area

[logistic_guy]

Calculate the area of the triangle whose vertices lie at the points $\displaystyle \langle 1,2,3\rangle, \langle -2,2,4\rangle, \text{and} \ \langle 7,-8,6\rangle$.

 

[The Highlander]

Calculate the area of the triangle whose vertices lie at the points $\displaystyle \langle 1,2,3\rangle, \langle -2,2,4\rangle, \text{and} \ \langle 7,-8,6\rangle$.

Where's your attempt???

(Just for the fun of it....)

​

 

[logistic_guy]

Thank you a lot professor Highlander for passing by. It is very difficult to get help in this forum when everybody thinks that you are playing around.

It is easy to get the lengths of the legs of the triangle. Suppose they are $\displaystyle 1,2, \text{and} \ 3$. Then, what? How to find the area? Don't tell me that you have a magical way to find the angles too

 

[The Highlander]

Thank you a lot professor Highlander for passing by. It is very difficult to get help in this forum when everybody thinks that you are playing around.

It is easy to get the lengths of the legs of the triangle. Suppose they are $\displaystyle 1,2, \text{and} \ 3$. Then, what? How to find the area? Don't tell me that you have a magical way to find the angles too

There is no magic involved in finding the lengths of the sides or, indeed, the angles of this triangle but you don't need the angles to find its area as, once you have the side lengths, Heron's formula may be used to find its area.

Now, can we see your attempt to solve the question?

 

[logistic_guy]

Heron's formula may be used to find its area.

I have never heard of it before. As professor Dave once said, everyday we learn something new!

Now, can we see your attempt to solve the question?

Here it is. I will use the basic distance formula.

$\displaystyle s_1 = \sqrt{(-2-1)^2 + (2-2)^2 + (4-3)^2} = \sqrt{10} \approx 3.16$

$\displaystyle s_2 = \sqrt{(7-1)^2 + (-8-2)^2 + (6-3)^2} = \sqrt{145} \approx 12.04$

$\displaystyle s_3 = \sqrt{(7+2)^2 + (-8-2)^2 + (6-4)^2} = \sqrt{185} \approx 13.60$

I have already told you, it is easy.

 

[The Highlander]

I have never heard of it before. As professor Dave once said, everyday we learn something new!

Then I suggest you Google it or just go straight to WIKIPEDIA.

Here it is. I will use the basic distance formula.
$\displaystyle s_1 = \sqrt{(-2-1)^2 + (2-2)^2 + (4-3)^2} = \sqrt{10} \approx 3.16$
$\displaystyle s_2 = \sqrt{(7-1)^2 + (-8-2)^2 + (6-3)^2} = \sqrt{145} \approx 12.04$
$\displaystyle s_3 = \sqrt{(7+2)^2 + (-8-2)^2 + (6-4)^2} = \sqrt{185} \approx 13.60$
I have already told you, it is easy.

Well, some might agree that what you've done is, indeed, the "easy" bit but what was asked for (in your OP) was the Area of the triangle, not just its side lengths!

So, to complete your attempt at the question you posed, it looks like you will have to learn about (and how to use) Heron's formula (unless you wish to adopt an altogether different approach).

The Wikipedia site (mentioned above) actually includes a calculator that, using your earlier 'results' for the side lengths gives 17.481 sq. units as the (approximate) area of your triangle but here's a little extension for you:-

Can you show that the area of the triangle is actually exactly 17.5 sq. units?
(Hint: Maybe don't find the (approximate) interim values of the side lengths?)

Hope that helps.

 

[logistic_guy]

I will calculate it with the help of $\displaystyle \text{Lady Alpha}$.

$\displaystyle A =$

$\displaystyle \sqrt{\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2}\left(\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2} - \sqrt{10}\right)\left(\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2} - \sqrt{145}\right)\left(\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2} - \sqrt{185}\right)}$

$\displaystyle = 17.500000000000004$



Thank you a lot professor Highlander. It seems that you applied the idea perfectly. BravoThe formula was ugly but it did the job. I am the Lord of mathematics and this is the first time to see it. It is very strange!

