how to test this?

[allegansveritatem]

Here is the problem:

Here is what I did with this:

Now, how would I prove this correct or otherwise? Another way to ask this question is: How to find the lengths of a and b for this triangle when all you know is that a line from the 90 degree angle perpendicular to the hypotenuse is 12cm. I guess you could say .5 ab =.5 12 times h and then choose, say, 5 for h, get 60 from their multiplication and say .5ab=60 and then let a=5 and b=6. When I follow this method I get p=5+6+sqrt(61) and inserting this into my formula I get h=5.74, but that can't be right. I mean, if a is 5 and b is 6 then h has to be something like 7.8. No? I seem to be confused here. Where am I going wrong with this reasoning? Maybe the formula I came up with is not correct but I can't see the flaw in it.

 

[lev888]

Here is the problem:
View attachment 29948
Here is what I did with this:
View attachment 29949
Now, how would I prove this correct or otherwise? Another way to ask this question is: How to find the lengths of a and b for this triangle when all you know is that a line from the 90 degree angle perpendicular to the hypotenuse is 12cm. I guess you could say .5 ab =.5 12 times h and then choose, say, 5 for h, get 60 from their multiplication and say .5ab=60 and then let a=5 and b=6. When I follow this method I get p=5+6+sqrt(61) and inserting this into my formula I get h=5.74, but that can't be right. I mean, if a is 5 and b is 6 then h has to be something like 7.8. No? I seem to be confused here. Where am I going wrong with this reasoning? Maybe the formula I came up with is not correct but I can't see the flaw in it.

You need to pick a and b that work with the problem's givens. If you know that the height of the triangle is 12, you can't pick any random values for a and b - e.g. 1 and 2 would obviously be too short.
One a/b pair which is easy to find is the case where a=b. You can use the small triangles where a is a hypotenuse to find a (the legs are 12 and 12).
To find other combinations I would try
1. Pick a value for h
2. Calculate the area
3. Area gives you the product ab
4. Combine this with the sum of squares and you have a system of 2 equations. If you picked a reasonable value for h it should have a solution or two.

 

[allegansveritatem]

You need to pick a and b that work with the problem's givens. If you know that the height of the triangle is 12, you can't pick any random values for a and b - e.g. 1 and 2 would obviously be too short.
One a/b pair which is easy to find is the case where a=b. You can use the small triangles where a is a hypotenuse to find a (the legs are 12 and 12).
To find other combinations I would try
1. Pick a value for h
2. Calculate the area
3. Area gives you the product ab
4. Combine this with the sum of squares and you have a system of 2 equations. If you picked a reasonable value for h it should have a solution or two.

Thanks for replying. I will copy your post and reread it tomorrow--too late to get my head around anything mathematical tonight...but I will quickly say that I did try at least to take this specific set of constraints into mind when looking for numbers to put on the sides of this figure...but , right, I made the mindlessly crazy choice of 5 cm for the hypotenuse and there is no way one can get a triangle with a 12 cm perpendicular line from the opposite angle to so short a hypotenuse. Why didn't I notice that before? Too scatterbrained, I reckon.

 

[allegansveritatem]

It just occurred to me that all I have to do to prove this is: Create a triangle where I know all the dimensions and plug in relevant data into my formula. At any rate I will work it out one way or the other today and post results here.

 

[allegansveritatem]

Today I first constructed a right triangle, all the dimensions of which I knew. From that I found what the length of a line running from the right angle perpendicular to the hypotenuse would be and plugged this information into my formula in an attempt to prove or disprove it. I seem to have succeeded in the latter. So, first I did this:

then I put this information into my formula thus:

and, as is obvious, I came a cropper. But I still can't find a flaw in the formula I have gone over it half a dozen times. Can anyone tell me what is wrong with the formula as it was derived?

