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# The length of the right triangle's leg is the diameter of

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The length of the right triangle's leg is the diameter of [#permalink]

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30 May 2004, 22:46
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The length of the right triangle's leg is the diameter of circle, which divides the hypotenuse into parts with ratio 1:4. If the length of the leg, that sircle is build on, is 1cm, what are the lengths of second leg and hypothenuse?

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30 May 2004, 23:17
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The length of the right triangle's leg is the diameter of circle, which divides the hypotenuse into parts with ratio 1:4. If the length of the leg, that circle is build on, is 1cm, what are the lengths of second leg and hypothenuse?

That's really simple:

there are 2 triangles with equal angles, thus if x is the length of th shortest part of hipotenuse, x/1 = 1/(5x) => x = 1/sqrt(5), other leg = 2, hipotenuse = sqrt(5).
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31 May 2004, 11:46
Emmanuel wrote:
The length of the right triangle's leg is the diameter of circle, which divides the hypotenuse into parts with ratio 1:4. If the length of the leg, that circle is build on, is 1cm, what are the lengths of second leg and hypothenuse?

That's really simple:

there are 2 triangles with equal angles, thus if x is the length of th shortest part of hipotenuse, x/1 = 1/(5x) => x = 1/sqrt(5), other leg = 2, hipotenuse = sqrt(5).

how's the second triangle formed?
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03 Jun 2004, 16:01
Base leg is the diameter of circle,so the center of the circle is the mid of the base.The circle also interecpts the hypo.Draw a line jiong this intersecting point on hypo and to the base mid point, this forms the second triangle which is similar to the original right angle triangle.notice the angles are same as the first one.
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03 Jun 2004, 16:55
[quote="GoalStanford"]Base leg is the diameter of circle,so the center of the circle is the mid of the base.The circle also interecpts the hypo.Draw a line jiong this intersecting point on hypo and to the base mid point, this forms the second triangle which is similar to the original right angle triangle.notice the angles are same as the first one.[/quote]

I'm not usre if these are similar triangles. becoz
if they are similar the sides should be in the same ratio.
so if the 2 triangles are similar the side that is the diameter to the circle would yield a ratio of 1:2 but the ratio on the hypotenuse would be 4: (4 + 1) which is not equal. I think to prove triangles to be similar the side have to be in the same ratio . Correct me if I'm wrong....
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03 Jun 2004, 17:38
My apologies,I think I gave wrong triangle there.I was talking about the triangle formed by the two vertices of base leg and the intersecting point on hypo.I will try explain here.

Original triangle ABC , angle B=90 degrees, BC is the base leg.

triangle BDC where D is the intersecting point on hypo.

Now ,
1.angle (ACB)=angle (DCF) no doubts here.you are actually referring to same angle.
2.angle (ABC)=90 degrees. Also, angle (BDC) will be 90 degrees from the cirle theory( angle in semicircle is right angle).

That is how angles are same in both the triangles.
03 Jun 2004, 17:38
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