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 $\triangle\text{ABC}$ and $\triangle\text{BDA}$
$\angle\text{BAC}=\angle\text{ADC}=90^\circ$
$\angle\text{B}=\angle\text{B}$ (Common)
$\triangle\text{ABC}\sim\triangle\text{BDA}$
$\frac{\text{AC}}{\text{AD}}=\frac{\text{BC}}{\text{AB}}\ ....(1)$
Using Pythagoras theorem in $\triangle\text{ABC}$ we get
$\text{BC}=\sqrt{(12)^2+(5)^2}$
$\Rightarrow\text{BC}=\sqrt{144+25}$
$\Rightarrow\text{BC}=\sqrt{169}$
$\Rightarrow\text{BC}={13}\text{cm}$
From $(1)$
$\Rightarrow\frac{12}{\text{AD}}=\frac{13}{5}$
$\Rightarrow\text{AD}=\frac{12\times5}{13}$
$\Rightarrow\text{AD}=\frac{60}{13}\text{cm}.$
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