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Look at this star!
 60 FPS video
9,043 Views • Nov 12, 2024 • Click to toggle off description
Metadata And Engagement

Views : 9,043
Genre: Education
License: Standard YouTube License
Uploaded At Nov 12, 2024 ^^


warning: returnyoutubedislikes may not be accurate, this is just an estiment ehe :3
Rating : 4.958 (4/379 LTDR)

98.96% of the users lieked the video!!
1.04% of the users dislieked the video!!
User score: 98.44- Masterpiece Video

RYD date created : 2024-11-22T20:16:05.06387Z
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46 Comments

Top Comments of this video!! :3

@xinpingdonohoe3978

1 week ago

I never was great at geometry, but I can still guess. The interior pentagon angle is 108°. This makes the repeated angle of the isosceles triangle 180°-108°=72°. The small angle is then 180°-2×72°=36°. 5 of them makes 5×36°=180°=π.

27 |

@hyperbolicandivote

3 days ago

If the star is regular a circle that contains all the vertices has tw adjacent vertices 72 degrees apart. So the angle across is half, or 36. 5*36 = 180.
Tap Tap

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@cosimobaldi03

1 week ago

Look how it shines for you...

5 |

@Greensea-j8t

1 week ago

Even if the pentagon inside is irregular the sum will be 180. This can be proved by using properties sum of angles of a triangle and vertically opposite angles between the 5 triangles.

2 |

@soyanshumohapatra

1 week ago

Angle of regular pentagon= 108°
It's opposite angle is also same
Angle base to the pentagon are both [360°- 2(108°)]/2 = 72°
Tip angle = 180° - 2×72° = 36°
Total = 5× 36° = 180°

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@janda1258

1 week ago

What is m?

2 |

@eess2396

5 days ago

难得做几何题啊😂

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@ibrahimalbayati2515

1 week ago

You can imagine it to be 180, try to stack them together centralized in a semicircle, all 5 would fit.

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@soyanshumohapatra

1 week ago

Hint: Each angle of a regular pentagon is 108°

1 |

@godofplay138

1 week ago

How can you say tha the interior pentagon is regular or not

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@unclouded

1 week ago

jojolion reference

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@hafizusamabhutta

1 week ago

180°, i watched your video

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@TheGabusan

1 week ago

Now do the star of David

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@deepanjansen2147

1 week ago

Wut if it's not regular pentagon ?

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@cyrusyeung8096

1 week ago

The required sum is 180°.
Let AD and BE intersect at X, and AC and BE intersect at Y.

Consider △BXD,
∠AXB = ∠B + ∠D (exterior ∠ theorem)

Consider △CYE,
∠AYE = ∠C + ∠E (exterior ∠ theorem)

Consider △AXY,
∠A + ∠AXB + ∠AYE = 180°
∠A + (∠B + ∠D) + (∠C + ∠E) = 180°


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