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triangle

by shashank.ism » Tue Feb 09, 2010 12:49 pm
In Triangle ABC, D and E are points on side AC such that AD:DE:EC is 2:1:1. BD and BE are
joined. Point F divides BD in the ratio 1:2 and point G divides BE in the ratio 2:1. Find the
ratio of the area of Triangles (AFG + BFG) to the area of the Triangle ABC.

2 : 5
3 : 10
1 : 3
1 : 4
2 : 7
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by firdaus117 » Sun Feb 21, 2010 11:23 am
To solve this question,we use the fact that if a point D on side AC of triangle ABC divides side into 1:2,it also divides the triangle into two triangles ABD & BDC in ratio 1:2.
Lets draw the figure(see the image attached)
Image
Let the area of tr. ABC be denoted by A.
Area tr.BEC=A/(1+1+2)
=A/4
Area tr.EGC=(Area tr. BEC)/3 (Since G divides BE in 2:1)
=A/12
Area tr.AGC=4*Area tr.EGC
=A/3
Also,Area tr.BGC=2/3 of Area tr.BEC
=A/6
Hence,Area tr.ABG=Area tr.ABC-(Area tr.BGC+Area tr.AGC)
=A/2
Area tr.ABF=1/3 of Area tr.ABD=A/6
area of Triangles (AFG + BFG)=Area tr.ABG-Area tr.ABF
=A/2 - A/6
=A/3
Hence,Required Ratio=1:3
However,I seriously doubt that such questions appear in GMAT. :)
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by shashank.ism » Sun Feb 21, 2010 11:56 am
firdaus117 wrote:To solve this question,we use the fact that if a point D on side AC of triangle ABC divides side into 1:2,it also divides the triangle into two triangles ABD & BDC in ratio 1:2.
Lets draw the figure(see the image attached)
Image
Let the area of tr. ABC be denoted by A.
Area tr.BEC=A/(1+1+2)
=A/4
Area tr.EGC=(Area tr. BEC)/3 (Since G divides BE in 2:1)
=A/12
Area tr.AGC=4*Area tr.EGC
=A/3
Also,Area tr.BGC=2/3 of Area tr.BEC
=A/6
Hence,Area tr.ABG=Area tr.ABC-(Area tr.BGC+Area tr.AGC)
=A/2
Area tr.ABF=1/3 of Area tr.ABD=A/6
area of Triangles (AFG + BFG)=Area tr.ABG-Area tr.ABF
=A/2 - A/6
=A/3
Hence,Required Ratio=1:3
However,I seriously doubt that such questions appear in GMAT. :)
Yeah firdauss.. it was question which I posted when i was very new to preparation..I now understand the level of question being asked in GMAT..so ignore that question..though u have solved it very well...
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