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Right angled triangle in a circle

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Right angled triangle in a circle [#permalink] New post 07 Feb 2012, 12:09
If a right angled triangle is inscribed in a circle, is it necessary for the hypotenuse of the right triangle to be the diameter of the circle? I know that the vice-versa is true.
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Re: Right angled triangle in a circle [#permalink] New post 07 Feb 2012, 12:53
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bsaikrishna wrote:
If a right angled triangle is inscribed in a circle, is it necessary for the hypotenuse of the right triangle to be the diameter of the circle? I know that the vice-versa is true.


A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle.

The reverse is also true: if the diameter of the circle is also the triangle’s side, then that triangle is a right triangle.

For more on this topic please check Circle chapter of Math Book: math-circles-87957.html

Hope it helps.
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Re: Right angled triangle in a circle [#permalink] New post 07 Feb 2012, 14:11
Thank you so much for all your replies. You've been a great help to me. :)
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Re: Right angled triangle in a circle [#permalink] New post 07 Feb 2012, 21:55
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bsaikrishna wrote:
If a right angled triangle is inscribed in a circle, is it necessary for the hypotenuse of the right triangle to be the diameter of the circle? I know that the vice-versa is true.



You can figure this out by drawing a diagram.
Attachment:
Ques3.jpg
Ques3.jpg [ 9.44 KiB | Viewed 9352 times ]

Look how the angle is increasing as you go higher up. Hence for every length of the minor arc, there is a unique inscribed and central angle. The right triangle's hypotenuse will be the largest length of the chord i.e. a diameter and its central angle will be 180 giving the inscribed angle as 90.

Also, length of arc = (Central angle/360) * 2*pi*r
Since inscribed angle is 90, central angle is 180.

length of arc = 180/360 * (2*pi*r)
length of arc = pi*r
i.e. you get a semi circle. So the chord (the hypotenuse of the right triangle) must be the diameter.
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Re: Right angled triangle in a circle   [#permalink] 07 Feb 2012, 21:55
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