# How do you find the area of a triangle whose vertices of triangle are (2,0,0) ; (0,3,0) ; (0,0,5)?

Apr 21, 2018

$\frac{19}{2}$

#### Explanation:

The most general way to find the area of a triangle given three or more dimensional coordinates is to use Archimedes' Theorem. A triangle with squared sides $A$, $B$, and $C$ has an area $a$ that satisfies:

$16 {a}^{2} = 4 A B - {\left(C - A - B\right)}^{2}$

Here we have

$A = {2}^{2} + {3}^{2} = 13$
$B = {2}^{2} + {5}^{2} = 29$
$C = {3}^{2} + {5}^{2} = 34$

$16 {a}^{2} = 4 \left(13\right) \left(29\right) - {\left(34 - 13 - 29\right)}^{2} = 1444$

$a = \sqrt{\frac{1444}{16}} = \frac{19}{2}$

Apr 21, 2018

Given the coordinates of the vertices of a triangle.

Use the points to make two vectors, $\vec{a}$, and $\vec{b}$

Compute the cross-product $\vec{c} = \vec{a} \times \vec{b}$

$\text{Area} = \frac{1}{2} | \vec{c} |$

#### Explanation:

Compute the vector from $\left(2 , 0 , 0\right)$ to $\left(0 , 3 , 0\right)$

$\vec{a} = \left(0 - 2\right) \hat{i} + \left(3 - 0\right) \hat{j} + \left(0 - 0\right) \hat{k}$

$\vec{a} = - 2 \hat{i} + 3 \hat{j}$

Compute the vector from $\left(2 , 0 , 0\right)$ to $\left(0 , 0 , 5\right)$

$\vec{b} = \left(0 - 2\right) \hat{i} + \left(0 - 0\right) \hat{j} + \left(5 - 0\right) \hat{k}$

$\vec{b} = - 2 \hat{i} + 5 \hat{k}$

This reference tells you how to compute the Cross product.

The cross product $\vec{a} \times \vec{b}$ is:

$\vec{c} = \left(- 2 \hat{i} + 3 \hat{j}\right) \times \left(- 2 \hat{i} + 5 \hat{k}\right)$

$\vec{c} = 15 \hat{i} + 10 \hat{j} + 6 \hat{k}$

The magnitude of $\vec{c}$ is:

$| \vec{c} | = \sqrt{{15}^{2} + {10}^{2} + {6}^{2}}$

$| \vec{c} | = 19$

The area of the triangle is half of the magnitude:

$\text{Area} = \frac{19}{2}$