Convex quadrilateral ABCD has AB = 9 and CD = 12. Diagonals AC and BD

Question

https://gmatclub.com/forum/download/file.php?id=47144 Convex quadrilateral ABCD has AB = 9 and CD = 12. Diagonals AC and BD intersect at E, AC = 14, and triangle AED and triangle BEC have equal areas. What is AE? (A) 9/2 (B) 50/11 (C) 21/4 (D) 17/3 (E) 6 d25c0c5968bc5ab64f9f91412801d2cdb764ce45.png

Shreshtha55 · Answer

chetan2u  can you please explain this question?

Archit3110 · Answer

∆ABE=∆DEC
AB/DC= 9/12 ; 3/4
so side EC in ∆DEC = 4 and side AE ∆ AEB= 3
AC ; 3:4 ratio ; 3/7 * 14; 6= AE
IMO E

Ypsychotic · Answer

AEB and CED are similar triangles since area of AED and BEC is equal. And Angle EAB=Angle CED(opposite angles)

Using similarity
Let AE=x, EC=14-x

\frac{x}{9}=\frac{14-x}{12} ,

Solve for x, x=6.

chetan2u · Answer

Firstly, convex quadrilateral does not mean AB and CD are parallel, and so is wrong to take ABE and CDE similar without any reasoning.  Convex means all angles are less than 180, and in GMAT a quadrilateral means all angles are less than 180.

YES, they are similar, but reasoning is-
Area of triangle AED and triangle BEC are equal. That is A(AED)=A(BEC), add A(ABE) to both sides. Thus, A(AED)+A(CDE)=A(BEC)+A(CDE), which becomes A(ACD)=A(BCD).
Both the triangles have same BASE, CD, so their height has to be EQUAL, and their height lies on AB, so both points on AB should be equidistant from CD. This is the reason why AB and CD are parallel.
Always write as per the equal angles, A=C, and so on.. so ABE~CDE so  \frac{AB}{CD}=\frac{AE}{CE} =  \frac{9}{12}=\frac{AC-CE}{CE} = \frac{3}{4}=\frac{14-CE}{CE} =3CE=56-4CE...7CE=56 or CE=8, and AE = 14-8=6

E

Fdambro294 · Answer

One way to solve the problem is to use the Area of the triangle formula:

Area of a Triangle = (1/2) (A) (B) (SIN of Included Angle)

Where A and B are Adjacent side lengths of the triangle and you take the SINE of the Angle included BETWEEN those two adjacent sides of the triangle

(1)
We are given that Area of Triangle AED is equal to the Area of Triangle BEC

Since the two interior Angles at vertex E for each triangle are Vertically Opposite Angles, they must be equal

Angle <AED = Angle <BEC

If the angles are equal, then if we were to take the SINE of each Angle, it would be equal

Call this (SIN X)

(2)
Set up the Area of the two equal triangles using the formula above

Area of triangle AED = (1/2) (AE) (ED) (SIN X)

Area of triangle BEC = (1/2) (BE) (CE) (SIN X)

Because we are told these two Areas are equal, we thus know:

(AE) (ED) = (BE) (CE) (equation 1)

(3) at this point, we can show that triangle AEB is similar to ~ triangle DEC

First, like above, the interior angles at Vertex E are Vertically Opposite Angles and thus Equal

Angle <AEB = Angle <DEC

And then rearranging equation 1 above we get:

AE / CE = BE / ED

Using Side-Angle-Side similarity ——-> triangle (AEB) is similar to triangle (CED):

The 2 corresponding side lengths are in proportion to each other (for each triangle these corresponding side lengths are the adjacent sides around the Interior Angle at Vertex E)

And

The included Angles are equal, vertically opposite angles: Angle <AEB = Angle <DEC

Therefore, the last corresponding side lengths of the similar triangles must be in the same proportion:

side AB of triangle (AEB) ———— corresponds to ——— side CD of triangle (CED)

Setting these corresponding sides of the similar triangle in proportion:

AE / CE = AB / CD

Since AB = 9 and CD = 12

AE / CE = 9 / 12 = 3 / 4

Since AE + CE = the diagonal AC

AE = (3/7) (Length of Diagonal AC)

We are given that AC = 14

AE = (3/7) (14) = 6

Answer

6

Posted from my mobile device

shivaniilall · Answer

Hello, Why have you written AC-CE in place of BE? Thanks

chetan2u · Answer

🙏. There was a typo, which is corrected now. 
It is AE = AC-CE

Convex quadrilateral ABCD has AB = 9 and CD = 12. Diagonals AC and BD

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Convex quadrilateral ABCD has AB = 9 and CD = 12. Diagonals AC and BD

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