Triangle area

Solar_blaze · Apr 21, 2015

I got a question with a triangle where they ask me to find the area
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

I know it's. ABC=H*side of triangle split by 2 but where to start?

Ishuda · Apr 21, 2015

Solar_blaze said:
I got a question with a triangle where they ask me to find the area
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

I know it's. ABC=H*side of triangle split by 2 but where to start?

I would start by computing the length of each side, L_AB as the length between point A and point B, L_BC as the length between point B and point C, L_CA as the length between point C and point A. I would then draw the equivalent triangle with length L_AB for the base and lengths L_BC and L_CA for the other two sides.

Can you continue from there? As a hint, you might have your old point B at the new location (0,0) and old point A at the new location (0, L_AB).

Steven G · Apr 21, 2015

Solar_blaze said:
I got a question with a triangle where they ask me to find the area
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

I know it's. ABC=H*side of triangle split by 2 but where to start?

I would use Heron's formula.

Solar_blaze · Apr 22, 2015

My text book only showed the S=height multiple by the side of tringle split by 2

stapel · Apr 23, 2015

Solar_blaze said:
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

...My text book only showed the S=height multiple by the side of tringle split by 2

Plot the points, so you know what you're working with.

Pick a side to be the base. Use the Distance Formula (here) to find the length of that side.

Find the slope of the line through the base (here). From this, find the slope of the perpendicular "height" line. Using this perpendicular slope and the coordinates of the vertex not on the base, find the straight-line equation (here) for the "height" line. Also, find the equation of the line containing the base.

Solve the system of equations (here) to find the point where the "height" line intersects the "base" line. Use the Distance Formula to find the distance between the off-base vertex and the base; that is, find the height.

Then apply your formula.

If you get stuck, please reply showing all of your work on the listed steps. Thank you!

Steven G · Apr 23, 2015

Solar_blaze said:
I got a question with a triangle where they ask me to find the area
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

I know it's. ABC=H*side of triangle split by 2 but where to start?

Why not just find the length of all 3 sides and then use Heron's formula? Isn't that what Heron's formula is for?!

Ishuda · Apr 23, 2015

Jomo said:
Why not just find the length of all 3 sides and then use Heron's formula? Isn't that what Heron's formula is for?!

Actually, there is a "better" way, see
http://math.stackexchange.com/quest...-if-the-coordinates-of-the-three-vertices-are
The formula there can be expressed as
\(\displaystyle \frac{1}{2}|(x_1 - x_3)\, (y_2 - y_3) - (x_2 - x_3)\, (y_1 - y_3)|\)
which is sometimes how you see it written. That formula is just the generalization of the way the formula is sometimes stated: Given the three points (0,0), (a,b), and (c,d) of the verticies of a triangle, the area of the triangle is given by
\(\displaystyle \frac{1}{2}|ad - bc|\)
The general solution just translates one of the points [(x₃, y₃)] to the origin, i.e. (x₃,y₃)-(x₃,y₃)=(0,0).

BTW: I was unaware of all of this before Jomo's post. Thanks, Jomo.

Steven G · Apr 23, 2015

Ishuda said:
Actually, there is a "better" way, see
http://math.stackexchange.com/quest...-if-the-coordinates-of-the-three-vertices-are
The formula there can be expressed as
\(\displaystyle \frac{1}{2}|(x_1 - x_3)\, (y_2 - y_3) - (x_2 - x_3)\, (y_1 - y_3)|\)
which is sometimes how you see it written. That formula is just the generalization of the way the formula is sometimes stated: Given the three points (0,0), (a,b), and (c,d) of the verticies of a triangle, the area of the triangle is given by
\(\displaystyle \frac{1}{2}|ad - bc|\)
The general solution just translates one of the points [(x₃, y₃)] to the origin, i.e. (x₃,y₃)-(x₃,y₃)=(0,0).

BTW: I was unaware of all of this before Jomo's post. Thanks, Jomo.

Thanks! That is interesting.

Steven G · Apr 23, 2015

Ishuda said:
BTW: I was unaware of all of this before Jomo's post. Thanks, Jomo.

I am Jomo and I too was unaware of all of before Ishuda's post. Thanks!
I guess that you too do not like Heron's formula which I guess motivated you to look for a better way to solve these problems. Heron's method is just too awkward, but this new method is awesome!

