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08 Aug 2012, 14:07
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In the figure below, if AC=12, DC=18, and DB=15, what is the length of AF?
A)8
B)9
C)10
D)11
E)12
Figure reference: https://24.media.tumblr.com/tumblr_m8gfp ... 1_1280.jpg
A)8
B)9
C)10
D)11
E)12
Figure reference: https://24.media.tumblr.com/tumblr_m8gfp ... 1_1280.jpg
08 Aug 2012, 15:11
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No able to view the link.
Kindly post the question as per rule with OA and in right forum.
Kindly post the question as per rule with OA and in right forum.
CMcAboy
Joined: 01 Mar 2009
Last visit: 10 Apr 2017
Posts: 175
Given Kudos: 23
Location: United States
Concentration: Strategy, Technology
Schools: Goizueta '16 (M$) Duke '16 (D) Tuck '16 (D) Johnson '16 (D) Darden '16 (D) Ross '16 (WL) McDonough '16 (A)
GMAT 1: 660 Q47 V34
GMAT 2: 680 Q46 V38
GPA: 3.2
WE:Consulting (Computer Software)
Schools: Goizueta '16 (M$) Duke '16 (D) Tuck '16 (D) Johnson '16 (D) Darden '16 (D) Ross '16 (WL) McDonough '16 (A)
GMAT 2: 680 Q46 V38
Posts: 175
08 Aug 2012, 15:41
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The answer is C) 10.
I thought this was a very difficult problem. I stared at this for a while before figuring it out. Make point X the point where DB and AF intersect. The angles of the four sided figure in the middle (BCFX), add up to 360. If there are two 90 degree angles, then the remaining angles, FXB and FCB, add up to 180. You will also note that angles FXB and FXD also add up to 180 as well as angles FXB and AXB.
FXB + FCB = 180
FXB + FXD = 180
FXB + AXB = 180
Therefore, Angles FCB, FXD, and AXB are all equal. Lets call their value Y for now. Now you have 3 triangles-- AXB, DXF, AFC, and DBC-- that all have angles that are 90 and Y degrees. They all have the same 3rd angle, which we can call Z for now. We do not need to know the value of Y or Z.
Now, from the above configuration lets extract 2 triangles: CFA and CBD. We can find their proportionality be comparing known values on proportional sides. Since we have values for both CA and CD and they proportional, lets use them. CB:CD = (12):(18) = 2:3. Therefore, the ratio of the sides of triangle CFA to CBD is 2:3. Now lets find AF. Since AF and BD are proportional at 2:3 and BD = 15, then AF = 15(2/3) = 30/3 = 10
I thought this was a very difficult problem. I stared at this for a while before figuring it out. Make point X the point where DB and AF intersect. The angles of the four sided figure in the middle (BCFX), add up to 360. If there are two 90 degree angles, then the remaining angles, FXB and FCB, add up to 180. You will also note that angles FXB and FXD also add up to 180 as well as angles FXB and AXB.
FXB + FCB = 180
FXB + FXD = 180
FXB + AXB = 180
Therefore, Angles FCB, FXD, and AXB are all equal. Lets call their value Y for now. Now you have 3 triangles-- AXB, DXF, AFC, and DBC-- that all have angles that are 90 and Y degrees. They all have the same 3rd angle, which we can call Z for now. We do not need to know the value of Y or Z.
Now, from the above configuration lets extract 2 triangles: CFA and CBD. We can find their proportionality be comparing known values on proportional sides. Since we have values for both CA and CD and they proportional, lets use them. CB:CD = (12):(18) = 2:3. Therefore, the ratio of the sides of triangle CFA to CBD is 2:3. Now lets find AF. Since AF and BD are proportional at 2:3 and BD = 15, then AF = 15(2/3) = 30/3 = 10
