Question

In Figure $AB=8cm$, $BC=6$cm, $AC=3$cm and the $\angle$ADC$={90}^{\circ }$. Calculate $CD$.

Answer 1

Let, $CD=x$ cm

Then from $\Delta$ ADC, using Pythagoras theorem ,

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

$\Rightarrow A{D}^2+x^2=3^2$

$\Rightarrow A{D}^2=(9-x^2)$.........(1)

In $\Delta$ ABD , using Pythagoras theorem,

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

$\Rightarrow 8^2=(6+x{)}^2+(9-x^2)$[Using equation (1)]

$\Rightarrow 64=36+x^2+12x+9-x^2$

$\Rightarrow 12x=64-45$

$\Rightarrow x=\frac{19}{12}$

$\therefore CD=$$\frac{19}{12}$ cm$=1.58\overline{3}$ cm