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geometry

by thephoenix » Tue Jan 26, 2010 1:51 am
In a triangle ABC,pt. D is on side AB nd point E is on side AC. BECD is a trapezium.. DE:BC=3:5. calculate the ratio of the area of
ADE and BCED ?

a)9/25
b)9/16
c)27/125
d)3/5
e)12/25
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by ajith » Tue Jan 26, 2010 2:28 am
thephoenix wrote:In a triangle ABC,pt. D is on side AB nd point E is on side AC. BECD is a trapezium.. DE:BC=3:5. calculate the ratio of the area of
ADE and BCED ?

a)9/25
b)9/16
c)27/125
d)3/5
e)12/25
B
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by bhumika.k.shah » Tue Jan 26, 2010 10:45 am
Ajith

thanks for the drawing and the explanation below it reminds me of school days :D

But could u please help me understand this = 1/(BC/DE*AF/AG -1)

how did u derive it ???

Thanks!
must say good work! :D
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by ajith » Tue Jan 26, 2010 11:09 am
bhumika.k.shah wrote:Ajith

thanks for the drawing and the explanation below it reminds me of school days :D

But could u please help me understand this = 1/(BC/DE*AF/AG -1)

how did u derive it ???

Thanks!
must say good work! :D
The area of a triangle = 1/2 base*height

area of triangle DEA = 1/2* DE* AG
are of triangle ABC = 1/2*BC*AF

Area of the trapezium = Area of ABC - Area of DEA

Ratio of area of DEA/ area of trapezium = DE*AG/ (BC*AF-DE*AG)

I was dividing both numerator and denominator by DE*AG

to get 1/(BC/DE*AF/AG -1)
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by bhumika.k.shah » Tue Jan 26, 2010 4:31 pm
Got it ! :D
ajith wrote:
bhumika.k.shah wrote:Ajith

thanks for the drawing and the explanation below it reminds me of school days :D

But could u please help me understand this = 1/(BC/DE*AF/AG -1)

how did u derive it ???

Thanks!
must say good work! :D
The area of a triangle = 1/2 base*height

area of triangle DEA = 1/2* DE* AG
are of triangle ABC = 1/2*BC*AF

Area of the trapezium = Area of ABC - Area of DEA

Ratio of area of DEA/ area of trapezium = DE*AG/ (BC*AF-DE*AG)

I was dividing both numerator and denominator by DE*AG

to get 1/(BC/DE*AF/AG -1)
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