Question

In the given figure, angle $ACB={90}^0=$ angle $ACD$. If $AB=10cm,BC=6cm\text{and}AD=17cm$, find:

(ii) $CD$

Answer 1

Since, $\Delta ABC$ is a right angled triangle with vertex C,


According to Pythagoras theorem,

$(Hypotenuse{)}^2$ = $(Perpendicular{)}^2$ + $(Base{)}^2$

$\Rightarrow (AB{)}^2$ = $(AC{)}^2$ + $(BC{)}^2$

$\Rightarrow (10{)}^2$ = $(AC{)}^2$ + $(6{)}^2$

$\Rightarrow 100=(AC{)}^2+36$

$\Rightarrow AC=\sqrt{100-36}$

$\Rightarrow AC=\sqrt{64}$

$\Rightarrow AC=8$

Hence, AC = 8 cm.


Since, $\Delta ACD$ is a right angled triangle with vertex C,

According to Pythagoras theorem,

$(Hypotenuse{)}^2$ = $(Perpendicular{)}^2$ + $(Base{)}^2$

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

$\Rightarrow (17{)}^2$ = $(8{)}^2$ + $(CD{)}^2$

$\Rightarrow 289=64+(CD{)}^2$

$\Rightarrow CD=\sqrt{289-64}$

$\Rightarrow CD=\sqrt{225}$

$\Rightarrow CD=15$

Hence, CD is 15 cm.