Area of a right-angled triangle

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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. :smile:
 
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chroot

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The area of any triangle is

[tex]\frac{1}{2} b h[/tex]

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

- Warren
 

HallsofIvy

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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!
 
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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

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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
 
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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.)
 
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chroot

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

- Warren
 
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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

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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

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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?
 
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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?
 
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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.
 
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matt grime

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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

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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

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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.
 
Lets calm down everyone. I think he means he knows the hypotenuse and one leg. If so, it's [tex]10^2=6^2+A^2[/tex]. Therefore, [tex]100=36-A^2[/tex] 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.
 
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Integral

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I don't think he is talking about side length, but rather perpendicular bisectors.
 
like the triangle is sitting on the hypotonuse???? is this a 45-45-90 triangle or can we not make that assumption?
 

chroot

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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
 
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.
 

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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.
 
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

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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. :wink:

- Warren
 
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

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

- Warren
 

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