Ishuda · Apr 23, 2015

Jomo said:
I am Jomo and I too was unaware of all of before Ishuda's post. Thanks!
I guess that you too do not like Heron's formula which I guess motivated you to look for a better way to solve these problems. Heron's method is just too awkward, but this new method is awesome!

It depends on what you are given. If you are given sides, the Heron's formula is good. Given vertex co-ordinates, the other is better. I just happened to run across the other when looking up Heron's formula.

Steven G · Apr 23, 2015

Ishuda said:
It depends on what you are given. If you are given sides, the Heron's formula is good. Given vertex co-ordinates, the other is better. I just happened to run across the other when looking up Heron's formula.

Actually you are correct about this. If you just have the length of the three sides then you don't necessarily have the coordinates. Once again you caught me being guilty of sloppy thinking.

Steven G · Apr 24, 2015

Solar_blaze said:
I got a question with a triangle where they ask me to find the area
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

I know it's. ABC=H*side of triangle split by 2 but where to start?

I was thinking about this problem and recall solving this problem in linear algebra when I taught the course. So armed with pen and paper (I hate pencils) I set out to derive the formula to find the area of a triangle just given the vertices using determinants. It actually was not that bad (I used trapezoids). I found that given three vertices of a triangle (a,b),(c,d) and (e,f) the area of this triangle is 1/2 determinant of [row 1 = a,b,1, row 2 =c,d,1, and row 3 = e,f,1]
The question is whether or not this reduces to what Ishuda got?
Translation of the three points yields (a', b'), (c', d') and (0,0). The the determinant
[row 1 = a', b', 1, row 2= c', d', 1, row 3 = 0,0,1] = det [ row 1= a', b', row 2 = c', d')] = a'd-b'c' which is exactly twice what Ishuda got.
I am 100% sure that Ishuda saw this before! Maybe one day I will know as much as Ishuda has forgotten.
After writing this post I see that I need to learn latex.

Steven G · Apr 24, 2015

Denis said:
thniknig ?

Have you been drinking again? Have you read my post on beer and math. The second link is quite nice if you never saw it before.

lookagain · Apr 25, 2015

Solar_blaze said:
I got a question with a triangle where they ask me to find the area
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

I know it's. ABC=H*side of triangle split by 2 but where to start?

Solar_blaze those other methods are needlessly complicated.

Plot the points. Extend BC until it meets the x-axis at (4, 0). Call that new point D.

Now you have two right triangles, ABD and ACD.

The desired area of triangle ABC = (area of ABD) minus (area of ACD).

The area of each right triangle is (1/2)*base*height, or, put another way, (1/2)*leg*leg.

As you will be able to see, there are no special calculations to get the respective leg lengths.

Ishuda · Apr 25, 2015

Denis said:
I get area of triangles (sides a,b,c) this way:
Area = ch
where:
h = SQRT(a^2 - k)
where:
k = [(a^2 - b^2 + c^2) / (2c)]^2

Don't like Heron's either !

Actually, starting from scratch, not knowing Heron's or the other this looks like what I had. I did a
h² = a² + (\(\displaystyle \alpha\) c)² = b² + [(\(\displaystyle 1\, -\, \alpha\)) c]²
to find \(\displaystyle \alpha\), then h, then used the standard \(\displaystyle \frac{1}{2}\, h\, c\) where c is the base.

Ishuda · Apr 25, 2015

Denis said:
yuk!!

Well, I'm a little slow sometimes and have to go the long way around.

lookagain · Apr 26, 2015

Solar_blaze said:
I got a question with a triangle where they ask me to find the area
It is known that coordinates B is (4,8) C is (4,2) and the there is A (0,0) they ask for area ABC?

I know it's. ABC=H*side of triangle split by 2 but where to start?

This approach should be one of the simplest.

After you sketch the diagram, note this this:

All triangles that have their bases on the line x = 4, and have their apex on x = 0,
have an altitude (height) of 4. The length of the base is the difference in the y-values
in this case, which is 6.

The area of any triangle is (1/2)*base*height.

So, the area of the triangle equals (1/2)(6)(4) = 12.

.

Triangle area

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