Question

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

Answer 1

Hint:In order to determine the area of $\Delta ABC$,using the formula$Area\,of\,\Delta = \dfrac{1}{2}|\vec a \times \vec b|$where $\vec a\,$and $\vec b\,$are the vectors from A to B and A to C vertices respectively to find the desired result.

Using heron’s formula

$
s = \dfrac{{a + b + c}}{2} \\
\\
$

Formula:
$Area\,of\,\Delta = \dfrac{1}{2}|\vec a \times \vec b|$

Complete step by step solution:
Given a triangle let it be $\Delta ABC$having vertices as $A(2,0,0),B(0,3,0)\,and\,C(0,0,5)$
Since, we are given the vertices in 3-dimensional plane,

So to calculate the area of $\Delta ABC$, we have to determine two vectors let them be
$\vec a\,$and $\vec b\,$

And their cross product $\vec c = \vec a \times \vec b$

The area of triangle will be
$Area = \dfrac{1}{2}|\vec c|$ -----------(1)

First computing the vector $\vec a\,$from vertices $A(2,0,0)$to$B(0,3,0)\,$

$
\vec a\, = (0 - 2)\hat i + (3 - 0)\hat j + (0 - 0)\hat k \\
\vec a\, = - 2\hat i + 3\hat j \\

$

computing the vector $\vec b\,$from vertices $A(2,0,0)$to$C(0,0,5)\,$

$
\vec b\, = (0 - 2)\hat i + (0 - 0)\hat j + (5 - 0)\hat k \\
\vec b\, = - 2\hat i + 5\hat k \\
$

Now calculating the cross product of $\vec a\,$and $\vec b\,$

$
\vec c = \vec a \times \vec b \\
\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 \\
$

Now finding the magnitude of $\vec c$as$\sqrt {{l^2} + {m^2} + {n^2}} $
\[

\left| {\vec c} \right| = \sqrt {{l^2} + {m^2} + {n^2}} \\

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

\left| {\vec c} \right| = \sqrt {225 + 100 + 36} \\

\left| {\vec c} \right| = \sqrt {361} \\

\left| {\vec c} \right| = 19 \\
\]

Now putting the value of \[\left| {\vec c} \right|\]in the equation (1)

$
Area = \dfrac{1}{2}|\vec c| \\
Area = \dfrac{1}{2}(19) \\
Area = \dfrac{{19}}{2}sq.units \\
$

Therefore, the area of $\Delta ABC$is equal to $\dfrac{{19}}{2}sq.units$.

Additional Information: 1. A triangle is a closed geometric shape having three no of edges and three vertices. A triangle having vertices named as A,B and C are denoted by $\Delta ABC$.

2.Area of Triangle in a 2-dimensional plane can be determined by following ways depending upon what type of information is given . If length of base and height is given then

Area of $\Delta ABC$=$\dfrac{1}{2}\left( {base \times height} \right)$
If length of all the sides are given then,

Area of $\Delta ABC$=$\sqrt {s(s - a)(s - b)(s - c)} $

If coordinates of the all 3 vertices in cartesian plane are given then,
Area of $\Delta ABC$$ = \left( {\dfrac{1}{2}} \right)\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right]$

Note: 1. Don’t Forgot to cross check the answer.
2.Use the correct formula and calculate the cross product carefully.
3. Note that the unit of area of the triangle is square units.