Question

In the given figure, ABC is a triangle. DE is parallel to and $\frac{AD}{DB}=\frac{3}{2}$

(i) Determine the ratio $\frac{AD}{AB}$.

(ii) What is the ratio of the areas of $\Delta DEF$ and $\Delta BFC?$

• A. (i) $\frac{3}{5}$, (ii) 9 : 25
• B. (i) $\frac{2}{5}$, (ii) 9 : 25
• C. (i) $\frac{2}{3}$, (ii) 4 : 9
• D. (i) $\frac{1}{2}$, (ii) 4 : 9

Answer 1

The correct option is A

(i) $\frac{3}{5}$, (ii) 9 : 25

$\left(i\right)\frac{AD}{DB}=\frac{3}{2}$

$\therefore \frac{DB}{AB}=\frac{2}{3}$

or $\frac{DB}{AD}+1=\frac{2}{3}+1$

or $\frac{DB+AD}{AD}=\frac{2+3}{3}$

or $\frac{AB}{AD}=\frac{5}{3}\Rightarrow \frac{AD}{AB}=\frac{3}{5}$

(ii) In $\Delta DEF$ and $\Delta CBF$

$\angle 1=\angle 2\left(\text{Alternate}\angle s\right)$

$\angle 3=\angle 4\left(\text{Alternate}\angle s\right)$

$\angle 5=\angle 6(\text{Vertically opp}.\angle s)$

$\therefore \Delta DEF\sim \Delta CBF$

$\frac{EF}{FB}=\frac{DE}{BC}=\frac{3}{5}$

As the ratio of the area of two similar triangles is equal to the ratio of the square of any corresponding sides.

$\therefore \frac{\text{Area of}\Delta DEF}{\text{Area of}\Delta BFC}=\frac{D{E}^2}{B{C}^2}$

$={\left(\frac{DE}{BC}\right)}^2={\left(\frac{3}{5}\right)}^2=\frac{9}{25}$