Question

In $\Delta ABC,N$ is any point on the median AM, where M is on the side BC. If $ar\left(ABC\right)=80c{m}^{2}andar\left(BMN\right)=12c{m}^2,$ then the area of $\Delta ABN$ is

• A.
  $56c{m}^2$
• B.
  $26c{m}^2$
• C.
  $28c{m}^2$
• D.
  $34c{m}^2$

Answer 1

The correct option is C

$28c{m}^2$
Given: $Ar\left(\Delta ABC\right)=80c{m}^{2}andAr\left(\Delta BMN\right)=12c{m}^2\text{Let AP be the height of}\Delta ABC.Now,Ar\left(\Delta ABC\right)=80c{m}^2\Rightarrow \frac{1}{2}\times BC\times AP=80\Rightarrow \frac{1}{2}\times 2BM\times AP=80\text{(BC = 2BM, AM is the median)}\Rightarrow BM\times AP=80\dots ..\left(i\right)\therefore Ar\left(\Delta ABM\right)=\frac{1}{2}\times AP\times BM=\frac{1}{2}\times 80\text{[From (i)]}=40c{m}^2\dots ..\left(ii\right)Now,Ar\left(\Delta ABN\right)=Ar\left(\Delta ABM\right)–Ar\left(\Delta BMN\right)=40–12\text{[From (ii)]}=28c{m}^2$

Hence, the correct answer is option (c).