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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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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.
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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using ∆ similarity we can say that ∆ Yellow area b*h/2 and that of blue ∆ bxhx/2 ; bh*x^2/2
we can re write as A2 = A1.x^2
Total blue ∆ area of region
figure can be divided into 17 ∆ of yellow and blue portions
for blue ∆
A+3A+9A+15A = 64
A = 64/32 ; 2
and total Yellow ∆ =
A+7A+11A+13A+17A ; 49*a ; 49*2 ; 98
OPTION C


Bunuel wrote:

A triangle is cut by 8 lines, each of which is parallel to the base of the triangle, as shown above. If the height of each of the 9 resulting segments is the same and the area of the blue segments is 64 cm^2, then what is the area of the yellow segments ?

A. 72
B. 81
C. 98
D. 102
E. 162


 


This question was provided by GMAT Club
for the GMAT Club World Cup Competition

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2022.png
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Attachment:
Screenshot 2022-07-12 at 9.16.45 PM.png
Screenshot 2022-07-12 at 9.16.45 PM.png [ 75.1 KiB | Viewed 3090 times ]


Given: A triangle is cut by 8 lines, each of which is parallel to the base of the triangle, as shown above.
Asked: If the height of each of the 9 resulting segments is the same and the area of the blue segments is 64 cm^2, then what is the area of the yellow segments ?

To find out length mid-point of segments, let us assume base of the triangle as L.
Length of mid point of segments starting from top = {L/18, 3L/18, 5L/18, 7L/18, 9L/18, 11L/18, 13L/18, 15L/18, 17L/18}

Area of blue segments = h/2 * L/18 (3+5+9+15) = 32hL/36 = 64
hL/36 = 2
Area of yellow segments = h/2 * L/18 (1+7+11+13+17) = 49hL/36 = 49*2 = 98

IMO C
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Hello everyone,

The correct answer is C.
I have attached the picture which shows the detailed explanation of the solution.
The answer is 98.
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Quote:
A triangle is cut by 8 lines, each of which is parallel to the base of the triangle, as shown above. If the height of each of the 9 resulting segments is the same and the area of the blue segments is 64 cm^2, then what is the area of the yellow segments?

A. 72
B. 81
C. 98
D. 102
E. 162


If the height is the same, then the base shall be in proportion with the whole triangle as well.

Say for the tiniest triangle, we compare it with the entire triangle. The height is 1/9th the height of the entire triangle, and since they are similar (as the lines drawn are parallel), the base is 1/9th of the base of the entire triangle.

So, the areas of the triangles, starting from the tiniest and ending with the entire triangle would be in the ratio of:
1:4:9:16:25:36:49:81

Therefore, the ratio of the area of the segments would be:
1:3:5:7:9:11:13:15:17.

On solving the sum of these ratios with the given info (sum of areas of blue segments is 64 cm^2), you get the answer as option C, i.e., 98
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Let's start with naming these trapezium (formed by the parallel line) as 1, 2, 3... till 9 (1 being the top & 9 being the bottom).

As per the question, the blue marked area, which is 2,3,5 & 8 as per the numbering, has an area of 64 unit squares.

The ratio of 1st triangle (small yellow colour from top) to 2nd triangle (small 1st blue colour from top) is 1:4. Use the triangle similarity to find this ratio. Similar, every triangle below the top triangle has a ratio of the square of consecutive integers.

Now, the Area of triangle contain trapezium 1,2,3..... will be in the ratio of 1:4:9:16....81, i.e. 1:2:3:4:5.....9 = 1:4:9:16:25:36....81

To find the area of the trapezium, subtract the area of the bigger one from the lower one. The area of the topmost triangle is 1, and the triangle below is 4. So the area of the trapezium (number 2) will be 4-1=3. Similarly, find the area of all trapezium.

Now the trapezium with blue colour will be 32 square unit, & yellow will be 49 square unit. Also, 32 square unit is 64 square cm, so 49 will be 98.

Hence C is the Answer
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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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
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Bunuel wrote:

A triangle is cut by 8 lines, each of which is parallel to the base of the triangle, as shown above. If the height of each of the 9 resulting segments is the same and the area of the blue segments is 64 cm^2, then what is the area of the yellow segments ?

A. 72
B. 81
C. 98
D. 102
E. 162


 


This question was provided by GMAT Club
for the GMAT Club World Cup Competition

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Attachment:
2022.png



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.
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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This is another way of solving this problem..

Hope this helps..

