Question

In the given figure, ABCD is a parallelogram. If C and N are the mid-points of DM and CB respectively and area of $\Delta DCNis30c{m}^2,$ then the $ar\left(ABCD\right)+ar\left(\Delta CNM\right)$ is equal to

• A.
  $60c{m}^2$
• B.
  $120c{m}^2$
• C.
  $135c{m}^2$
• D.
  $150c{m}^2$

Answer 1

The correct option is D

$150c{m}^2$
Given: $Ar\left(\Delta DCN\right)=30c{m}^{2}LetPN\perp MD.Now,Ar\left(\Delta CNM\right)=\frac{1}{2}\times CM\times NP=\frac{1}{2}\times CD\times NP(\because CM=CD)=Ar\left(\Delta DCN\right)=30c{m}^2...\left(i\right)\text{Draw NG || AB.}\text{Since N is the mid-point of BC.}\therefore \text{G is also the mid-point of AD.}\therefore Ar\left(ABCD\right)=2Ar\left(GNCD\right)=2\left[2Ar\right(\Delta DCN\left)\right](\therefore \Delta DCN\text{and parallelogram GNCD lie}\text{on same base DC and between the same parallels}\text{GN and DC)}=4Ar(\Delta DCN)=4\times 30(Given)=120c{m}^2\dots ..(ii)\therefore Ar(ABCD)+Ar(\Delta CNM)=120+30\text{[From (i) and (ii)]}=150c{m}^2$
Hence, the correct answer is option (d).