Question

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

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

Answer 1

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

As we know if any two right triangles have an acute angle measure in common, the ratios of their corresponding side lengths would be equal.

In ΔABC and ΔFEC:



$\angle ABC=\angle FEC\text{.... Each}{90}^{\circ }$

$\angle ACB=\angle FCE\text{(Common)}$

$\text{Thus, we see:}\frac{AC}{FC}=\frac{AB}{FE}$

Putting values we get:

$\Rightarrow \frac{4+3}{4}=\frac{\sqrt{21}}{FE}(\because AC=AF+FC)$

$\Rightarrow \frac{7}{4}=\frac{\sqrt{21}}{FE}$

$\Rightarrow FE=\frac{4\times \sqrt{7}\times \sqrt{3}}{7}$

$\Rightarrow FE=4\sqrt{\frac{3}{7}}yd$

Hence, the correct answer is option (d).