# Very simple triangle area problem

I have a problem with a triangle question. The question is: Imagine a triangle with the points $A(-5, 0), B(-3,-7), C(2,-2)$, where the sides $AB$ and $AC$ are equal. What is the area of this triangle?

My solution is simply to draw a bounding box around the triangle with the base $2-(-5) = 7$ and the height of $7$ ($-7$ from $B$). This gives me the area of $\frac{49}{2}$. The answer should be $\frac{45}{2}$ according to my text book, what am I doing wrong?

Thank you.

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(1) Consider a triangle with vertices (0,0), (9,10), (10,9). According to your method the area of this triangle should be 50, but if you draw it it's clearly much smaller. So obviously your bounding-box idea isn't working. Think back to where the bounding-box idea came from, and try to figure out what condition you need for the bounding box to work like you want it to. (2) You can use the bounding box, but you can't just divide the area by two. Think about splitting the bounding box into several triangles--the one you are interested in and a few others whose areas are easy to calculate. – Jonathan Christensen Dec 20 '12 at 15:56
Thank you for your comment. I see my mistake and what I did wrong. – Lukas Arvidsson Dec 20 '12 at 15:59

$$7^2 - \frac{7\times 2}{ 2} -\frac{ 5 \times 5}{ 2} -\frac{2 \times 7 }{ 2} = \frac{45}{2}$$

This will be less than half the rectangle unless two of the vertices are at corners.

Method 5 here shows the calculation should be $\dfrac{7\times 7}{ 2} -\dfrac{ 2 \times 2}{ 2}$ or more generally $\dfrac{fg}{ 2} -\dfrac{vw}{ 2}$.

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Thank you for showing me how it is done :) I was just way too quick to assume that the bounding box method would work, don't really know why. Very helpful! – Lukas Arvidsson Dec 20 '12 at 16:03

Your approach is obviously wrong. Imagine a "collapsed triangle" $A(0,0)$, $B(10,10)$, $C(10,10)$. The area is obviously zero, but your method gives $50$.

The correct approach is to remember that length of a vector product of 2 vectors is the area of the rhomboid they form. The trangle has is half as large.

Therefore $x=B-A=(2,-7,0)$ and $y=C-A=(7,-2,0)$ (we have to consider the vectors in 3D). Then $x\times y=(0,0,45)$, it's length is obviously $45$ and therefore the area is $45/2$.

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Thank you! I totally see my mistake now, thank you for the answer! – Lukas Arvidsson Dec 20 '12 at 15:57

If you have three points $(x_a,y_a)$, $(x_b,y_b)$, and $(x_c,y_c)$ you can simply compute the (signed) triangles area by $$\frac{1}{2}\det\; \begin{pmatrix} x_a & x_b & x_v \\ y_a & y_b & y_v \\ 1 & 1 & 1 \end{pmatrix}.$$

In your case it is $$\frac{1}{2}\det\; \begin{pmatrix} -5 & -3 & 2 \\ 0 & -7 & -2 \\ 1 & 1 & 1 \end{pmatrix}=(35+6-10+14)/2=45/2.$$

The formula generalizes nicely to higher dimensions.

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