+618 
In the diagram, $ABCD$ and $EFGD$ are squares each of area 16. If $H$ is the midpoint of both $BC$ and $EF$, find the total area of polygon $ABHFGD$.
+33616 The side of each square must be of length 4 ( to give areas of 16). The length of HC is therefore 2, as is that of BH.
Imagine a line from H to D. Triangle HCD has area (1/2)*2*4 = 4. Similarly triangle HDE has area 4. Hence area of polygon HCDE is 8.
Area of polygon ABHED is therefore 16 - 8 = 8
Add this to the area of square DEFG to get a total area of 16 + 8 = 24 for polygon ABHFGD.
+33616 The side of each square must be of length 4 ( to give areas of 16). The length of HC is therefore 2, as is that of BH.
Imagine a line from H to D. Triangle HCD has area (1/2)*2*4 = 4. Similarly triangle HDE has area 4. Hence area of polygon HCDE is 8.
Area of polygon ABHED is therefore 16 - 8 = 8
Add this to the area of square DEFG to get a total area of 16 + 8 = 24 for polygon ABHFGD.
