How to find areas of similar triangles

a) We are to find area of $\bigtriangleup{ABC}$

This is quite simple, all we need to do is to use the idea of similar triangles, to do this, simply extract the $\bigtriangleup{AMN}$ from $\bigtriangleup{ABC}$ so that we now have two similar triangles.

From this, it is evident that $AM=6cm$ and $AB=6+4=10cm$. $AM$ and $AB$ are corresponding sides therefore

$\frac {AM}{AB}=\frac{6}{10}=\frac{3}{5}$ is the scale factor. 

Next we move to obtaining the ratio of the areas of the triangles known as area factor, 

Area factor=$(scaler factor)^2=(\frac{3}{5})^2=\frac{9}{25}$

From the area factor we now obtain the area for $\bigtriangleup{ABC}$:

$\frac{Area of \bigtriangleup{AMN}}{Area of \bigtriangleup{ABC}}=\frac{9}{25}$

Area of $\bigtriangleup{AMN}=12cm^2$ is given from the figure, 

$\frac{12}{Area of \bigtriangleup{ABC}}=\frac{9}{25}$

$Area of \bigtriangleup{ABC}=\frac{12\times{25}}{9}=\frac{300}{9}=33.3cm^2$

b) Area of $MNCB$, $MNCB$ is not a triangle therefore we obtain its area from the area of the triangles in the shape. 

Since the area of $\bigtriangleup{AMN}=12cm^2$ and the area of $\bigtriangleup{ABC}=33.3cm^2$ then we can obtain the area of $MNCB$ by simply subtracting area of the smaller triangle from the area of the bigger triangle. 

=$33. 3-12=21.3cm^2$

Reference:

New concept mathematics, JSS III, exercise 13.1, for Nigeria schools