Find the area of the shaded region

[chwala]

Homework Statement

Consider the diagram below, find the area of the shaded region

Relevant Equations

similarity-congruency

this is another question that i saw on the internet...

 

[chwala]

This is my attempt, in the first attempt i tried to see whether i could be able to find the shaded region by way of calculations.
In the second attempt, i tried to use ratio,...

 

[chwala]

i made a mistake in the area formula, the dimensions are not correct , let me re-do it...

 

[chwala]

the dimensions ought to be $3.72cm$ and not $1.5cm$ that is the dimensions for each equilateral triangle...

 

[chwala]

clearly this is not correct...i am missing something here...

 

[Lnewqban]

AD should be 3.72 x 3.5.

 

[chwala]

AD should be 3.72 x 3.5.

Aaaaargh the last triangle...I will re do the working again...and at same time appreciate any alternative way of doing this.

 

[Lnewqban]

Aaaaargh the last triangle...I will re do the working again...and at same time appreciate any alternative way of doing this.

I see 7 triangles, total area is 7 times 6 units of area: 42 units.
The diagonal line divides in half the shape formed by those 7 triangles: 21 units each big triangle.

Then, we remove the top triangle and investigate any symmetry among sides.
The top side measures 3 entire small-triangle sides.
In a symmetrical way, two of each truncated sides have the same dimension.

What could we do with that information?
Could we divide the area of the top big angle in two in order to obtain the value of the green area?

 

[chwala]

I see 7 triangles, total area is 7 times 6 units of area: 42 units.
The diagonal line divides in half the shape formed by those 7 triangles: 21 units each big triangle.

Then, we remove the top triangle and investigate any symmetry among sides.
The top side measures 3 entire small-triangle sides.
In a symmetrical way, two of each truncated sides have the same dimension.

What could we do with that information?
Could we divide the area of the top big angle in two in order to obtain the value of the green area?

Interesting...I am an analytical person, these geometry shapes and checking for congruence or similarity I don't like it lol

 

[Lnewqban]

Interesting...I am an analytical person, these geometry shapes and checking for congruence or similarity I don't like it lol

Those are only two of the ways to reach the same conclusion.

If you also see 7 equilateral triangles, you would agree that the the diagonal line only devides the area of 6 of them.

Because in a symmetrical way, two of each truncated sides have the same dimension, we could imaginarily cut those shapes off from the big triangle formed by the 6 consecutive triangles and glue two matching sides together, obtaining three original triangles that way.

Therefore, the area of the green-colored parts is 6 units x 3 triangles = 18 units.

 

[hmmm27]

Therefore, the area of the green-colored parts is 6 units x 3 triangles = 18 units.

Well, that's obviously wrong, or I read you wrong. Each of your two 'big triangles' is already 18 units. The green area encompasses a portion of that : looks like less than half.

(Not that I have anything positive to add).

 

[MidgetDwarf]

Interesting...I am an analytical person, these geometry shapes and checking for congruence or similarity I don't like it lol

I don't mean to judge. But surely you can take 5 min out of your day to understand the definition of similar triangles, there properties, and later similarity between different geometric objects? It will save you a lot of time this route, you learn more math, and spend less time using computations you don't need to do.

 

[docnet]

Interesting...I am an analytical person, these geometry shapes and checking for congruence or similarity I don't like it lol

another member might have suggested this. but one could also set up a coordinate system and set up a handful of integrals.. not that anyone would WANT to do that lol

 

[chwala]

my other attempt...is the area of triangle $AFT$=$4.396cm^2$ or i am getting it wrong, are my angles correct?

 

[chwala]

I don't mean to judge. But surely you can take 5 min out of your day to understand the definition of similar triangles, there properties, and later similarity between different geometric objects? It will save you a lot of time this route, you learn more math, and spend less time using computations you don't need to do.

noted, i will try boss...

 

[Lnewqban]

Well, that's obviously wrong, or I read you wrong. Each of your two 'big triangles' is already 18 units. The green area encompasses a portion of that : looks like less than half.

(Not that I have anything positive to add).

It was wrong, indeed.
Thank you.

After calculating the areas of different triangles, and using ratios among similar triangles, I have this new result:
The area of the green-colored parts is 4.5 + 2.0 + 0.5 = 7.0 units square.

 

[Frabjous]

After calculating the areas of different triangles, and using ratios among similar triangles, I have this new result:
The area of the green-colored parts is 4.5 + 2.0 + 0.5 = 7.0 units square.

Would you provide some hints on what you did? I cannot get anywhere without resorting to analytic geometry.

 

[Lnewqban]

Would you provide some hints on what you did? I cannot get anywhere without resorting to analytic geometry.

I used the similar triangles idea mentioned by @MidgetDwarf in post #12 above.

Please, see:
https://www.mathsisfun.com/geometry/triangles-similar.html

https://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html

Calculated the length of sides of any of the equilateral triangles.
$A=\sqrt{3}b^2/4=6$
$b=3.722~units$

Triangle AMN is similar to triangle ACD / Ratio = 4
Calculated length of CD
Calculated length of $BC = b - CD$

Knowing that angle ABD is 60°, calculated area of triangle ABC, using
$A=(0.5)(AB)(BC)(sin~60)=4.5~units^2$

Triangles AGH and AKL are similar to triangle AMN.
Triangles EFG and IJK are similar to triangle ABC.

Repeated process for calculating areas of triangles EFG ($2.0~units^2$) and IJK ($0.5~units^2$).
Added individual areas for a total of $7.0~units^2$.

 

[chwala]

Thanks guys...

i saw this answer below on the internet; I am trying to follow it...I thought i should share it here,

what i do not understand is "why is he squaring'
$\frac{A_1}{A_2}$=$\frac{0.75^2}{0.5^2}$ is it really necessary to square?i do not think so,
MY understanding is that $(ASF)=(LSF)^2$
therefore, $LSF$ of $\frac{A_1}{A_2}$=0.75$×$2$=\frac{3}{2}$
from here it follows that $(ASF)=\frac{9}{4}$

 

[Keith_McClary]

MY understanding is that

I don't know what you mean by ASF and LSF.
The area of similar triangles is proportional to the square of the length of corresponding sides.

 

[chwala]

$(ASF)$= Area Scale Factor & $LSF$= Linear Scale Factor...and of course,
$(VSF)$=Volume Scale Factor.
$(VSF)=(LSF)^3$

 

[chwala]

i was able now to use sine rule to find the areas;find attached...

what i was missing in this question was how to come up with the linear scale dimensions,
$0.75$, $0.5$ and $0.25$ for the 1st, 2nd and 3rd triangles respectively...

 

[Keith_McClary]

linear scale dimensions for the 1st, 2nd and 3rd triangles respectively...

But once you have those, it is obvious that the answer is
$$6 \times \frac{3}{4} \times (1 + (\frac{2}{3})^2 + (\frac{1}{3})^2 )$$

 

[Zgort]

 

[hmmm27]

Does the spare triangle make any sense ?

 

[SammyS]

View attachment 278132

While it is true that the green area above the diagonal is equal to the white area below the diagonal, there is no reasoning given to show that the green area above the diagonal equals the white area above.
In other words, the is no reasoning given to show that half of each half is green and half of each half is white. In fact that's not true.