Question

In the given figure, ABCD is a parallelogram, if M is any point that lies on AD when produced, then which of the following is not true?

• A.
  $ar\left(\Delta ABD\right)=ar\left(\Delta ACB\right)$
• B.
  $ar\left(\Delta ABC\right)=\frac{1}{2}ar\left(ABCD\right)$
  
• C.
  $ar\left(\Delta AMC\right)=2ar\left(\Delta ADC\right)$
• D.
  $ar\left(MBD\right)=ar\left(\Delta DCM\right)$

Answer 1

The correct option is C

$ar\left(\Delta AMC\right)=2ar\left(\Delta ADC\right)$
We know that triangles on the same base and between the same parallels are equal in area.

$\Delta ABDand\Delta ABC$lie on the same base AB and between the same parallels AB and DC.

$\therefore Ar\left(\Delta ABD\right)=Ar\left(\Delta ABC\right)Also,\Delta MBDand\Delta DMC$ lie on the same base MD and between the same parallels AM and BC.

$\therefore Ar\left(\Delta MBD\right)=Ar\left(\Delta DCM\right)$

Now, we know that if a parallelogram and a triangle lie on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Parallelogram ABCD and $\Delta ABC$ lie on the same base AB and between the same parallels AB and CD.



$\therefore Ar\left(\Delta ABC\right)=\frac{1}{2}\left(ABCD\right)$

Let N be a point on AD when produced, such that $CN\perp AD.$
Now, $Ar\left(\Delta AMC\right)=\frac{1}{2}\times AM\times CNandAr\left(\Delta ADC\right)=\frac{1}{2}\times AD\times CN$

However, since $AM\ne 2AD.$

Therefore, $Ar\left(\Delta AMC\right)\ne 2Ar\left(\Delta ADC\right).$

Hence, the correct answer is option (c)