Question

In a $△ABC,$ let $D,E$ and $F$ be the points on lines $BC,CA$ and $AB$ which are concurrent. If $\frac{CE}{EA}=\frac{3}{5},\frac{BD}{DC}=\frac{5}{7}$ and side $AB$ has a length of $14\text{cm},$ then the lengths of $AF$ and $FB$ will be respectively :

• A. $9\text{cm},5\text{cm}$
• B. $7\text{cm},7\text{cm}$
• C. $3\text{cm},7\text{cm}$
• D. $9.8\text{cm},4.2\text{cm}$

Answer 1

The correct option is D $9.8\text{cm},4.2\text{cm}$

If $AD,BE,CF$ are concurrent lines meeting at the point $O$(assuming) of a $△ABC$ then
$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$
$\Rightarrow \frac{AF}{FB}\times \frac{5}{7}\times \frac{3}{5}=1$
$\Rightarrow \frac{AF}{FB}=\frac{7}{3}$

Given : $AB=14\text{cm}$
$\Rightarrow AF=7\times 1.4\text{cm}=9.8\text{cm}$
$\Rightarrow FB=3\times 1.4\text{cm}=4.2\text{cm}$