# Area of a right-angled triangle

what do you mean by ""just not a right one?""

basically, if you say that a triangle with a hypotenuse is a triangle that has one angle as right or measure is 90 degrees.

On the otherhand, a pythagorean triple would be possible such that the two measures of the sides of the triangle would be 6 and 10. In this case, the other side would measure 8. in short, the sides would be the triple (6,8,10).

anyway, its nice to have a good conversation, i mean exchange of ideas with you.

chroot
Staff Emeritus
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What I mean is, Chen is right. You cannot construct a right triangle with hypotenuse 10, and height (measured perpendicular from the hypotenuse) of 6.

The largest angle you can get is 79.6 degrees.

Try it for yourself, see if you agree with me.

- Warren

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anyway, i have just stated some of the possibilities of the existence of such triangle. anyway thank you very much. nothing lost!

chroot
Staff Emeritus
Gold Member
Originally posted by oen_maclaude
anyway, i have just stated some of the possibilities of the existence of such triangle. anyway thank you very much. nothing lost!
Are you asserting that such a triangle actually does exist?

- Warren

NateTG
Homework Helper
oen_maclaude:

This is about a right triangle with an altitude of 6 from the hypotenuse of length 10. The 10-8-6 tripple is quite obvious.

It's really simple to show that a right triangle with a Hypotenuse of 10 cannot have an height of 6 from that hypotenuse. The maximum height is 5 and is achieved with a 45-45-90 triangle.

If you construct a circle that contains the three vertices of the triangle, then the hypotenuse will be a diameter since the triangle is a right triangle. With a hypotenuse of 10, the radius is 5 and that is clearly the most distant point on the circle from the hypotenuse.

Alternatively, think of the right triangle as half of a rectangle. The distance from a diagonal to a vertex is at most 1/2 the length of the diagonal since the diagonals of a rectangle bisect each other. Since the height is the minimum distance from the edge to the vertex, it's length is at most 1/2 the length of the hypotenuse.

yes i do understand now.

(to chroot) however, the fact that the possibility that a certain triangle of .such dimension will be existing only that the height is not perpendicular to the hypotenuse

chroot
Staff Emeritus
Gold Member
Originally posted by oen_maclaude
only that the height is not perpendicular to the hypotenuse[/SIZE]
Right -- and what good is that? That isn't the problem at hand.

- Warren

I apologize if I'm intruding, but there must be some important distinction I'm missing. I would appreicate it if somebody would clarify for me. I'm working with these definitions and ideas:

A right triangle is a triangle with a right angle (90 degrees)

The hypotenuse of a right triangle is the side opposite the right angle

The two remaining sides are called legs.

Either of the legs may arbitrarily be called the base, and then the length of the remaining leg is called the height (customarily the base is drawn horizontally and as the other leg is at a 90 degree angle, the other leg is drawn vertically, again, by custom we draw this leg so that it leaves the corner of the right angle by travelling up-- thus the height coincides with our usual idea of height)

Any triangle drawn in a circle with one edge forming the diameter of aforesaid circle is a right triangle, and further, the hypotenuse is the diameter, and the opposite angle is the right angle.
Imagine you have such a circle/triangle... by moving the pt that corresponds to the right angle closer and closer to either of the endpts you create a triangle with one very short leg and one leg that is arbitrarily close the length of the hypotenuse... by rotating the circle and the inscribed triangle we can make that long edge correspond to the height, so the restriction on the height is that the height may not be equal to (or longer) then the hypotenuse.

The ratio of edges on a 45,45,90 triangle is not 1 to 1 to 2 since 1^2 + 1^2 is not 2^2 It is 1 to 1 to sqrt(2).

I think that's everything... of course I'm assuming we're doing all this on a flat surface... all bets are off if we're on some two manifold.

chroot
Staff Emeritus
Gold Member
curiousbystander,