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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

Hint

The area of a triangle is half of the area of parallelogram

- Thread starter Petrus
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- Thread starter
- #1

- Feb 21, 2013

- 739

Hint

The area of a triangle is half of the area of parallelogram

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- #2

- Jan 26, 2012

- 4,205

The vector cross product, I believe (or the magnitude thereof) gives the area of the corresponding parallelogram. Hence, divide by two to get the answer. I get

$$[ \langle -4,1,0 \rangle- \langle 1,4,0 \rangle] \times [ \langle -4,1,0 \rangle- \langle -5,10,0 \rangle]= \langle -5,-3,0 \rangle \times \langle 1,-9,0 \rangle$$

$$= \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}\\ -5 &-3 &0 \\ 1 &-9 &0\end{matrix} \right| = \hat{k}((-5)(-9)-(-3)(1))= \hat{k}(45+3)=48 \hat{k}.$$

Hence, the area of the triangle is $24$ units squared.

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- #3

Using the formula I developed in this topic, we find:

\(\displaystyle A=\frac{1}{2}|(-5+4)(4-1)-(1+4)(10-1)|=\frac{48}{2}=24\)

If you get really desperate, you could work out the length of each side of the triangle by evaluating the distance between each set of points, and then use Heron's Formula.

Hint

The area of a triangle is half of the area of parallelogram

- Mar 15, 2012

- 8