 

[Dr.Peterson]

Today I first constructed a right triangle, all the dimensions of which I knew. From that I found what the length of a line running from the right angle perpendicular to the hypotenuse would be and plugged this information into my formula in an attempt to prove or disprove it. I seem to have succeeded in the latter. So, first I did this:
View attachment 29960
then I put this information into my formula thus:
View attachment 29962
and, as is obvious, I came a cropper. But I still can't find a flaw in the formula I have gone over it half a dozen times. Can anyone tell me what is wrong with the formula as it was derived?

Your formula is correct!

Redo your calculations here; and show us the details. (For instance, what is the numerical value of your perimeter?)

Why haven't you tried what was suggested, namely to use a 45-45-90 (right isosceles) triangle as your test case? Then h is 24, and a and b are 12sqrt(2). If you don't see why, draw an accurate picture.

Or, you could use a nicer triangle, like a 3-4-5.

 

[allegansveritatem]

Your formula is correct!

Redo your calculations here; and show us the details. (For instance, what is the numerical value of your perimeter?)

Why haven't you tried what was suggested, namely to use a 45-45-90 (right isosceles) triangle as your test case? Then h is 24, and a and b are 12sqrt(2). If you don't see why, draw an accurate picture.

Or, you could use a nicer triangle, like a 3-4-5.

Thanks for checking. I will try the isosceles and post again. I should have done that earlier but...

 

[allegansveritatem]

so, I followed the advice of better heads than mine and used an isosceles right triangle with a height of 12cm and computed with sin and cos the rest and came up with:

It doesn't look like one, admittedly, but that is an isosceles there and by its help I found that my hard earned formula is correct. Happy me. Many Thanks to both contributors to this thread.

 

[lev888]

so, I followed the advice of better heads than mine and used an isosceles right triangle with a height of 12cm and computed with sin and cos the rest and came up with:
View attachment 29975
It doesn't look like one, admittedly, but that is an isosceles there and by its help I found that my hard earned formula is correct. Happy me. Many Thanks to both contributors to this thread.

Don't really need trigonometry here. Pythagoras' theorem should've been enough.

 

[allegansveritatem]

well, what I had was a right isosceles triangle of which I only know the length of a line drawn from the 90 degree angle perpendicularly to the hypotenuse. I didn't know the length of the base or the length of the sides. I knew the angles and the side of one of two equal right triangles made from the original isosceles triangle. No? How would I know side a, for instance, without dividing 12 by the cosine of 45 degrees? But I am not well versed in geometry so...I am not saying it can't be done with Pythagoras' theorem but it is not occurring to me just how. No doubt, tonight while I lay trying to sleep at 3am it will suddenly occur to me how to do it but...maybe I will be lucky and can sleep tonight.

 

[lev888]

well, what I had was a right isosceles triangle of which I only know the length of a line drawn from the 90 degree angle perpendicularly to the hypotenuse. I didn't know the length of the base or the length of the sides. I knew the angles and the side of one of two equal right triangles made from the original isosceles triangle. No? How would I know side a, for instance, without dividing 12 by the cosine of 45 degrees? But I am not well versed in geometry so...I am not saying it can't be done with Pythagoras' theorem but it is not occurring to me just how. No doubt, tonight while I lay trying to sleep at 3am it will suddenly occur to me how to do it but...maybe I will be lucky and can sleep tonight.

Spoiler: Explanation

The height in the right isosceles triangle bisects the angle, so the 2 smaller triangles have base angles 45 degrees, which makes them also isosceles. The 2 legs are 12. Apply Pythagoras' theorem to find a.

 

[allegansveritatem]

Spoiler: Explanation

The height in the right isosceles triangle bisects the angle, so the 2 smaller triangles have base angles 45 degrees, which makes them also isosceles. The 2 legs are 12. Apply Pythagoras' theorem to find a.

This morning a 3 am I lay in the throes of insomnia and began, as is my habit at such times, to think about math problems. I looked with my mind's eye at the triangle I had been working with in this problem and realized that the height was exactly half the base. This seemed to me to be significant. And I also noticed that the two triangles formed by the line perpendicular to the hypotenuse of the large triangle, the mother triangle if you will, were themselves isosceles triangles with dimensions exactly half those of the mother. So I cried Eureka and so...no spoiler alert needed! haha.