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Geometry DS

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by srcc25anu » Fri Apr 19, 2013 2:51 pm
If angle ABC is 30 degrees, what is the area of triangle BCE?

1.Angle CDF is 120 degrees, lines L and M are parallel, and AC = 6, BC = 12, and EC = 2AC

2.Angle DCG is 60 degrees, angle CDG is 30 degrees, angle FDG = 90, and GC = 6, CD = 12 and EC = 12

OA: D
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by neha24 » Fri Apr 19, 2013 11:30 pm
well it has to be D
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by neha24 » Fri Apr 19, 2013 11:34 pm
st 1 can be thought as follows :in triangle EBC BC = EC =12 that means angles CEB = angle CBE thant concludes us to the fact that angle ABE is also 30 degree !!
so now u know the base as 12 and u know the ht and hence we can find the area of triangle EBC

st 2 is rather simple : all u r given is the area of a triangle b/w two two parallel lines with base as 6 and then u are asked the area of a triangle b/w those same parallel lines with base as 12 .honestly i didn't calculate st 2 .logically we know we can get the area !!
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by Anju@Gurome » Sat Apr 20, 2013 12:48 am
Deleted.
Last edited by Anju@Gurome on Sat Apr 20, 2013 2:49 am, edited 3 times in total.
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by rseafsf » Sat Apr 20, 2013 1:55 am
can anyone elaborate on statement 1...how did you prove AB to be the height

@Neha ...how did u prove st 1 : "can be thought as follows :in triangle EBC BC = EC =12 that means angles CEB = angle CBE thant concludes us to the fact that angle ABE is also 30 degree !"
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by neha24 » Sat Apr 20, 2013 2:41 am
@Neha ...how did u prove st 1 : "can be thought as follows :in triangle EBC BC = EC =12 that means angles CEB = angle CBE thant concludes us to the fact that angle ABE is also 30 degree !"
AB bisects the side CE doesn't mean AB bisects the angle CBE.
This is a common misconception.

Refer to the diagram below.
i guess i did major part of calculation on my paper and didnt write it here !!
when i know that BC = EC =12 then in triangle ABC we have the following : (6/sin30) = (12/ sin BAC)
this gives us sin BAC = 1 ------> angle BAC =90 degree ----> angle ABE = 30
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by Anju@Gurome » Sat Apr 20, 2013 2:53 am
neha24 wrote:i guess i did major part of calculation on my paper and didnt write it here !!
when i know that BC = EC =12 then in triangle ABC we have the following : (6/sin30) = (12/ sin BAC)
this gives us sin BAC = 1 ------> angle BAC =90 degree ----> angle ABE = 30
You skipped the most important part in your earlier post.
Sorry for assuming that you made a mistake.
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by rseafsf » Sat Apr 20, 2013 4:20 am
thank u neha
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