calculator that works out area?

[JonJ678]

You stated that for a right angle triangle, my definition of height wouldn't work out. So I drew a right angle triangle for you.

Subdividing it into two was the "proof" that half long side times "jon's height" gives the area, as "1/2 base times height" is obviously true for a right angled triangle and you didn't seem to agree. My point is that height has to be defined like this, or the equation doesn't give the area.

Would you feel better if I redrew it without the 90 degree angle, picked a side arbitrarily and then split it into two right angled triangles? Touch has pointed out that it doesn't matter which side you start from, I'm not sure if "any side times perpendicular distance to corner" is easier to remember than "long side times perpendicular distance to corner" or not. The second seems simpler to me.

 

[touch]

You stated that for a right angle triangle, my definition of height wouldn't work out. So I drew a right angle triangle for you.

Subdividing it into two was the "proof" that half long side times "jon's height" gives the area, as "1/2 base times height" is obviously true for a right angled triangle and you didn't seem to agree. My point is that height has to be defined like this, or the equation doesn't give the area.

I didnt mean to say that your definition of height wouldnt work, i just misunderstood what you were saying. When you said "height has to be consistently defined", i thot you were saying that there is only 1 correct height of a triangle and have been trying to point out that there is 3 correct heights to a triangle.

And my post about not really dividing the triangle up was in reply to whitecrook saying
"Now how do I work out the area of the 2 smaller right angle triangles? Divide them into smaller triangles?
When does it stop ?"

 

[JonJ678]

I was replying to Whitecrook, not to you. A shorter reply would probably have been that you stop subdividing as soon as you only have right angled triangles to deal with.

I definitely meant consistent as in always perpendicular to a side and passing through the opposite corner. I suppose I could have led with "assuming without loss of generality that the height is taken perpendicular to the longest side" but I don't think all that much rigour is needed here. I hope not at least, or it will swiftly become apparent that I am not a mathematician.

 

[Klo]

The chances are the calculator you want (if it exists) will not be allowed in the exam. Check to see which calculators are allowed.

Plus, I've forgotten my GCSE in Maths (4 years a go), however I think there is a page filled with formula at the beginning.

 

[whitecrook]

You stated that for a right angle triangle, my definition of height wouldn't work out. So I drew a right angle triangle for you.

No I didn't

Subdividing it into two was the "proof" that half long side times "jon's height" gives the area, as "1/2 base times height" is obviously true for a right angled triangle and you didn't seem to agree. My point is that height has to be defined like this, or the equation doesn't give the area.

Your 'proof' begs the question in that it uses the same method to prove the thing it's trying to prove?

Would you feel better if I redrew it without the 90 degree angle, picked a side arbitrarily and then split it into two right angled triangles? Touch has pointed out that it doesn't matter which side you start from, I'm not sure if "any side times perpendicular distance to corner" is easier to remember than "long side times perpendicular distance to corner" or not. The second seems simpler to me.

I'm not arguing, I agree with the method to calculate area of a triangle. It works for all triangles. You are defining height as the longest (or 'one of the 2', or 'either') side which meets the other (base) at 90 degrees.

When you start with a right angled triangle this is already defined for you. there is no need to redefine it. Height is already there on the page no matter which way up the triangle is. So the end formula is the same 0.5*base*height.

 

[Stupot_]

JonJ678
I'm used to maths being obsessively precise so I'm not sure how to define terms. Are you familiar with the following words?

Perpendicular (angle of 90 degrees between the lines)
Parallel (the lines will never meet however much longer you make them)
Perimeter (the total distance around the shape)
Circumference (perimeter of a circle)
Radius (distance from centre of a circle to the edge)
Pi (a number, equal to 3.142. Turns up all over the place when circles are involved)

Ah, that's a fantastic summary for GCSE maths. I think tangents might come up as well but I can't remember.

Quoting this for the lols

And lol at you lot trying to teach each other how to measure triangles.......! 5 stars!

 

[JonJ678]

You have a point there Stupot. Cheers

@White, I'm fairly confident you're deliberately provoking me now. Were you to read the previous posts you would not be confused. That is not how I am defining height, as I am capable of understanding that not all triangles are right angled triangles. Assuming Euclidean geometry and a triangle composed of three straight lines, "height" is the magnitude of the non-zero vector which perpendicularly intersects the arbitrarily chosen "base" and also intersects exactly one vertex. Not as any edge which is at 90 degrees to the base, as for many triangles there is no such edge.

 

[touch]

And lol at you lot trying to teach each other how to measure triangles.......! 5 stars!

Haha, i think the problem is that we all know how to do the maths perfectly well but cant explain it very well and have all ended up inadvertently trolling each other

 

[JonJ678]

Arguing over definitions of height probably isn't going to help the OP, but I really did start with the best of intentions

 

[whitecrook]

You have a point there Stupot. Cheers

@White, I'm fairly confident you're deliberately provoking me now. Were you to read the previous posts you would not be confused. That is not how I am defining height, as I am capable of understanding that not all triangles are right angled triangles. Assuming Euclidean geometry and a triangle composed of three straight lines, "height" is the magnitude of the non-zero vector which perpendicularly intersects the arbitrarily chosen "base" and also intersects exactly one vertex. Not as any edge which is at 90 degrees to the base, as for many triangles there is no such edge.

I see you've edited this once or twice... I totally agree with what you have said there!
Can't you see that by doing this the perpendicular line by definition makes a 90 degree angle and therefore you have 'created' a right angled triangle (2 infact, call them 'sub triangles' if you will? Your formula then takes the length of this newly created line (ie. the height). This line is the line which intersects at 90degrees i.e. perpendicular. In the case of a right angled triangle to start with you don't need to do this!! Why bother calculating the area of 2 right angled triangles when you can calculate the area of one i.e the original one?


I feel we may be going in circles. Or should that be triangles.?

 

[JonJ678]

Edited it a good four or five times, it started off quite impolite

The whole point is to create two right angled triangles of known "height", the areas of which are readily calculated and summed. Yes, I see that if the triangle is right angled this process is not required. Do you see that it is a justification for applying this definition of height to non-right angled triangles in order to apply 0.5bh? Or that your definition of height cannot apply to non right angled triangles?

 

[whitecrook]

Edited it a good four or five times, it started off quite impolite

The whole point is to create two right angled triangles of known "height", the areas of which are readily calculated and summed. Yes, I see that if the triangle is right angled this process is not required.

that's what i've been saying all along

Do you see that it is a justification for applying this definition of height to non-right angled triangles in order to apply 0.5bh?

yes, i never said otherwise

Or that your definition of height cannot apply to non right angled triangles?

I wrote a bunch of stuff here, but this whole thing is circular. suffice to say 'my definition of height' does not apply to non right angled triangles because non right angled triangles do not have a right angle. I never claimed otherwise.