Official Solution:

In the diagram above, ABCD is a parallelogram and the areas of yellow regions are 8, 10, 72, and 79. What is the area of the red triangle?
A. \(7\)
B. \(8\)
C. \(9\)
D. \(10\)
E. \(12\)
Recall that the area of a parallelogram is \(bh\), where \(b\) is the base of the parallelogram and \(h\) is its height (altitude to the base).
Consider the blue triangle below:
It's area is \(\frac{1}{2}*AD*(height \ to \ AD)\) but \(AD*(height \ to \ AD)\) is the area of the parallelogram, so the area of the blue triangle is half of the area of the parallelogram.
Now, consider two green triangles. Their combined area is
\(\frac{1}{2}*AE*(height \ to \ AB)+\frac{1}{2}*EB*(height \ to \ AB) =\)
\(\frac{1}{2}*(height \ to \ AB)*(AE+EB)= \frac{1}{2}*(height \ to \ AB)*AB\). The same way as above, \((height \ to \ AB)*AB\) is the area of the parallelogram, so the combined area of two green triangle is half of the area of the parallelogram.
Thus, the area of blue triangle is equal to the combined area of two green triangles (both are half of the area of the parallelogram):
\(x+79+y+10=x+(red \ region)+72+y+8\);
\((red \ region)=89-80=9\).
Answer: C