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Re: In the figure, what is the area of triangle ADB [#permalink]
mrk9414 wrote:
Attachment:
figure geomentry.png


In the figure, what is the area of triangle ADB given that the area of triangle ACE is 4 and the area of triangle CDE is 3?

Give your answer as a fraction.

Show: :: OA
\(\frac{28}{3}\)



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


thanks in advance for solving
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Re: In the figure, what is the area of triangle ADB [#permalink]
Carcass , anybody,

Please help how to solve this
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Re: In the figure, what is the area of triangle ADB [#permalink]
Carcass , anybody, GreenlightTestPrep
Please help how to solve this
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Re: In the figure, what is the area of triangle ADB [#permalink]
1
Ahasunhabib999 wrote:
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
Image

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
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In the figure, what is the area of triangle ADB [#permalink]
GreenlightTestPrep wrote:
Ahasunhabib999 wrote:
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
Image

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
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Re: In the figure, what is the area of triangle ADB [#permalink]
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Re: In the figure, what is the area of triangle ADB [#permalink]
Explanation for this question?
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In the figure, what is the area of triangle ADB [#permalink]
4
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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Re: In the figure, what is the area of triangle ADB [#permalink]
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Re: In the figure, what is the area of triangle ADB [#permalink]
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