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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Using concept of similar triangles,

Area (small triangle) / Area (larger triangle) = (1/2) ^ 2 (since height of larger triangle is twice of smaller triangle)

Thus, Area (small triangle) = 1 m
Area (larger triangle) = 4 m
Area (balance polygon) = 4 m - 1 m = 3m

Hence, Areas of triangles (smallest to largest) = 1m, 4m, 9m, 16m, 25m, 36m, 49m, 64m, 81m
Area of 8 portions of largest triangle (top to bottom) = 1m, 3m, 5m, 7m, 9m, 11m, 13m,15m,17m

Given, Sum of Blue Areas = 3m + 5m+ 9m + 15m = 64 cm^2
Hence, Sum of Yellow Areas = 1m + 7m+ 11m + 13m + 17m = 49m = 98 cm^2

CORRECT OPTION - C
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Since the height is same, say h
the lines dividing the triangle into 9 regions are in proportion to each other as they are parallel to base of triangle
we can number the lines as 11', 22',33',....88'
if 11' = 1, then 22' = 2 and so on..
Area of Blue region is
1/2 * h * (11'+22'+22'+33'+44'+55'+77'+88') = 64
solving we get h = 4
area of Yellow region is
4/2*(11'+33'+44'+55'+66'+66'+77'+88'+BC) =2*(1+3+4+5+6+6+7+8+9) = 49*2 =98

Answer C
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Bunuel wrote:

A triangle is cut by 8 lines, each of which is parallel to the base of the triangle, as shown above. If the height of each of the 9 resulting segments is the same and the area of the blue segments is 64 cm^2, then what is the area of the yellow segments ?

A. 72
B. 81
C. 98
D. 102
E. 162


 


This question was provided by GMAT Club
for the GMAT Club World Cup Competition

Compete, Get Better, Win prizes and more

 




Attachment:
The attachment 2022.png is no longer available


Please refer to the attachment for the solution of this problem.
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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IMO C

Since the lines are equally spaced and parallel to base of the larger triangle ABC, all the inner triangles are similar to triangle ABC.

Area of yellow segments = Total Area (ABC) - Area of blue segments. -----(1)

Given Sum of blue segments = 64 --- (2)
Also sum of blue segments= Segment (2 +3) + Segment (5) + Segment (8) ----(3)
Segment (2+3) = Segment (1+2+3) - Segment (1) = (3/9)^2 ABC - (1/9)^2 ABC
[By the property that the Ratio of Area of small triangle to the Bigger triangle = Ratio of side square = Ratio of height square.]
Segment (5) = (5/9)^2 ABC - (4/9)^2 ABC
Segment (8) = (3/9)^2 ABC - (1/9)^2 ABC

From equation 2 & 3
We will get Total Area of ABC as 162

From equation 1 Area of yellow segments = 162-64 = 98
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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My answer is C

Posted from my mobile device
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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let area of top most segment be A
the next segment will be 3A
the following segment will 5A so on
A+3A+5A+7A+9A+11A+13A+15A+17A is the total area od the triangle

3A+5A+9A+15A = 64
A=2

Area of yellow segments = A+7A+11A+13A+17A
=49A
=98

Hence C
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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Bunuel wrote:

A triangle is cut by 8 lines, each of which is parallel to the base of the triangle, as shown above. If the height of each of the 9 resulting segments is the same and the area of the blue segments is 64 cm^2, then what is the area of the yellow segments?

A. 72
B. 81
C. 98
D. 102
E. 162


The nine triangles' areas are ratio to 1 : 4 : 9 : 16 : 25 : 36 : 49 : 64 : 81
The area of the blue segments in terms of the smallest yellow triangle's area is: (4-1) + (9-4) + (25-16) + (64-49) = 32
=> The area of the smallest yellow triangle is: 64 / 32 = 2 cm^2
=> The area of the yellow segments is: 2 * (81-32) = 98 cm^2

Choose C
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Re: GMAT Club World Cup 2022 (DAY 2): A triangle is cut by 8 lines, each [#permalink]
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From the figure it's clear that the 8 parallel lines cut the main triangle in 9 small triangles.
If we consider the base of the main triangle (biggest one) to be b1 and go on above it and consider the bases so formed to be b2, b3, b4, b5, b6, b7, b8, and b9. Then from properties of similar triangle we get a relationship among the bases:
b2 = (8/9) b1
b3 = (7/8) b2 = (7/9) b1
b4 = (6/7) b3 = (6/9) b1
....
And similarly, b9 = (1/9) b1
We can find individual area of the yellow and blue segments from the trapizium area formula : (a+b)*(h/2)
As height is same for all segments,
Applying this and substituting values in case of the blue segments we can get the value of (b1*h)/2 = 18
Substituting this value for the segments of the yellow colour we get area of yellow segments =
(49/9)* 18 = 98
Hence answer choice C

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