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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 $ABHFGD$.

 Aug 25, 2017

Best Answer 

 #1
avatar+27795 
+2

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.

 Aug 25, 2017
 #1
avatar+27795 
+2
Best Answer

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.

Alan Aug 25, 2017

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