Questions   
Sort: 
Apr 23, 2018
 #3
avatar+2439 
+3

Geometers detest making assumptions; it is generally frowned upon because the idea to construct sound logic using previous theorems and postulate to formulate new ideas. However, rectangles cannot have lengths--sides can. Because of this, I have had to dissect the given information and have had to make an educated guess as to where the given information fits. I will let this diagram demonstrate this:

 

 

We can solve for the remaining side length, \(BC\), by understanding the perimeter formula for a rectange. 

 

\(2(AB+BC)=\text{Perimeter}_{ABCD}\) Let's plug in the known values and solve for the missing one.
\(2(20\text{cm}+BC)=60\text{cm}\) Instead of distributig the 2, we can divide it from both sides. 
\(20\text{cm}+BC=30\text{cm}\) Subtract 20cm from both sides to finish the algebra.
\(BC=10\text{cm}\)  
   

 

We know that \(EFGH\) is similar to \(ABCD\), which means that the sides are proportional. We can use this information to our advantage. 

 

\(\frac{AB}{BC}=\frac{EF}{FG}\) Yet again, substitute in the known values and solve for the unknown.
\(\frac{20\text{cm}}{10\text{cm}}=\frac{32\text{cm}}{FG}\) Let's simplify the left hand side of this proportion so that the calculation is easier.
\(2=\frac{32\text{cm}}{FG}\) We can use cross-multiplication here to clear all fractions.
\(2FG=32\text{cm}\) Divide by 2 from both sides.
\(FG=16\text{cm}\) We are not done yet! We must find the area of the rectange. That's what the question is asking for, after all!
\(\text{Area}_{EFGH}=EF\cdot FG\) Plug those values in.
\(\text{Area}_{EFGH}=32\text{cm}*16\text{cm}=512\text{cm}^2\) This is your final answer!
Apr 23, 2018
 #2
avatar+2448 
+1
Apr 23, 2018
 #4
avatar
+2
Apr 23, 2018
 #3
avatar+12525 
+2
Apr 23, 2018

4 Online Users

avatar
avatar