 

[The Highlander]

I will calculate it with the help of $\displaystyle \text{Lady Alpha}$.
$\displaystyle A =\sqrt{\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2}\left(\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2} - \sqrt{10}\right)\left(\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2} - \sqrt{145}\right)\left(\frac{\sqrt{10} + \sqrt{145} + \sqrt{185}}{2} - \sqrt{185}\right)}$
$\displaystyle = 17.500000000000004$
Thank you a lot professor Highlander. It seems that you applied the idea perfectly. BravoThe formula was ugly but it did the job. I am the Lord of mathematics and this is the first time to see it. It is very strange!

Sorry, but that isn't what I asked you to show! (Nor does it comply with the "Answers" you appear to be providing!)

17.500000000000004 ≠ 17.5​

(It often pays to read everything!)

And who (tf) is Alpha???

 

[logistic_guy]

And who (tf) is Alpha???

WolframAlpha.

17.500000000000004 ≠ 17.5​

In my world they are equal.

(Nor does it comply with the "Answers"

Omit the first two. They were for something else!

Sorry, but that isn't what I asked you to show!

What was the work you asked?

 

[The Highlander]

In my world they are equal.

Then you really need to join the rest of us in the real world!

Omit the first two. They were for something else!

I was ignoring them; I was simply focussing on the third answer (17.5), ie: the correct answer to the OP.

What was the work you asked?

*** I asked you to show that the area of the triangle is exactly 17.5 sq. units (Not 17.500000000000004 sq. units or any other approximation!) ***

(BTW: If you're getting WolframAlpha to do the work for you then you can't be taking any credit for it yourself!)

 

[logistic_guy]

Then you really need to join the rest of us in the real world!

I was ignoring them; I was simply focussing on the third answer (17.5), ie: the correct answer to the OP.

*** I asked you to show that the area of the triangle is exactly 17.5 sq. units (Not 17.500000000000004 sq. units or any other approximation!) ***

(BTW: If you're getting WolframAlpha to do the work for you then you can't be taking any credit for it yourself!)

lol
Are you asking me to simplify the big square root with all the mess inside it?

 

[The Highlander]

lol
Are you asking me to simplify the big square root with all the mess inside it?

No, I was suggesting that you assimilate all the information I directed you toward and, instead of just plugging everything into WolframAlpha (to get its approximate evaluation of the string of square roots you entered into the formula), I hinted that you should only need to employ the radicand values for the side lengths and, thereby, demonstrate some fairly simple algebraic manipulation that obviated the need for any approximations to arrive at the exact value for the triangle's area as 17.5 sq. units.

Forum protocol is that the OP is given a few days to try to solve their problem themselves but, if you can't do it, then I'm sure someone will post a solution by the end of the week.

(I'll do it myself if nobody else bothers. )

 

[logistic_guy]

No, I was suggesting that you assimilate all the information I directed you toward and, instead of just plugging everything into WolframAlpha (to get its approximate evaluation of the string of square roots you entered into the formula), I hinted that you should only need to employ the radicand values for the side lengths and, thereby, demonstrate some fairly simple algebraic manipulation that obviated the need for any approximations to arrive at the exact value for the triangle's area as 17.5 sq. units.

Forum protocol is that the OP is given a few days to try to solve their problem themselves but, if you can't do it, then I'm sure someone will post a solution by the end of the week.

(I'll do it myself if nobody else bothers. )

I think that I got what I needed. If you want to add extra gradients, you are welcome! Thank you a lot again for the formula.

 

[jonah2.0]

Beer drenched reaction follows.

... The formula was ugly but it did the job. ...

I say nay. Heron's formula is anything but what you claim it to be. It be one of the most beautiful formulas ever. Wearing bitter goggles as you do has dulled your appreciation for beauty.

... I am the Lord of mathematics and this is the first time to see it. It is very strange! ...

What's even more strange is that you're under the impression that you're a "Lord of mathematics" considering your ignorance of Heron's formula and other gaps in your basic geometric knowledge. I have yet to hear somebody with a PhD in mathematics boasting he's a "Lord of mathematics" and yet here you are, someone with serious gaps in his knowledge of basic math, bragging that he's a "Lord of mathematics".
Who was it that granted you that dubious title anyway? Did you suddenly woke up one day and crowned yourself that grand title? You must have been smoking some really good stuff to walk around with such a dumb title.

 

[The Highlander]

I think that I got what I needed. If you want to add extra gradients, you are welcome! Thank you a lot again for the formula.

No, I have to disagree! I don't believe you did get what you "needed". What you needed to get was the correct answer (as provided in the "Answers to Exercises" you posted), ie: exactly 17.5!

