A Very Nice Geometry Problem | You should be able to solve this!!

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A Very Nice Geometry Problem | You should be able to solve this!!
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Опубликовано:

15 сен 2024

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Комментарии : 13

@jarikosonen4079 7 месяцев назад

The circle is there, but it is not necessary needed to be cyclic case...

@Irtsak 3 месяца назад

Corollary : *The ratio of the areas of two triangles, with equal heights is equal to the ratio of their bases.* Triangles AOB,AOD have common height => (AOB)/(AOD)=OB/OD => *9/X=OB/OD* (1) Also BOC,OCD have common height => (BOC)/(OCD)= OB/OD => *Y/4=OB/D* (2) (1),(2) => 9/X=Y/4 => XY=36

@pi5355 7 месяцев назад

Top side=2a Bottom side=2b Top height=h1 Bottom height=h2 ah1=9 bh2=4 h1/a=h2/b X=2a(h1+h2)/2-2ah1/2=ah2 y=ah2=bh1 xy=ah2bh1=ah1bh2=9×4=36

@Zina308 7 дней назад

x/9 = DO/OB, 4/y =DO/OB. x/9= 4/y, xy= 36.

@ДмитрийИвашкевич-я8т 7 месяцев назад

x/9=d/b=4/y xy=36

@user-ii5vr3um9r 7 месяцев назад

X:9=4:Y, XY=36

@1Konu1Zoru 7 месяцев назад

no need 8 minutes .. just need 5 seconds to solve this :)

@misterenter-iz7rz 7 месяцев назад

9=1/2 absin s, 4=1/2 cd sin s, X=1/2 da sin s, Y=1/2 bc sin s, XY=1/4 abcd sin^2 s=9×4=36.😊

@howardaltman7212 7 месяцев назад

It's nice that you do these different ways on different videos so that we can see a variety of approaches. On some other videos, you've done these by finding the area of 2 triangles with colinear bases and the same vertex (same height).

@tarunmajumdar6166 7 месяцев назад

No need for circle. Multiplication of opposite areas for any quadratarial is equal.

@RAG981 7 месяцев назад

It is actually a nice result, that the product of the areas of opposite triangles is the same. And so obvious a solution.

@josegiovannygio3679 7 месяцев назад

Sou do Brasil , gosto muito dos seus vídeos de geometria e graças a você melhorei bastante meu desempenho na matemática, desejo sucesso pra vcê e seu canal !!!!

@MathBooster 7 месяцев назад

Thank you 😊