Is the area of quadrilateral ABCD equal to area of quadrilateral AECD?

in your justification above, the statement B is not just giving us the X co-ordinate but also telling us the locus of that point.
if we follow that the base of triangle AEC will remain same as before(8 Units) and altitude will be (6 Units), constant distance of that line from Y Axis .
This will give us constant area irrespective of the location of point on the line.
Hence, C is the answer

My approach to the problem will be same as given by mynamegoeson. For m it's difficult to understand how any co-ordinate of E will lead to the same area.



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kumarparitosh123 Please refer to this.
It might be helpfull
https://www.teachoo.com/4295/1052/Theor ... /Theorems/

also please try drawing the quadrilateral on co-ordinate axis if you have doubt wrt base begin same for all triangles

isn't true that E can be anywhere on the line x=6? E can be at the same point as B, or it can be at any other point at x=6...

Thanks Saurabh..
Helpful for me.



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When we get co-ordinate in the question .
It is always better to plot the co-ordinate and then solve the question . The task at hand is very easy then .

St1. only ABCD points are given . NO mention of E . Not sufficient
St2 . Only point E is given as ( 6,0) ==> x=6

together ,
the point B and E are same . Hence the ABCD = AECD
C ans .

Regards,
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Stmt1 and Stmt2 alone are not sufficient.
On combining, We get 2 triangles for each quadrilateral
ABCD has ABC and ACD
AECD has AEC and ACD
Now, we have to see if area of ABC and AEC are same.
ABC and AEC has same base AC and same height 6 from AC to line x=6 so area will be same of both triangle.
Answer is C.

Just to clarify my thought to E.

Point E may lie on a line where x=6, but the Y coordinate could be different. Am I missing something?

Yes, Y coordinate vary but area of triangle will remain the same as Area of triangle is \(\frac{1}{2}*BASE*HEIGHT\)
Base is common AC and Height is same distance from AC to x=6.

Hope I am clear.

Amazing question......relies on 2 rules and understanding the figure.

Rule 1: there exists only one value for the perpendicular height between 2 parallel lines

Rule 2: given that 2 triangles have the same base, the ratio of the areas of those triangles will depend on the ratio of the respective Heights drawn to the common Base


S1 and s2 don’t give you enough information to solve the question.


However, after filling in the points for A , B, C, and D

You will see that Point E must lie on the vertical line that passes through X = 6 right through Point B (6 , 0)

Placing E on the line X = 6 at a point in between (6 , 20) and (6 , -12) you can see from the figure that both quadrilateral ABCD and AECD will have as common the triangular area given by Points A, D and C along with the Y Axis.


The rest of the area of the quadrilateral ABCD will be given by the triangular area ABC and the Y Axis in the first quadrant. The Base will exist on the Y Axis from Points A to C

The height will be given by the perpendicular distance between the 2 parallel lines of the Y Axis and Vertical Line of X = 6




Further, the rest of the Area of Quadrilateral AECD will be given by the Triangular region in quadrant 1 defined by Point E on line x = 6 and the Base AC located on the Y Axis.


Thus, both triangular regions ABC and AEC will share the common Base on the Y Axis from points A to C and both have their 3rd point on the Parallel Line X = 6:

Point B will be at (6 , 0)

And

Point E will be somewhere between (6 , 20) and (6 , -12)


Because no matter where you put point E on the parallel line X = 6 the perpendicular height between the Y Axis and X = 6 will remain the same, both triangular areas will have the same base and the same height and therefore the same area.

C- the quadrilaterals will have the same area

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Can you explain this "Point E will be somewhere between (6 , 20) and (6 , -12)" perhaps with a drawing? I drew it out and I see no way the two quadrilaterals can have the same area unless E is located at (6,0).