Since you have now made it clear that you don't propose to do anything further on this problem, I will just illustrate how it could have been tackled to get the right answer.

If you had bothered to read even part way down the website I recommended to you, you would have seen that Heron's formula can also be written in terms of just the side lengths instead of using the semiperimeter.

If we call the three sides of the triangle a, b & c respectively, then (as you already calculated above)....

$\displaystyle a=\sqrt{10}\qquad b=\sqrt{145}\qquad \text{and}\qquad c=\sqrt{185}\quad$ (units of length).​

And one of the ways the formula may be rearranged (to use only the side lengths and not the semiperimeter) is...

$\displaystyle \qquad\;\;A=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$​

​

$\displaystyle \implies A=\frac{1}{4}\sqrt{4\times10\times145-(10+145-185)^2}$​

​

$\displaystyle \qquad\quad\;\;=\frac{1}{4}\sqrt{5800-(-30)^2}=\frac{1}{4}\sqrt{5800-900}$​

​

$\displaystyle \qquad\quad\;\;=\frac{1}{4}\sqrt{4900}=\frac{1}{4}\times70$​

​

$\displaystyle \qquad\quad\;\;=\underline{\underline{17.5}}\;\text{sq. units (exactly!)}\quad\text{Q.E.D.}$​

If you do (truly) want to learn anything new, then perhaps you will follow the advice offered rather than rejecting it in favour of your own little world view?
(Where 17.500000000000004 = 17.5 Doh!)

 

[logistic_guy]

No, I have to disagree! I don't believe you did get what you "needed". What you needed to get was the correct answer (as provided in the "Answers to Exercises" you posted), ie: exactly 17.5!

Since you have now made it clear that you don't propose to do anything further on this problem, I will just illustrate how it could have been tackled to get the right answer.

If you had bothered to read even part way down the website I recommended to you, you would have seen that Heron's formula can also be written in terms of just the side lengths instead of using the semiperimeter.

If we call the three sides of the triangle a, b & c respectively, then (as you already calculated above)....

$\displaystyle a=\sqrt{10}\qquad b=\sqrt{145}\qquad \text{and}\qquad c=\sqrt{185}\quad$ (units of length).​

And one of the ways the formula may be rearranged (to use only the side lengths and not the semiperimeter) is...

$\displaystyle \qquad\;\;A=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$​

​

$\displaystyle \implies A=\frac{1}{4}\sqrt{4\times10\times145-(10+145-185)^2}$​

​

$\displaystyle \qquad\quad\;\;=\frac{1}{4}\sqrt{5800-(-30)^2}=\frac{1}{4}\sqrt{5800-900}$​

​

$\displaystyle \qquad\quad\;\;=\frac{1}{4}\sqrt{4900}=\frac{1}{4}\times70$​

​

$\displaystyle \qquad\quad\;\;=\underline{\underline{17.5}}\;\text{sq. units (exactly!)}\quad\text{Q.E.D.}$​

If you do (truly) want to learn anything new, then perhaps you will follow the advice offered rather than rejecting it in favour of your own little world view?
(Where 17.500000000000004 = 17.5 Doh!)

I am the laziest person in the universe. I didn't continue reading $\displaystyle \text{Wikipedia}$ because I knew the answer was $\displaystyle 17.5$ and wanted to use any method to get it. Therefore, I got enough with the Heron's formula. But I admit that you have done it brilliantly in the latest post. Thank you a lot for the interest, effort, and time.

Beer drenched reaction follows.

I say nay. Heron's formula is anything but what you claim it to be. It be one of the most beautiful formulas ever. Wearing bitter goggles as you do has dulled your appreciation for beauty.

What's even more strange is that you're under the impression that you're a "Lord of mathematics" considering your ignorance of Heron's formula and other gaps in your basic geometric knowledge. I have yet to hear somebody with a PhD in mathematics boasting he's a "Lord of mathematics" and yet here you are, someone with serious gaps in his knowledge of basic math, bragging that he's a "Lord of mathematics".
Who was it that granted you that dubious title anyway? Did you suddenly woke up one day and crowned yourself that grand title? You must have been smoking some really good stuff to walk around with such a dumb title.

Here is the misconception. The "Lord" in my dictionary is someone who thinks that he has solved a problem in every field. So he may be called a Jack of all trades, master of none.

If you explored all the seas in the world, it doesn’t necessarily mean you’ve seen all the fish.