MCQ
In a $\triangle\text{ABC},\ \angle\text{A}=90^\circ, AB = 5\ cm$ and $AC = 12\ cm.$ If $\text{AD}\perp\text{BC},$ then $AD =$
- A$\frac{13}{2}\text{cm}.$
- ✓$\frac{60}{13}\text{cm}.$
- C$\frac{13}{60}\text{cm}.$
- D$\frac{2\sqrt{15}}{13}\text{cm}.$
✓
Answer
Correct option: B.
$\frac{60}{13}\text{cm}.$
In$\triangle\text{ABC},$
$\angle\text{A}=90^\circ, AB = 5\ cm, AC = 12\ cm$
$\text{AD}\perp\text{BC}$
$\text{BC}^2=\text{AB}^2+\text{AC}^2 ($Pythagoras Theorem$)$
$=(5)^2+(12)^2$
$=25+144$
$=169$
$=(13)^2$
$\therefore\text{BC}=13\text{cm}$
Now area of $\triangle\text{ABC}=\frac{1}{2}\text{AB}\times\text{AC}$
$=\frac{1}{2}\times5\times12$
$=30\text{cm}^2$
and also area of $\triangle\text{ABC}=\frac{1}{2}\text{BC}\times\text{AD}$
$\Rightarrow30=\frac{1}{2}\times13\times\text{AD}$
$\Rightarrow\text{AD}=\frac{30\times2}{13}=\frac{60}{13}\text{cm}.$
$\angle\text{A}=90^\circ, AB = 5\ cm, AC = 12\ cm$
$\text{AD}\perp\text{BC}$
$\text{BC}^2=\text{AB}^2+\text{AC}^2 ($Pythagoras Theorem$)$
$=(5)^2+(12)^2$
$=25+144$
$=169$
$=(13)^2$
$\therefore\text{BC}=13\text{cm}$
Now area of $\triangle\text{ABC}=\frac{1}{2}\text{AB}\times\text{AC}$
$=\frac{1}{2}\times5\times12$
$=30\text{cm}^2$
and also area of $\triangle\text{ABC}=\frac{1}{2}\text{BC}\times\text{AD}$
$\Rightarrow30=\frac{1}{2}\times13\times\text{AD}$
$\Rightarrow\text{AD}=\frac{30\times2}{13}=\frac{60}{13}\text{cm}.$
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