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Quadrilateral $CDEF$ is a parallelogram. Its area is $36$ square units. Points $G$ and $H$ are the midpoints of sides $CD$ and $EF,$ respectively. What is the area of triangle $CDJ?$

Guest Oct 28, 2018

Best Answer 

 #1
avatar+963 
+3

\(\because \angle DHE=\angle FHJ \) (vertical angles) 

\(\because EH=HF\) (midpoint)

\(\because \overline{DE}\parallel\overline{FJ} \Rightarrow \angle EDJ =\angle FJD\)  (alternating interior angles)

\(\therefore \triangle HDE\cong\triangle HJF\) (AAS Congruency) 

\(\therefore [HDE]=[HJF] \Rightarrow [CDHF]+[HDE]=[CDHF]+[HJF]\)

\([CDJ]=[CDEF]\)

 

\(\boxed{36}\)

 

I hope this helped,

 

Gavin. 

GYanggg  Oct 28, 2018
 #1
avatar+963 
+3
Best Answer

\(\because \angle DHE=\angle FHJ \) (vertical angles) 

\(\because EH=HF\) (midpoint)

\(\because \overline{DE}\parallel\overline{FJ} \Rightarrow \angle EDJ =\angle FJD\)  (alternating interior angles)

\(\therefore \triangle HDE\cong\triangle HJF\) (AAS Congruency) 

\(\therefore [HDE]=[HJF] \Rightarrow [CDHF]+[HDE]=[CDHF]+[HJF]\)

\([CDJ]=[CDEF]\)

 

\(\boxed{36}\)

 

I hope this helped,

 

Gavin. 

GYanggg  Oct 28, 2018

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