/sci/ - Science & Math » Thread #13038620

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nub geometry question

Anonymous Tue 27 Apr 07:54:25 2021 No.13038620 View Reply Original Report

Quoted By: >>13038793 >>13038797 >>13038865 >>13038954

I have an equilateral triangle. Would the area be different if one of the sides is not a straight line?
Still same length tho.

Anonymous

Anonymous Tue 27 Apr 2021 08:43:51 No.13038793 Report

Quoted By: >>13038822 >>13038916

>>13038620
You dont need math to know that if you overlap the last two triangles you dont have the same area.

Anonymous

Anonymous Tue 27 Apr 2021 08:44:15 No.13038797 Report

Quoted By: >>13038809

>>13038620
>Still same length tho
Must be a straight line then

Anonymous

Anonymous Tue 27 Apr 2021 08:46:41 No.13038809 Report

Quoted By: >>13038865

>>13038797
The diagram are not at scale so the angles of the strIght lines are different.

Anonymous

Anonymous Tue 27 Apr 2021 08:49:07 No.13038819 Report

Quoted By: >>13039006

the concave one should tell you the answer.

Anonymous

Anonymous Tue 27 Apr 2021 08:50:04 No.13038822 Report

Quoted By: >>13038832

>>13038793
Retard, if all sides are the same length, the angle at the top is not the same, i.e. they are not to scale. You couldn't tell by overlapping the triangles wether or not they have the same area.

Anonymous

Anonymous Tue 27 Apr 2021 08:51:17 No.13038832 Report

Quoted By:

>>13038822
nvm, if the curve is the same the top angle is the same, I'm retarded

Anonymous

Anonymous Tue 27 Apr 2021 08:56:57 No.13038865 Report

Quoted By: >>13039031

>>13038620
>>13038809
there is only one possible triangle configuration for three 1-lines, so the angles must be the same at any scale for all three triangles

Anonymous

View Same Google iqdb SauceNAO question2.png, 10KiB, 656x504

Anonymous Tue 27 Apr 2021 09:09:37 No.13038916 Report

Quoted By: >>13039008

>>13038793
disregard the third triangle for a sec. Consider the first two triangles; Since the two ends of the curved line of the second triangle are closer than if it were straight(yellow points). Would the area gained from the curved line(Green) equal the area lost when the angle between the two straight lines decreases(Red)?

Anonymous

Anonymous Tue 27 Apr 2021 09:16:15 No.13038954 Report

Quoted By: >>13038963

>>13038620
You can't curve the line keeping it a length of 1 without changing the top angle.

Anonymous

Anonymous Tue 27 Apr 2021 09:17:50 No.13038963 Report

Quoted By: >>13038975

>>13038954
yes, you can, just make the top out of steel and bottom out of rubber

Anonymous

Anonymous Tue 27 Apr 2021 09:20:36 No.13038975 Report

Quoted By:

>>13038963
Then the length of the bottom will be greater than one.

Anonymous

Anonymous Tue 27 Apr 2021 09:25:59 No.13039006 Report

Quoted By:

>>13038819
Im unsure how the concave side case would affect the angle between the two straight sides. My intuition tells me it would increase to accommodate the area lost from the inward bowing of the curved edge. Pretty much the inverse of the other pic I posted.

Anonymous

Anonymous Tue 27 Apr 2021 09:26:14 No.13039008 Report

Quoted By: >>13039013 >>13039026

>>13038916
The area of the triangle is $\sqrt{3}/4$ . Consider deforming the triangle into a circle with the same circumference of 3. Then the area would be
$A = \pi r^2 = \pi (3/2\pi)^2 = \frac{3}{4\pi} \neq \frac{\sqrt{3}}{4}$
Thus area is not conserved under constant circumference.

Anonymous

Anonymous Tue 27 Apr 2021 09:27:43 No.13039013 Report

Quoted By:

>>13039008
Its $\frac{9}{4\pi}$ but the point remains.

Anonymous

Anonymous Tue 27 Apr 2021 09:30:40 No.13039026 Report

Quoted By:

>>13039008
Your knowledge is akin to a god...TY. may the universe bless you good sir

Anonymous

Anonymous Tue 27 Apr 2021 09:31:24 No.13039031 Report

Quoted By: >>13039036

>>13038865
wait nevermind

Anonymous

Anonymous Tue 27 Apr 2021 09:32:14 No.13039036 Report

Quoted By:

>>13039031
no take backs!

Anonymous

Anonymous Tue 27 Apr 2021 09:57:03 No.13039120 Report

Quoted By:

Area of the first figure is simply the area of a unit equilateral triangle which is
$\frac{\sqrt{3}}{4}$
Area of the second figure, given that the arc is of a circle is the ratio between the arc and the total circumference times the total area,which is
$\frac{1}{2\pi}\pi (1)^2$

$\frac{1}{2}$

Area of the third figure is the area of the equilateral triangle minus then in parenthesis, the area of the second figure minus the area of the first figure which is
$\frac{\sqrt{3}}{4}-(\frac{1}{2}-\frac{\sqrt{3}}{4})$

$\frac{\sqrt{3}}{2}-\frac{1}{2}$

$\frac{\sqrt{3}-1}{2}$

the first figure has the largest area and the third figure has the smallest

Anonymous

View Same Google iqdb SauceNAO Segment-and-Area-of-a-Segment-Of (...).png, 15KiB, 750x467

Anonymous Tue 27 Apr 2021 13:57:59 No.13039835 Report

Quoted By: >>13040168

this is trivial but long to prove. the principle here is that the length of the arc starting from a to b is longer than than than the chord starting from a to b so you assume that the lower side is a part of a bigger circle and calculate the area of that segment then add it to a triangle with lower side not equal to one and get more area

Anonymous

Anonymous Tue 27 Apr 2021 15:39:08 No.13040168 Report

Quoted By:

>>13039835
Okay wait help me out here. If we take a circle with r=1, and take and angle of 1 rad, wouldn't the surface of that slice of the circle just be 1/2?

And the surface of an equilateral triangle with sides of length 1 is ?3 ÷ 4

So the area of the third shape in OP's pic would be (?3 - 1) ÷ 2

...right?

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