could someone please solve this for me? as it needs some good skills and logic to solve it.


thanks in advance for solving

Carcass , anybody,

Please help how to solve this

Carcass , anybody, GreenlightTestPrep
Please help how to solve this

The question is just a mirrored image of a question on GC https://gmatclub.com/forum/in-the-figur ... l#p2059207

First recognize that ∆ADC and ∆BCD both have the same base.
Also, if we let h = the height of ∆ADC, then ∆BCD also has height h


area of triangle = (base)(height)/2
So, if ∆ADC and ∆BCD have the same base and the same height, then they must have the same area.

So, if the area of ∆ADC is 28, then the area of ∆BCD must also be 28


Answer: D

Thank you for your quick response.

sir but mirroring the image also changes the base of the figure which will be more challenging to solve, would really kind of you if you shed some light on it

I was stuck in this question for a while and ended up googling the solution,
https://math.stackexchange.com/questions/3611594/quadrilateral-geometry-find-the-area-of-all-the-triangles-in-a-trapezoid

Explanation for this question?

Given: \(ar(ACE) = 4, ar(CED) = 3\)

\(ar(ACD) = ar(CED) + ar(DEB) = 4 + 3 = 7\)

Dropping a perpendicular from point \(D \) on base \(CD\), taking the point of intersection as \(X\),
\(ar(CED) = \frac{1}{2} * EX * CD\)......\((2)\)
Also,
\(ar(ACD) = \frac{1}{2} * AC * CD\)......\((3)\)
Dividing \((2)\) by \((3)\),
\(\frac{EX}{AC} = \frac{3}{7}\).....\frac{(4)[}{fraction]

Now, to find the area of triangle \(AEB\), we drop a perpendicular from point \(E \) to base \(AB\), taking the point of intersection as \(Y\),
\(ar(AEB) = [fraction]1/2}AB * EY \)
\(AC = EX + EY \)
Thus, \(EY = AC - EX = AC - \frac{3}{7}AC\).....(from equation \(4\))
\(EY = \frac{4}{7}AC\)

\(ar(AEB) = \frac{1}{2} * AB * \frac{4}{7}AC\)
\(= \frac{4}{7}ar(ABC)\) ....... (since \(ar(ABC) = \frac{1}{2} AB * AC \))

\(ar(ABC) = ar(AEC) + ar(AEB) = 4 + \frac{4}{7}ar(ABC)\)
\(\frac{3}{7}ar(ABC) = 4\)
\(ar(ABC) = \frac{28}{3}\)

Now, triangles \(ABC \) and \(ADB \) have the same base \(AB \) and same height \(AC\), so they have the same area.
Thus, \(ar(ADB) = \frac{28}{3}\)

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