Calculating Area of a Triangle in Excel?

  • Thread starter Thread starter lindyoppa
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lindyoppa

Hello.

I have a graph that is plotting 3 points in excel. These 3 points for
a triangle. (So, I know the "x and y" coordinates for each point)

I'd like to find out the formula of the triangle.

I know generally the area formula is
A=Sq. root[S(S-a)(S-b)(S-c)]
Where S=(a+b+c+)/2
and a,b,c are length of sides.

The distance formula b/t 2 pts is
d=Sq. root [(x'-x)+(y'-y)

Can anyone please help me put this into one formula in excel?
thanks :)

please email (e-mail address removed)
 
Assuming your data is in A1:C3:
If Column A contains your x values, Column B contains your y values, and
Column C contains the constant 1, then I believe the area of a triangle can
be calculated like this:

=MDETERM(A1:C3)/2

Hope I said this correctly. :>)
 
If it's the area you are trying to calculate, I belive it can also be done
like this. ( I hope this is correct).

=(x3*(y1 - y2) + x1*(y2 - y3) + x2*(y3 - y1))/2

--
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =


Dana DeLouis said:
Assuming your data is in A1:C3:
If Column A contains your x values, Column B contains your y values, and
Column C contains the constant 1, then I believe the area of a triangle can
be calculated like this:

=MDETERM(A1:C3)/2

Hope I said this correctly. :>)

--
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =


lindyoppa > said:
Hello.

I have a graph that is plotting 3 points in excel. These 3 points form
a triangle. (So, I know the "x and y" coordinates for each point)

I'd like to find out the formula of the triangle.

I know generally the area formula is
A=Sq. root[S(S-a)(S-b)(S-c)]
Where S=(a+b+c+)/2
and a,b,c are length of sides.

The distance formula b/t 2 pts is
d=Sq. root [(x'-x)+(y'-y)

Can anyone please help me put this into one formula in excel?
thanks :)

please email (e-mail address removed)
Click to expand...
Click to expand...
 
Hi!

I'm never too keen on taking a formula on trust especially if there i
an intuitive way of getting at the problem...

Think of your triangle (draw it, indeed) relative to x and y axes and
to make it easy, put it entirely in the positive quadrant (x>0,y>0).

Drop perpendiculars from the triangle's vertices to the x-axis.

Now each side of the triangle is a side of a trapezium. These trapezi
hold the key because the tiangle is what is left when you add two an
take away the third (if the apex is above the base). Simila
considerations if the apex is below the base.

Area of trapezium : base *(average height) leads to (x2-x1)*(y2-y1)/2
Put it together from there...

Well: sorry about the lecture but it takes me back 50 years ;-)

Al
 
Ooops!!!
Did I really write that?
The average height of the trapezium is obviously (y1+y2)/2
not (y1-y2)/2.

Sorry!

Al
 
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