Question

Find the area of the triangle whose vertices are $A(1,1,1),B(1,2,3)$ and $C(2,3,3)$

Answer 1

$A=(1,1,1),B=(1,2,3),C=(2,3,3)$

$a$ = length of $AB=\sqrt{(1-1{)}^2+(2-1{)}^2+(3-1{)}^2}=\sqrt{5}$

$b$= length of $BC=\sqrt{(2-1{)}^2+(3-2{)}^2+(3-3{)}^2}=\sqrt{2}$

$c$= length of $AC=\sqrt{(2-1{)}^2+(3-1{)}^2+(3-1{)}^2}=3$

Semi-perimeter $s=\frac{a+b+c}{2}=\frac{3+\sqrt{2}+\sqrt{5}}{2}$

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

$=\sqrt{\left(\frac{3+\sqrt{2}+\sqrt{5}}{2}\right)\times \left(\frac{\sqrt{2}+\sqrt{5}-3}{2}\right)\times \left(\frac{3+\sqrt{5}-\sqrt{2}}{2}\right)\times \left(\frac{3+\sqrt{2}-\sqrt{5}}{2}\right)}$

$=\frac{1}{4}\times \sqrt{(3+(\sqrt{2}+\sqrt{5}\left)\right)\times \left(\right(\sqrt{2}+\sqrt{5})-3)\times \left[\right(3+\sqrt{5}-\sqrt{2})\times (3-(\sqrt{5}-\sqrt{2})\left)\right]}$
$=\frac{1}{4}\times \sqrt{\left(\right(\sqrt{2}+\sqrt{5}{)}^2-3^2\left)\right(3^2-(\sqrt{5}-\sqrt{2}{)}^2)}$

$=\frac{1}{4}\times \sqrt{(2+5+2\sqrt{10}-9)\times (9-(7-2\sqrt{10}\left)\right)}$

$=\frac{1}{4}\times \sqrt{(2\sqrt{10}-2)\times (2+2\sqrt{10})}$

$=\frac{1}{4}\sqrt{(2\sqrt{10}{)}^2-2^2)}$

$=\frac{1}{4}\times \sqrt{40-4}$

$=\frac{1}{4}\times \sqrt{36}$

$=\frac{6}{4}=1.5sq.units$