Area of a right-angled triangle

[Chen]

In a given right-angled triangle, the length of the hypotenuse is 10 and the length of one of the heights is 6. What is the area of the triangle?

a) 12
b) 30
c) 24
d) 24 or 30

Please explain your answer.

 

[chroot]

The area of any triangle is

$$\frac{1}{2} b h$$

where b is the length of the base of the triangle, and h is its height.

- Warren

 

[HallsofIvy]

I always get a kick out of people who just quote a problem. Do we get points for a correct answer?

A (almost!*) always, Chroots answer is correct- the area of any triangle is (1/2)bh where b is the "base", one of the sides of the triangle and h is the "height", measured perpendicular to the base.

In a right triangle, the two legs are perpendicular so you can use one as base and the other as height.
Use the Pythagorean theorem to find the length of the other leg of the triangle, then apply "(1/2)bh".


* I, myself, never make mistenk(s!

 

[Chen]

Ahh, but I do make mistakes. The problem should read "the length of one of the heights is 6". What would be the correct answer then? (Bearing in mind that the given height could be the height to the hypotenuse.)

 

[chroot]

The triangle has more than one height?

Listen, we've told you how to do this problem, but we're not going to do it for you. Try it, it's not hard.

- Warren

 

[Chen]

I do know how to solve this problem and already have, years ago. I am just trying to see if you can, but clearly I overestimated you.

And yes, every triangle - the thing with three sides, you know - has three heights. Not just one. But three. (And it's this little factoid that makes the problem more than just a 1st grade exercise.)

 

[chroot]

No, if you first select one side as a base, a triangle has only one height. Sorry.

- Warren

 

[Chen]

Originally posted by chroot
No, if you first select one side as a base, a triangle has only one height. Sorry.

Where have I selected a base? All the problem says is that the length of one of (the three) heights in this triangle is 6.

 

[chroot]

We don't really have time here for your little riddles. Sorry. At the very least, if you're "challenging" us with a riddle, label it as such. We thought you were a schoolchild who didn't understand how to find the area of triangles.

- Warren

 

[Integral]

Originally posted by Chen
Where have I selected a base? All the problem says is that the length of one of (the three) heights in this triangle is 6.

Your initial statement completely specifies the triangle, where is the difficulty?

The triangle you describe is a 3-4-5 triangle, one of the most useful combinations of all right triangles. Now tell us why finding this area should be difficult?

 

[Chen]

Originally posted by Integral
Your initial statement completely specifies the triangle, where is the difficulty?

The triangle you describe is a 3-4-5 triangle, one of the most useful combinations of all right triangles. Now tell us why finding this area should be difficult?

Is it not possible that the height whose length is 6, is perpendicular to the hypotenuse of this triangle?

 

[Chen]

Originally posted by chroot
We don't really have time here for your little riddles. Sorry. At the very least, if you're "challenging" us with a riddle, label it as such. We thought you were a schoolchild who didn't understand how to find the area of triangles.

I did not realize that the answer to a question depends on the inquirer. And if you really thought I were a schoolchild you could at least pretend to be nice about it, because if I really were a 10 year old your replies would have been insulting to me. Luckily I have passed that age and I am just saddened by the fact that it is a mentor - who should set an example for others - that speaks like this.

 

[matt grime]

Originally posted by Chen
Is it not possible that the height whose length is 6, is perpendicular to the hypotenuse of this triangle?

So you've got one those infinitely large triangles lying around have you?

 

[chroot]

Originally posted by Chen
And if you really thought I were a schoolchild you could at least pretend to be nice about it

Your question, as originally posted, was "what is the area of a right triangle with one leg 6 feet and hypotenuse 10 feet?", which is a schoolchild's question.

I am just saddened by the fact that it is a mentor - who should set an example for others - that speaks like this.

Oh no.

- Warren

 

[Integral]

Perhaps a link to a drawing to this "triangle" would clear up the questions. It is said a drawing is worth a 1000 words. Perhaps english is not your native tounge and you do not realize what you are saying.

 

[KingNothing]

Lets calm down everyone. I think he means he knows the hypotenuse and one leg. If so, it's $$10^2=6^2+A^2$$. Therefore, $$100=36-A^2$$ and 64=A^2 and A=8. 8 is your other leg. 8*6=48, so your area is 24. It's pretty late here, so I hope I didnt skip anything, its been so long since I even touched a problem like this.

Please stop fighting, people. There's no reason to act negative towards others here.

That's a good theory, Integral. His profile says he is from Israel, and if it is true, it makes sense to me to say some things incorrectly.

 

[Integral]

I don't think he is talking about side length, but rather perpendicular bisectors.

 

[lastlaugh]

like the triangle is sitting on the hypotonuse? is this a 45-45-90 triangle or can we not make that assumption?

 

[chroot]

Either 24 or 30. If the triangle is sitting on either of its legs, it's 24. If the triangle is sitting on its hypotenuse, it's 30.

- Warren

 

[oen_maclaude]

i guess, everybody had his/her contribution. i hope that chen would have understood it in anyway.

However, if chen haven't pictured the situation, then there is still a need for us to draw the figure to help him/her out.

 

[Chen]

Actually, both of you (oen_maclaude and chroot) are wrong... the 6 height cannot be perpendicular to the hypotenuse. And that is really all I wanted to hear in the first place, but I got my answer somewhere else already.

 

[oen_maclaude]

well i guess i have not read any detail on the characteristics of the height of the triangle. i think that it is also correct to say that the height of the triangle may as well be perpendicular to the hypotenuse. I hope that you have seen my drawing or illustration on the possible cases that would happen given those information.

if none of those were correct, then you might as well teach me more of geometry specially on the right triangle.

 

[chroot]

Chen's correct, it's not possible for a right triangle with hypotenuse 10 units to have a 6 unit height. I didn't think to check that.

- Warren

 

[oen_maclaude]

well, i might have overlooked on the triangle. however, is it really true that you cannot have a height of 6, if the hypotenuse is 10? as far as i know, there is a possibility that there would be such triangle formed.

 

[chroot]

You can form many such triangles -- just not a right one.

- Warren

 

[oen_maclaude]

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]

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

 

[oen_maclaude]

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

 

[chroot]

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]

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.

 

[oen_maclaude]

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]

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

 

[curiousbystander]

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 traveling 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]

curiousbystander,

I'm not sure I saw any question in your post. Please read NateTG's post carefully.

- Warren

 

[oen_maclaude]

yup! i agree with chroot! look back to what NateTG's post. but is was just ok that you really just dropped by to give us your insights.