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  4. What is the area of the triangle whose vertices are (0,0,0) (3, 4, 0) and (3, 4, 6) ?

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Q. What is the area of the triangle whose vertices are (0,0,0) (3, 4, 0) and (3, 4, 6) ?

Straight Lines

Solution:

Let A (0, 0, 0), B (3, 4, 0) and C (3, 4, 6)
Area, $(\Delta ) = \sqrt{(\Delta x^2 + \Delta y^2 + \Delta^2})$
Now,
$\Delta x = \frac{1}{2} \begin{vmatrix}y_{1}&z_{1}&1\\ y_{2}&z_{2}&1\\ y_{3}&z_{3}&1\end{vmatrix} = \frac{1}{2} \begin{vmatrix}0&0&1\\ 4 &0&1\\ 4&6&1\end{vmatrix} $
$= \frac{1}{2}\left[1\left(24\right)\right] = 12 $
Similarly,
$\Delta y = \frac{1}{2} \begin{vmatrix}0&0&1\\ 0 &3&1\\ 6 &3&1\end{vmatrix} = \frac{1}{2}\left[1\left(-8\right)\right]=-9 $
and $\Delta z = \frac{1}{2}\begin{vmatrix}0&0&1\\ 3&4&1\\ 3&4&1\end{vmatrix} = \frac{1}{2}\left[12-12\right]=0$
Area of $ \Delta = \sqrt{12^{2} + \left(-9\right)^{2} +\left(0\right)^{2}} $
$= \sqrt{144+81} = \sqrt{225} = 15 $ square units