Question

In the above figure $\Delta ABC$ is a right triangle $\angle A={90}^{\circ }$ and CD is the median then $B{C}^2-C{D}^2$ = ____

• A. $\frac{1}{2}A{B}^2$
• B. $\frac{3}{2}A{B}^2$
• C. $\frac{3}{4}A{B}^2$
• D. $\frac{1}{4}A{B}^2$

Answer 1

The correct option is C $\frac{3}{4}A{B}^2$

In $△ABC$ using Pythagoras theorem

$(AB{)}^2+(AC{)}^2=(BC{)}^2.......(i)$

$CD$ is the median

$\therefore AB=2AD$

Similarly in $△ACD$

$(AD{)}^2+(AC{)}^2=(CD{)}^2$

$(\frac{AB}{2}{)}^2+(AC{)}^2=(CD{)}^2$

$\frac{(AB{)}^2}{4}+(AC{)}^2=(CD{)}^2.......\left(ii\right)$

Subtracting $\left(ii\right)$ from $\left(i\right)$

$(BC{)}^2+(CD{)}^2=\frac{3(AB{)}^2}{4}$

So option $C$ is correct/