Rearranging Heron's formula

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Simonsky

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Unfortunately I can't use latex yet so advanced apologies for messy notation:

heron's Formula: A = sqroot: s(s-a)(s-b)(s-c) where s =a=b+c/2 (semiperimeter)

So the question says : rearrange formula for A to make c the subject.

This is what I tried:

Remove sqroot: s^2 = s(s-a)(s-b)(s-c)

then s^2/S(s-a)(s-b) = s-c

then s^2/s(s-a)(s-b) -s = -c

then reverse signs: -s^2/s(s-a)(s-b) +s = c

I suspect I've messed this up-but my book doesn't have the answer for this so I can't check.

Any pointers to where I've gone wrong-gratefully received.
 
Denis said:
A = sqrt[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2 : brackets REQUIRED!

Short cut: let u = s(s-a)(s-b) ; then:
A = sqrt[u(s-c)]
A^2 = u(s-c)

Finish it...
Click to expand...

So: A^2/u =(s-c)

then: A^2/u - s = -c

then c = -A^2/u + s which is: c = -A^2/s(s-a)(s-b) + s/1 (that's the same as I had above without the mistake of confusing s for A!)

I suppose I could work out the brackets: c = -A^2/s^3 - s^2b - as^2 = abs + s

not sure atr all that I've got this....
 
Simonsky said:
heron's Formula: A = sqroot: s(s-a)(s-b)(s-c) where s =a=b+c/2 (semiperimeter)

So the question says : rearrange formula for A to make c the subject.

This is what I tried:

Remove sqroot: s^2 = s(s-a)(s-b)(s-c)

then s^2/S(s-a)(s-b) = s-c

then s^2/s(s-a)(s-b) -s = -c

then reverse signs: -s^2/s(s-a)(s-b) +s = c
Click to expand...

The big problem is that s depends on c, so you can't just solve the equation as given. You can't have s in the final answer.

You have A = sqrt[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2. Replacing s with its value,

A = sqrt[(a+b+c)((a+b+c)-2a)((a+b+c)-2b)((a+b+c)-2c)/16]

Do you follow that? You can simplify further, but then solving for c looks like it will be a mess. After squaring both sides, you will have a fourth-degree equation in c.

However, I can see some ways to organize the work of expanding that polynomial to keep it simple and avoid errors; and you end up with what is actually a second-degree polynomial in c^2, which can be solved.

See what you can do with those ideas.
 
Dr.Peterson said:
The big problem is that s depends on c, so you can't just solve the equation as given. You can't have s in the final answer.

You have A = sqrt[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2. Replacing s with its value,

A = sqrt[(a+b+c)((a+b+c)-2a)((a+b+c)-2b)((a+b+c)-2c)/16]

Do you follow that? You can simplify further, but then solving for c looks like it will be a mess. After squaring both sides, you will have a fourth-degree equation in c.

However, I can see some ways to organize the work of expanding that polynomial to keep it simple and avoid errors; and you end up with what is actually a second-degree polynomial in c^2, which can be solved.

See what you can do with those ideas.
Click to expand...


Thanks Dr. Peterson. I can see you've got rid of the fraction by multiplying by two but I'm not sure where the /16 comes from. By the way, this appears to be another question that demands knowledge well beyond where the course has reached. I've already contacted the publishers (Oxford University Press) about the previous case I posted. At this point in the book, even 2nd degree polynomials haven't been taught yet!

For your interest here's the page attached, it is question 7;

Fifth degree.jpg
 
Simonsky said:
Thanks Dr. Peterson. I can see you've got rid of the fraction by multiplying by two but I'm not sure where the /16 comes from. By the way, this appears to be another question that demands knowledge well beyond where the course has reached. I've already contacted the publishers (Oxford University Press) about the previous case I posted. At this point in the book, even 2nd degree polynomials haven't been taught yet!
Click to expand...

Yes, I did several steps at once, perhaps to give you a little challenge (or just to save myself some writing). An intermediate step might have been this:

A = sqrt[(a+b+c)/2((a+b+c)/2 - a)((a+b+c)/2 - b)((a+b+c)/2 - c)]
= sqrt[(a+b+c)/2 ((a+b+c)-2a)/2 ((a+b+c)-2b)/2 ((a+b+c)-2c)/2]
= ...

Can you see what I did now? I multiplied by 2, in 4 places.

