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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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    Thank you for your comment. I see my mistake and what I did wrong.2012-12-20

3 Answers 3

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

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!2012-12-20
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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!2012-12-20
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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.