In a , =90^ , AB = 5 cm and AC = 12 cm. If AD , then AD = - Vidyadip

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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