HOT Competition 4 Sep/8AM: The figure above is made of 7 identical squ

Question

https://gmatclub.com/forum/download/file.php?id=57267 The figure above is made of 7 identical squares of side length of x cm. If the area of yellow region is 56 cm^2, what it the value of x? A. 1 B. 2 C. 3 D. 4 E. 5 M37-92 M37-92 Untitled2.png

Bunuel · Accepted Answer

Official Solution:

https://gmatclub.com/tests-beta/resources/import/img/M37-92.png 
 
The figure above is made of 7 identical squares of side length of  x \ cm . If the area of yellow region is  56 \ cm^2 , what it the value of  x ?

A.  1 
B.  2 
C.  3 
D.  4 
E.  5

Divide the yellow region into blue and red triangles as shown below:
 
  https://gmatclub.com/tests-beta/resources/import/img/M37-92-2.png 
 
The area of the blue triangle is  \frac{1}{2}*base*height = \frac{1}{2}*(2x)*(2x)=2x^2 . 
 
The area of the red triangle is  \frac{1}{2}*base*height = \frac{1}{2}*(3x)*(x)=1.5x^2 . 
 
Thus, the area of the yellow region is  2x^2+1.5x^2=3.5x^2 . 
 
Since given that the area of yellow region is  56 \ cm^2 , then  3.5x^2=56 ; 
 
 \frac{7}{2}*x^2=56 ;
 
 x^2=16 ; 
 
 x=4 .

Answer: D

kush09 · Answer

Solution:
On rearranging the yellow portion, and trying to consolidate them, you will find Yellow portion accounts for 3.5 squares.

As side of square is x, area will be x^2, then
=> 3.5 * x^2 = 56
=> x^2 = 16 
=> x = 4.

IMO answer is D

Superman249 · Answer

IMO ans is D

Refer diagram, marked Area in yellow regions.

As small square are identical,

We have One full area marked yellow; let its area be "a"--------------------(1)
We have two half triangle each with area = "a/2"--------------------(2)
The yellow region in the rightmost corner and left most corner to combine to make the area "a/2"- -----------------(3)
The yellow region in bottom two square in between combine to make the area "a"-------(4)
  
adding all 4= a+2*(a/2)+a/2+a (eq 1,2,3 &4 in order)

we have 
3 .5a=56
=>a=16

a=x^2
=>x=4

Rogermoris · Answer

The idea is very simple, the yellow region is formed by squares and triangles.

So, Let's start with figure 1, area = x^2 - (1)

Figure 2, area =  \frac{1}{2}*b*h  =  0.5x^2  -(2)
Similarly figure 3 has the same area =  0.5x^2  -(3)

Now, we look at figure 4 that compromises of 4 squares. We shall evaluate the area of the region that is not shaded in yellow.

So, that gives us 2 triangles, each with height of x but base of 4x and x respectively.

Total area of yellow region in bottom 4 squares =  4x^2-(\frac{1}{2}*x*4x + \frac{1}{2}*x*x)   -(4)

Adding the areas from (1), (2), (3) and (4), 
The question says that the area of shaded region is 56, hence equating the two terms.

\frac{7}{2}*x^2 = 56

x = \sqrt{16} 
x = 4

Mona2019 · Answer

The yellow region comprises of the area of 3.5 squares .
If a is the side of the square 
Then 7/2a^2= 56
So a is 4 
Answer is D

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jhavyom · Answer

The figure above is made of 7 identical squares of side length of x cm. If the area of yellow region is 56 cm^2, what it the value of x?

A. 1
B. 2
C. 3
D. 4
E. 5

EatMyDosa · Answer

Given :

1. 7 identical  square s, each of  side length  =  x  cm

2.  Area  (yellow region) =  56  cm^2

In the diagram  shown in the  attached screenshot ,

Area  ( shaded region , yellow) =  Area  (7   identical squares  ) –  Area  ( Region 1 + Region 2 + Region 3 + Region 4 ) ----- (1)

Area (7 identical squares)  = 7x^2 --------------- (2)

Region 1 = Region 2 = Region 3  = 1/2 * (Area of Square)

= 1/2 * x^2

Therefore, combined area of three regions = 3*1/2*x^2 -------------- (3)

Area (Region 4)  =  Area of triangle (ABC)

Height of triangle  ABC = Side AB =  x 
 
Base of triangle  ABC = Side BC = x + x + x + x =  4x

So, Area of triangle (ABC) = 1/2 * x * 4x = 2x^2 ---------------- (4)

Substituting values from equation (2), (3), and (4) in equation (1), we get

56 = 7x^2 – (3/2 * x^2 + 2x^2)

= 7x^2 – (7/2)* x^2

= (7/2)* x^2

On simplifying, we get

x^2 = 16
  x = 4   (since x= side length, so -ve value x= -4 is rejected)

Option D  (correct)

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HOT Competition 4 Sep/8AM: The figure above is made of 7 identical squ

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