Question

In the given figure, ABC is a right triangle right-angled at B, EF  ⊥  BC. If AF  =  3 cm, AB = $\sqrt{21}$cm and FC = 4 cm, then EF is equal to

• A. $4\sqrt{3}cm$
• B. $7\sqrt{\frac{3}{4}}cm$
  
• C. $4\sqrt{7}cm$
• D. $4\sqrt{\frac{3}{7}}cm$
  

Answer 1

The correct option is D $4\sqrt{\frac{3}{7}}cm$


In ΔABC and ΔFEC



$\angle ABC=\angle FEC$ (Each ${90}^0$)
$\angle ACB=\angle FCE$ (Common)
$\therefore \Delta ABC\sim \Delta FEC$ (AA similarity)
Thus, $\frac{AC}{FC}=\frac{AB}{FE}$
$\Rightarrow \frac{4+3}{4}=\frac{\sqrt{21}}{FE}$
$(\because AC=AF+FC)$
$\Rightarrow \frac{7}{4}=\frac{\sqrt{21}}{FE}$
$\Rightarrow FE=\frac{4\times \sqrt{7}\times \sqrt{3}}{7}$
$\Rightarrow FE=4\sqrt{\frac{3}{7}}cm$
Hence, the correct answer is option (d).