Question

In $\Delta ABC,m\angle B<90$ and $\overline{AD}\perp \overline{BC}$. Also $AB=9,BC=6$ and $BD=2$ then find $AC$.

Answer 1

By Pythagoras theorem in $\Delta ADB$,

$A{B}^2=B{D}^2+A{D}^2$

$\Rightarrow A{D}^2=A{B}^{2}0-B{D}^2$

Similarly in $\Delta ACD,A{C}^2=A{D}^2+D{C}^2$

$\Rightarrow A{C}^2=A{B}^2-B{D}^2+D{C}^2$

$=81-4+16=93$

$\Rightarrow AC=\sqrt{93}cm$.