Question

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

Answer 1

Since, $\Delta ABC$ is a right- angled triangle with vertex $C$,​ According to Pythagoras Theorem,
​
$({\text{Hypotenuse)}}^2={\text{(Perpendicular)}}^2+{\text{(Base)}}^2$

$(AB{)}^2=(AC{)}^2+(BC{)}^2$
$(10{)}^2=(AC{)}^2+(6{)}^2$
$100=(AC{)}^2+36$
$AC=\sqrt{100-36}$
$AC=\sqrt{64}$
$AC=8cm$
Hence, $AC=8cm$.
According to Pythagoras Theorem, ​
$({\text{Hypotenuse)}}^2={\text{(Perpendicular)}}^2+{\text{(Base)}}^2$
$(AD{)}^2=(AC{)}^2+(CD{)}^2$
$(17{)}^2=(8{)}^2+(CD{)}^2$
$289=64+(CD{)}^2$
$CD=\sqrt{289-64}$
$CD=\sqrt{225}$
$CD=15cm$
Hence, $CD\text{is}15cm$