Question

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

• A. 12 square units
• B. 15 square units
• C. 30 square units
• D. 36 square units

Answer 1

Answer: (B)

15 square units

Answer 2

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