Question

In $△ABC,\angle BAC={90}^o$ and AD is the altitude. If $AB=2\sqrt{5},BD=4$. Then find BC and AC.

Answer 1

$A{B}^2=BC.BD$ [using the corollary]
$(2\sqrt{5}{)}^2=BC\times 4$

$\therefore \frac{20}{4}=BC$

$\therefore BC=5$
Now, $CD=BC-BD=5-4=1$
$A{C}^2=BC.CD$

$⟹A{C}^2=5\times 1$

$\therefore AC=\sqrt{5}$