Question

In the given figure, $\angle ABC={90}^o$ and $BD\perp AC$. If BD = 8 cm, AD = 4 cm, find CD.

Answer 1

Sol:
Consider ΔABD,

$A{B}^2=A{D}^2+B{D}^2=8^2+4^2=64+16=80AB=4\sqrt{5}$

Consider ΔBCD,

Let BD = x

$B{C}^2=B{D}^2+C{D}^2=8^2+x^2=64+x^{2}BC=\sqrt{(64+x^2)}$

Consider ΔABC,
$A{C}^2=B{C}^2+A{B}^2(4+x{)}^2=(4\sqrt{5}{)}^2+(\sqrt{(64+x^2)}{)}^2\Rightarrow 16+x^2+8x=80+64+x^2\Rightarrow 16+8x=80+64\Rightarrow 16+8x=144\Rightarrow 8x=128\Rightarrow x=16$

Therefore, the length of CD is 16 cm.