Question

If $A,B,C,D$ are four points in space such that the area of a triangle $ABC$ is $4$ sq units then find the value of $|AB\ast CD+BC\ast AD+CA\ast BD|.$

• A. $40$
• B. $20$
• C. $10$
• D. $noneofthese$

Answer 1

The correct option is A $40$

$A(0,2),B(0,0),C(4,0)$

Area of triangle $=\frac{1}{2}b\times h$

$4=\frac{1}{2}b\times h$

$\Rightarrow 8=b\times h$

$\therefore 8=AB\times BC$

Assume $AB=2,BC=4$

$AB=\sqrt{0^2+2^2}=2$

$BC=\sqrt{4^2+0^2}=4$

$AC=\sqrt{4^2+2^2}=\sqrt{20}$

$BD=\sqrt{4^2+2^2}=\sqrt{20}$

$AD=\sqrt{4^2+0^2}=4$

$DC=\sqrt{0^2+2^2}=2$

$|AB\times CD+BC\times AD+CA\times BD|$

$=|2\times 2+4\times 4+\sqrt{20}\times \sqrt{20}|$

$=|4+16+20|$

$=40$ units.