Question

In a parallelogram $ABCD,\left|AB\right|=a,\left|AD\right|=b$ and $\left|AC\right|=c.$ Then, $DB.AB$ has the value

• A. $\frac{3a^2+b^2-c^2}{2}$
• B. $\frac{a^2+3b^2-c^2}{2}$
• C. $\frac{a^2-b^2+3c^2}{2}$
• D. $\frac{a^2+3b^2+c^2}{2}$

Answer 1

The correct option is A $\frac{3a^2+b^2-c^2}{2}$

$\because DB=DA+AB$ or, $DA=DB-AB$

$\therefore {\left(DA\right)}^2={\left(DB\right)}^2+{\left(AB\right)}^2-2DB.AB$

In parallelogram $2(a^2+b^2)=c^2+{DB}^2$

$\therefore {\left(DB\right)}^2=2a^2+2b^2-c^2$

$\therefore$ From (1), $b^2=2a^2+{2b}^2-c^2+a^2-2AB.DB$

$\therefore AB.DB=\frac{3a^2+b^2-c^2}{2}$.