Hello, Sarah!

$\displaystyle A, B,\text{ and }C$ are midpoints of the sides of $\displaystyle \Delta PQR$.

$\displaystyle M, N,\text{ and }O$ are the midpoints of the sides of $\displaystyle \Delta ABC$.

The interiors of $\displaystyle \Delta PAB,\:\Delta AQC,\:\Delta BCR,\text{ and }\Delta MNO$ are shaded.

What fraction of the interior of $\displaystyle \Delta PQR$ is shaded? Code:

P
*
/:\
/:::\
/:::::\
/:::::::\
/::::O::::\
A* - - * - - *B
/:\ /:\ /:\
/:::\ /:::\ /:::\
/::::M*- - -*N::::\
/:::::::\ /:::::::\
/:::::::::\-/:::::::::\
* - - - - - * - - - - - *
Q C R

It can be shown that:

. . When the midpoints of the sides of a triangle are joined,

. . the triangle is divided into four congruent triangles.

Each triangle has one-fourth of the area.

Joining the midpoints of the side of $\displaystyle \Delta PQR$

. . we see that $\displaystyle \frac{3}{4}\text{ of }\Delta PQR$ is shaded.

The unshaded triangle, $\displaystyle \Delta ABC$, has $\displaystyle \frac{1}{4}$ the area of $\displaystyle \Delta PQR.$

. . and $\displaystyle \frac{1}{4}$ of it is shaded.

That is: .$\displaystyle \Delta MNO \:=\:\left(\frac{1}{4}\right)\left(\frac{1}{4}\rig ht)\:=\:\frac{1}{16}\, \text{ of }\Delta PQR$

Therefore, the total shaded area is: .$\displaystyle \frac{3}{4} + \frac{1}{16} \:=\:\boxed{\frac{13}{16}}\,\text{ of }\Delta PQR.$