GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each

There's nothing to stop us from redrawing the figure as an isosceles right triangle.
Let's call it 9x9 units.
The top row is yellow. Base is 1 and height is 1, so area is 0.5 units.
Next row is blue. It's a 1x1 square plus a 1x1 triangle, so area is 1.5 units.
We can keep going adding 1 to the area for each row that we go down. For each row, the area is equal to the row number minus 0.5 units.

Yellow is rows 1, 4, 6, 7, 9, which means area is 0.5+3.5+5.5+6.5+8.5 = 24.5 units
Blue is rows 2, 3, 5, 8, which means area is 1.5+2.5+4.5+7.5 = 16 units

We are told that blue is 64cm^2, so each unit is 4cm^2.
Yellow is therefore 4*24.5 = 98.

Answer choice C.

Ok, So, I approached it like this:
Assuming the area of the top most yellow triangle = 1 unit.
The blue layer below will have 3 such small triangles in it and below that there will be 5 such triangles in the blue region and so on.
So, like a pyramid
Layer 1 = 1 unit
Layer 2 = 3
Layer 3 = 5
Layer 4 = 7
Layer 5 = 9
Layer 6 = 11
Layer 7 = 13
Layer 8 = 15
Layer 9 = 17

Now, blue region is layers 2,3,5 & 8; for this, sum of the unit triangles (area of total blue region) = 3+5+9+15 = 32
And yellow region is layers 1,4,6,7 & 9; for this, sum of the unit triangles (area of total yello region) = 1+7+11+13+17 = 49

The ratio of the area will be Blue: Yellow = 32:49
Now, the Total area of the blue region is 64 cm^2
So, total area of the yellow region will be 49*2 = 98 cm^2
Answer C

With reference to the figure attached -

We know that each triangle is similar as the bases are parallel. So the area of the triangles are inter-related.

Let the area of \(\triangle ABC\) = x

\(\frac{Area(\triangle ADE) }{ Area(\triangle ABC)} = (\frac{1}{2})^2\)

Area(\(\triangle ADE\)) = 4x

Area of quadrilateral BCDE = 4x - x = 3x

Similarly, lets calculate the area of \(\triangle AFG\)

\(\frac{Area(\triangle AFG) }{ Area(\triangle ADE)} = (\frac{3}{2})^2\)

Area(\(\triangle AFG\)) = 9x

Area of quadrilateral FDEG = 9x- 4x = 5x

So we can see that the area of each quadrilateral is 2x in addition to the previous quadrilateral. Hence the area of each quadrilateral can be found as shown.

Area of blue portion : 32x = 64 \(unit^2\)

x= 2 unit

Area of yellow portion: 49x = 49 * 2 = 98 \(unit^2\)

Option C

The correct option is C.

Intuitive Approach.
I suspect the total area of the triangle should be divided by 9, since there are 9 layers and the areas of the 4 blue layers is an integer and all the options have integer values.
Option A+64 = 72+64 = 136 Not evenly divisible by 9
Option B+64 = 81+64 = 145 Not evenly divisible by 9
Option C+64 = 98+64 = 162 Evenly divisible by 9, it is probably the answer.
Option D+64 = 102+64 = 166 Not evenly divisible by 9
Option E+64 = 162+64 = 226 Not evenly divisible by 9


Arithmetic Approach.
The blue trapezium layers are layers, 2,3,5 and 8 counting from the top.
Assume each layer is x in height. Assume the base of the larger big triangle is 9.

Therefore the bottom base of each of the layer, starting from top layer is 1,2,3,4,5,6,7,8,9.

The area of the 4 blue trapeziums using the trapezium formula is (1.5+2.5+4.5+7.5)x=64
Therefore, 16x=64 , x (height) =4. This is only an assumption. There are many different ways to solve the question and many different assumptions.

Calculate the area of the first yellow triangle using area of triangle formula = 0.5*4 =2
The yellow trapezium layers are layers, 4,6,7 and 9 counting from the top.
Calculate the areas of the remaining yellow trapezoids using areas of trapezium formula = (3.5+5.5+6.5+8.5)*4 = 24*4 =96
Add yellow triangle area to yellow trapezium areas = 2+96 =98.
Therefor option C is the answer.

Please refer to the attachment for the solution of this problem.

We can use the similarity of the triangles since the hight of all segments is the same. I demonstrated in the picture below in more detail.