Solution to Problem 6: The Area of The Right Triangle
Uploader Comments (vorojtsov)
Top Comments
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why did you watch it then?
All Comments (51)
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@vorojtsov he just proved that it isn't a right triangle. so this "right triangle" doesnt exist because it doesn't have a 90 degree angle.
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Or even easier. Draw a semicircle using the "hypotenuse" as a diameter. The radius of that circle is 5, and all right triangles formed with that line as a hypotenuse would have the opposing vertex on the circle (I think if you don't know this, it's easy enough to show). But if the opposite vertex of *your* triangle is 6 units above the hypotenuse, it cannot lie on the circle, and thus, cannot form a right angle.
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and if we generalize the equation :
x² - 10x + 36 = 0 by a relation between
the Base(10 in the example) and Height(6 in the example) we find
x² - B² x+ H² = 0 and only one solution
when B²=4H² so if Base = 10 the Height must be equal 5 and if the Height equal 6 then Base must be 12 (generally B=2H) that it's false in the example you shown so this right triangle don't exists.
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I am ashamed. That was very enlightening.
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you look like you are a smart guy! but jesus the way you speak english! gosh!!!!
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???
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say what
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cool. i didnt even understand the problem at the beginning but well, its not that difficult :)
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con't... In this larger circle (actually it's an oval, because the base of the triangle is still 10) we have the same properties as before. We can move the point around the hyperbolic shape of the oval, and still get a triangle.
We cannot however, get the shape drawn in the video because, like before, as X increases, Y decreases, and in the video, the second triangle is exactly the same height as the first inscribed in the oval. Imagine a triangle where SideA=2, SideB=2 and SideC=20,000.
this is not the same triangle of the question...
rakounopaido 3 years ago
why not?
vorojtsov 3 years ago 3