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what is the answer to this pls

 

Problem:

ABCD is a kite such that AB=CD and CD=DA. P is the centroid of triangleABC and Q and is the centroid of triangleCDA. We know [ABCD]=60. Find [PCQA]

 Jan 18, 2019
 #1
avatar+103915 
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If CD = DA, then AB = BC  [ not AB = CD ]  since a kite has two distinct [different ]  pairs of equal adjacent sides 

 

[If AB = CD, we would have a square or a rhombus ]   

 

Let the diagonals be AC and BD

 

And let their intersection = E

 

Since P is a centroid, then EP = 1/3  of EB

And similarly, EQ = 1/3 of ED

 

So.....the area of the kite =   AC * EB / 2   +  AC * ED / 2  =  (1/2) [ AC * EB + AC * ED ]

 

And the area of triangle ABC = AC * (1/3)EB / 2  =  AC * EB / 6

 

And the area of triangle ADC = AC * (1/3)ED / 2  = AC * EB / 6

 

So...the area of   PCQA =   area of triangle ABC + area of triangle ADC  =  (1/6) [ AC * EB + AC * ED] =

 

(1/3) (1/2) [ AC * EB + AC * ED ]  =

 

(1/3) area of kite  =   20

 

 

cool  cool cool

 Jan 18, 2019
edited by CPhill  Jan 18, 2019
edited by CPhill  Jan 18, 2019
edited by CPhill  Jan 18, 2019
edited by CPhill  Jan 18, 2019

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