What is the area of the shaded region? Problem 2 seems to have two ways of going about it.
Way 1
Assume the whole shape is a triangle and the unshaded region is a trapezoid. Subtract the trapezoid's area from the triangle's area.
Way 2
Assume both shaded regions are triangles. Add the shaded triangles' areas. 
I'm fairly certain at least Triangle 1's area is correct because finding the area of the trapezoid containing Triangle 1 and the unshaded trapezoid yields the same area as when combining the area of Triangle 1 determined in Way 2 and the area of the trapezoid determined in Way 1: A=(a+b)*h/2 = ((12+13)+15)*11/2 = 220.
Question
Why doesn't Way 2 provide the correct answer?

 A: It's the second solution which is correct, and the first one is wrong!
The error hides in the assumption the whole figure is a triangle. If it was, the big triangle would be similar to the smaller one on the right side, hence the proportion would hold
$$\frac{12+13}{11+21}=\frac{15}{21}$$
However, it does not, as
$$\frac{12+13}{11+21}=0.78125 > 0.7142857 \approx \frac{15}{21}$$
and the big figure is a concave quadrangle.
A: You numerical calculation is correct, indeed for the first


*

*whole: $\frac12 \cdot 32 \cdot 25 =400$

*trapezium: $\frac12 \cdot (15+12) \cdot 11 = 148.5$


then $A=251.5$.
For the second


*

*triangle 1: $\frac12 \cdot 15 \cdot 21 = 157.5$

*triangle 1: $\frac12 \cdot 13 \cdot 11 = 71.5$


then $A=229$.
The discreapncy depends upon the fact that the given of the problems are wrong, indeed observe that the segment with length 15 is not orthogonal to the base indeed
$$\frac{15}{21}=\frac{5}{7}\approx 0.71423\neq \frac{25}{32}=0.78125$$
A: The first triangle is not a right triangle, so determining the base and height for the formula $A=\frac12(base)(height)$ is a little more delicate.
Indeed, for the base of triangle $1$, one needs to draw a right triangle with the base of triangle $1$ being the hypotenuse of the "new" triangle. Then we can calculate the base to have length:
$$
\sqrt{3^2+11^2}=\sqrt{130}.
$$
The height is a little more challenging since we don't know the angles of the given triangles, but the length of the third side can easily be calculated using a similar technique, and one can then apply Heron's Formula to calculate the area of triangle $1$.
