 Seven identical square are drawn in the rectangle as shown above. What is the area of one square ? A. 4 B. 9 C. 16 D. 25 E. 36 Untitled11.png

Bunuel · Accepted Answer

Official Solution:

https://gmatclub.com/tests-beta/resources/import/img/M37-29.png 
 
Seven identical square are drawn in the rectangle as shown above. What is the area of one square ?

A.  4 
B.  9 
C.  16 
D.  25 
E.  36

Consider little right triangle at the bottom right corner of the rectangle. Say its legs are  x  and  y  and the hypotenuse is  z . Notice that the hypotenuse coincides with the side of the square. Place (draw) such right triangles on the sides of other squares as shown in the image below:
 
  https://gmatclub.com/tests-beta/resources/import/img/M37-29-2.png 
 
Notice that the vertical width of the rectangle equals to  y+x+x+x+y=17  and horizontal length equals to  x+y++y+x+y+y+y=26 
 
Solving  3x+2y=17  and  2x+5y=26  gives  x=3  and  y=4 . Thus,  z=\sqrt{4^2+3^2}=5 .
 
Therefore, the area of one square is  5^2=25

Answer: D

cs15811581 · Answer

A quick way would be estimation

Take the rightmost squares. Green and blue are in straight line and then a purple one on their right. Of course the squares diagonals are not parallel to side 17, but the options are far apart to allow taking them as parallel.

If we take sides of squares as x, then their diagonal will be x \sqrt{2} .

So the side 17 is approximately the size of 2 and half diagonals.
 
Thus  2.5*x\sqrt{2}=17…….x=\frac{17\sqrt{2}}{5} 
Area= x^2=\frac{289*2}{25}=11.6*2=23

Closest is 25.

D

Lizaza · Answer

Oh this one is so nice! What you need to do is notice a little triangle 'cut out' in the lower right corner. Basically, if we name two catets x and z, it leaves us with the hypothenuse of y, which also equals the side of the square. Because this right triangle rests on the hypothenuse, we can simply recreate it many times in each of the little squares.

I show it on the attached image. Basically, if we do so, we create an unbroken chain going from one side of the rectangle to the opposite one, both vertically and horizontally, and we can formulate two equations of width and height based on x and z. I tried to show the horizontal chain in green, and the vertical one in blue.

Now let's calculate:
17 = x + z + z + z + x = 2x + 3z
26 = z + x + x + z + x + x + x = 2z + 5x
Let's substitute z from first equation and put into second:
z =  \frac{17-2x }{ 3} 
26 = 5x +  \frac{2(17-2x) }{ 3}  =  \frac{11x + 34 }{ 3} 
11x = 78 - 34 = 44
x = 4
z =  \frac{17-2*4 }{ 3}  = 3

Therefore, y = 5 (which is the square side)
Square area equals 5*5 = 25.  Answer D.

Lenel · Answer

Consider the bottom right triangle, its hypotenuse is the side of each square (7 identical squares)
Let the hypotenuse be  x , and sides be  y  and  z 
 x^2  is the area of the square
 z^2 + y^2 = x^2

From the image attached we can find that for width of the rectangle
 y + z+z+z+y = 17 
 2y+3z = 17

From the image attached we can find that for length of the rectangle
 z + y+y+z+y + y+y= 26 
 5y+2z = 26

Equating  2y+3z = 17  and  5y+2z = 26 
 y=4  and  z=3

From the right angle triangle,  x^2 = z^2 + y^2  
  x^2 = 3^2 + 4^2  
  x^2=25 
Area of each square  x^2 = 25

Answer : D

Kratosgmat · Answer

A great question that tests spatial arrangement. here is my 2 cents

consider the small triangle on the bottom right corner
Assume
longer leg is y
smaller leg is x
hypotenuse (which is side of sq) is z

3x+2y=17
5y+2x=26

solving we get x=3 and y = 4 
hence z is 5 and area of one sq is 25

sujoykrdatta · Answer

If we join the diagonal of the rectangle: the top left corner and the bottom right corner: 
Length of the diagonal = (26²+17²)^0.5 = 31.1

If we assume each side of the squares as a, the diagonal 31 is approximately 6 such sides (i.e. 6s) starting from the top left corner till you reach the hypotenuse of the small corner triangle on the right. 
Assuming that both legs of this triangle are almost the same, the perpendicular from the 90⁰ vertex to the hypotenuse is s/√2 = 0.7s

Thus, we have: 6s + 0.7s = 31.1
=> s = 31.1/6.7 = 4.6
=> Area = 4.6² = 21.2 approx

Closest option is 25

Answer D

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GMAT CLUB OLYMPICS: Seven identical square are drawn in the rectangle

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