I imagine the authors forgot that s depends on c, and wanted you to just solve the equation itself for c, treating s as an independent variable. That's really unforgivable, in that poor students would be happy with it, but good students would struggle to do it right.

You might find it interesting to pick one of the examples, after finding A, and replace one of the sides with x and try solving for that. You will notice, first, that if you allow yourself to use the knowledge of what s is, then you can solve for x just from the definition of s, not using A at all; and, second, that if you replace s with its expression (e.g. s = (4 + 6 + x)/2), you can simplify Heron's formula and have something a little less intimidating to actually solve. (I haven't tried it to see how big a challenge it would be; you would obviously have to try to recall what you have learned in the past, which may be more than you want to do just now.)
 
Dr.Peterson said:
Yes, I did several steps at once, perhaps to give you a little challenge (or just to save myself some writing). An intermediate step might have been this:

A = sqrt[(a+b+c)/2((a+b+c)/2 - a)((a+b+c)/2 - b)((a+b+c)/2 - c)]
= sqrt[(a+b+c)/2 ((a+b+c)-2a)/2 ((a+b+c)-2b)/2 ((a+b+c)-2c)/2]
= ...

Can you see what I did now? I multiplied by 2, in 4 places.

I imagine the authors forgot that s depends on c, and wanted you to just solve the equation itself for c, treating s as an independent variable. That's really unforgivable, in that poor students would be happy with it, but good students would struggle to do it right.

You might find it interesting to pick one of the examples, after finding A, and replace one of the sides with x and try solving for that. You will notice, first, that if you allow yourself to use the knowledge of what s is, then you can solve for x just from the definition of s, not using A at all; and, second, that if you replace s with its expression (e.g. s = (4 + 6 + x)/2), you can simplify Heron's formula and have something a little less intimidating to actually solve. (I haven't tried it to see how big a challenge it would be; you would obviously have to try to recall what you have learned in the past, which may be more than you want to do just now.)
Click to expand...

Thanks Dr. Peterson -I'm still mystified by the /16 to be honest....I can see you've multiplied each bracket by 2 so it is no longer a fraction, so why is the further division by 16 needed if you simply multiply both sides by 2? Ah... I now get it: the brackets multiply to you get 2^4, so you are not multiplying BOTH sides just collecting up the divisions at the end to get a cleaner notation....have I got that?
 
Simonsky said:
Thanks Dr. Peterson -I'm still mystified by the /16 to be honest....I can see you've multiplied each bracket by 2 so it is no longer a fraction, so why is the further division by 16 needed if you simply multiply both sides by 2? Ah... I now get it: the brackets multiply to you get 2^4, so you are not multiplying BOTH sides just collecting up the divisions at the end to get a cleaner notation....have I got that?
Click to expand...

Right -- it's just a matter of multiplying together four fractions, each with denominator 2, which were obtained by combining fractions using common denominators:

\(\displaystyle A = \sqrt{\dfrac{a+b+c}{2}\left(\dfrac{a+b+c}{2} - a\right)\left(\dfrac{a+b+c}{2} - b\right)\left(\dfrac{a+b+c}{2} - c\right)}\) \(\displaystyle = \sqrt{\dfrac{a+b+c}{2}\cdot \dfrac{a+b+c-2a}{2}\cdot \dfrac{a+b+c-2b}{2}\cdot \dfrac{a+b+c-2c}{2}} \)

We aren't solving (multiplying by 2) but simplifying.
 
Denis said:
This seems to work:
c = SQRT[a^2 + b^2 + 2k] where k = SQRT[(ab)^2 - 4A^2]

Try it on Heronian triangle a,b,c = 4,13,15 (area = 24)
Click to expand...

Yes, that's just what I got by applying the quadratic formula to my quadratic in c^2. Not quite as ugly as I expected, actually: \(\displaystyle \sqrt{a^2 + b^2 \pm 2\sqrt{a^2b^2 - 4A^2}}\).

But note that there are two solutions; in your example I get c = 15 or \(\displaystyle \sqrt{145}\). It does check out!

The trick to keeping the work simple is to take advantage of several differences of squares.
 
Denis said:
Should there be another "+-", at the front of that?

Also the notation: a>0 and b>0 ?

Btw, got help at another site with this....
Click to expand...

You'd add another plus-or-minus at the front only if you allow side lengths to be negative. That would mean you live in a different universe.

But then, why would you require a and b to be positive, but not c?

In our reality, all are understood to be positive. I wouldn't think that needs to be stated explicitly.
 
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