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

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

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

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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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

Given:

1. 7 identical squares, 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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