3420.27 – Triangle Area

In parallelogram ABCD, M and N are midpoints of sides AD and BC, respectively. CM and DN intersect at point O. DN intersects AB at P, and CM intersects AB at Q. If the area of parallelogram ABCD is 24 sq cm, find the area of triangle QPO.

Solution

Master plan:

$\triangle QPO = \square ABNM + \triangle MNO + \triangle QAM + \triangle BPN.$

Here goes:

\begin{aligned} ABNM &=& \frac{1}{2} ABCD \\ \triangle MNO &=& \frac{1}{2}\triangle MNC, \text{ (half the base MC, same altitude)} \\ &=& \frac{1}{2}(\frac{1}{2}MNCD) = \frac{1}{2}(\frac{1}{2}(\frac{1}{2}ABCD)) = \frac{1}{8}ABCD, \\ \triangle QAM &=& \triangle CDM = \frac{1}{2}(MNCD) = \frac{1}{2}(\frac{1}{2}ABCD ) = \frac{1}{4} ABCD. \end{aligned}

Likewise, $\triangle PBN = \frac{1}{4} ABCD$ . So (see the top)

\begin{aligned} \triangle QPO &=& \frac{1}{2} ABCD + \frac{1}{8} ABCD + \frac{1}{4} ABCD +\frac{1}{4} ABCD \\ &=& \frac{9}{8}ABCD = \frac{9}{8} \cdot 24 = 27 \text{ sq cm}. \end{aligned}