+42 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]
+128399 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
