Question

In the given figure, ABC is a triangle and FD is a median of $\Delta FBC.IfDE||BF,AB||DFandareaof\Delta ABC=48c{m}^2,thenar\left(\Delta ADF\right)+ar\left(\Delta BDE\right)is$

• A.
  $6c{m}^2$
• B.
  $12c{m}^2$
• C.
  $18c{m}^2$
• D.
  $32c{m}^2$

Answer 1

The correct option is C

$18c{m}^2$


Since, FD is a median of $\Delta FBC.\therefore BD=CDAlso,BF||DEin\Delta BCF.\therefore CE=EF\text{(By converse of mid-point theorem)}Now,in\Delta ABC,AB||DFAnd,BD=CD\therefore AF=CF\text{(By converse of mid-point theorem)}\Rightarrow F\text{is the median of}\Delta ADC.Now,areaof\Delta ABCis48c{m}^2.$

And, we know that median divides the triangle into two equal halves.

$\therefore Ar\left(\Delta ADC\right)=\frac{1}{2}\times Ar\left(\Delta ABC\right)=\frac{1}{2}\times 48=24c{m}^{2}Also,Ar\left(\Delta ADF\right)=\frac{1}{2}\times Ar\left(\Delta ADC\right)(\because \text{F is the mid-point of AC})=\frac{1}{2}\times 24=12c{m}^2$

We know that triangles on the same base and between the same parallels are equal in area.
$\therefore Ar\left(\Delta BDE\right)=Ar\left(\Delta DEF\right)(\because BF||DE)=\frac{1}{2}\times Ar\left(\Delta DFC\right)\text{(DE is the median of}\Delta DFC)=\frac{1}{2}\times Ar(\Delta ADF\left)\text{(DF is the median of}\Delta ADC\right)=\frac{1}{2}\times 12=6c{m}^2\Rightarrow Ar\left(\Delta ADF\right)+Ar\left(\Delta BDE\right)=12+6=18c{m}^2$

Hence, the correct answer is option (c).