Let us draw the figure according to the available information,

In $\triangle$ AMP and $\triangle$ CMR

$\angle$ MAP = $\angle$ MCR

$\angle$ AMP = $\angle$ CMR

Therefore, we can say that $\triangle$ AMP $\sim$ $\triangle$ CMR

Hence, we can say that $\dfrac{AP}{CR}=\dfrac{MP}{MR}$

$\Rightarrow$ $\dfrac{AB/3}{AB/2}=\dfrac{MP}{MR}$

$\Rightarrow$ $MR=\dfrac{3}{2}*MP$

Therefore, we can say that $\Rightarrow$ $MR=\dfrac{3}{5}*RP$ ... (1)

Similarly, in $\triangle$ ANQ and $\triangle$ CNR

$\angle$ NAQ = $\angle$ NCR

$\angle$ ANQ = $\angle$ CNR

Therefore, we can say that $\triangle$ ANQ $\sim$ $\triangle$ CNR

Hence, we can say that $\dfrac{AQ}{CR}=\dfrac{NQ}{NR}$

$\Rightarrow$ $\dfrac{2AB/3}{AB/2}=\dfrac{NQ}{NR}$

$\Rightarrow$ $NR=\dfrac{3}{4}*NQ$

Therefore, we can say that $\Rightarrow$ $NR=\dfrac{3}{7}*RQ$ ... (2)

In $\triangle$ RMN and $\triangle$ RPQ

$\dfrac{\text{Area of triangle RMN}}{\text{Area of triangle RPQ}} = \dfrac{0.5*RM*RN*sinMRN}{0.5*RP*RQ*sinPRQ}$

$\text{Area of triangle RMN}=\dfrac{3}{5}*\dfrac{3}{7}*\text{Area of triangle RPQ}$

We know that, Area of triangle RPQ = 1/6*Area of rectangle ABCD = 1/6*90 = 15 sq. units

$\Rightarrow$ $\text{Area of triangle RMN}=\dfrac{3}{5}*\dfrac{3}{7}*15$ = $\dfrac{27}{7}$ sq. units

Hence, the area of the quadrilateral PQNM = 15 - $\dfrac{27}{7}$ = $\dfrac{78}{7}=11\dfrac{1}{7}$ sq. units. Therefore, option D is the correct answer.

Consider a rectangle ABCD of area 90 units. The points P and Q trisect AB, and R bisects CD. Thediagonal AC intersects the line segments PR and QR at M and N respectively. What is the area of the quadrilateral PQNM?

Consider a rectangle ABCD of area 90 units. The points P and Q trisect AB, and R bisects CD. The
diagonal AC intersects the line segments PR and QR at M and N respectively. What is the area of the quadrilateral PQNM?