Question
In figure, Triangle $\text{ABC}$ is a right angled triangle at $B$. Given that $AB = 9\ cm, AC = 15\ cm$ and $D, E$ are the mid$-$points of the sides $AB$ and $AC$ respectively, calculate
$i.$ The length of $BC$
$ii.$ The area of $\triangle\text{ADE}.$

Answer


$i.$ In $\triangle\text{ABC},\angle\text{B}=90^\circ,$
By using Pythagoras theorem
$AC^2 = AB^2 + BC^2$
$\Rightarrow15^2=9^2+\text{BC}^2$
$\Rightarrow\text{BC}=\sqrt{15^2-9^2}$
$\Rightarrow\text{BC}=\sqrt{225-81}$
$\Rightarrow\text{BC}=\sqrt{144}=12\text{ cm}$
$ii.$ In $\triangle\text{ABC},$
$D$ and $E$ are mid$-$points of $AB$ and $AC$
$\therefore\text{DE}\|\text{BC},=\frac{1}{2}\text{BC} [$By mid$−$point theorem$]$
$\text{AD}=\text{DB}=\frac{\text{AB}}{2}=\frac{9}{2}=4.5\text{ cm}$
$[\therefore$ $D$ is the mid$−$point of $AB]$
Area of $\triangle\text{ADE}=\frac{1}{2}\times\text{AD}\times\text{DE}$
$=\frac{1}{2}\times4.5\times6$
$=13.5\text{ cm}^2$

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