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Q109. In triangle ABC, pt. X is the midpt. of side AC and

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Q109. In triangle ABC, pt. X is the midpt. of side AC and [#permalink]

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New post 01 Jul 2009, 11:13
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Q109.
In triangle ABC, pt. X is the midpt. of side AC and pt. Y is the midpt. of side BC. If pt. R is the midpt. of line segment XC and if pt. S is the midpt. of the line segment YC, what is the area of trianglular region RCS?

1) Area of triangle region ABX is 32
2) The length of one of the altitudes of triangle ABC is 8.

Plz help me solve this ..

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Re: DS Q109, OG 12 [#permalink]

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New post 01 Jul 2009, 11:52
it should be in math forum... but my guess is A - 1) is sufficient...

Sure you dont know the B/H to form the area of 32 but the area of the triangle RCS uses the same basic Base and Height (BXH), just in a very smaller ratio to the original area. I think there is a triangle ratio rule or something that you can use from the knowledge of the midpoints.... it's been a while
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Re: DS Q109, OG 12 [#permalink]

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New post 01 Jul 2009, 11:55
a.
1 is suff.
2 is insuff.

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Re: DS Q109, OG 12 [#permalink]

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New post 01 Jul 2009, 11:57
yeah ur right... A is the answer.. n i too guessed it .. but i want to know the exact concept behind it...

As u said sumthing related to concept of midpoints n all...

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Re: DS Q109, OG 12 [#permalink]

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New post 01 Jul 2009, 12:11
When you have 1 midpoint and draw it to the opposing vertex you get a 1:2 ratio of the original area. Now you do that with 3 midpoints in the beginning you should get 6 equal pieces. Now you have the last 2 midpoints which is splitting the 1/3 of the total area apart into 6 pieces. you want 2 of the 6 pieces for the area.
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Re: DS Q109, OG 12 [#permalink]

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New post 01 Jul 2009, 12:17
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Their are 2 concepts involved.

1) A line connecting midpoint of any side and the opposite vertex will divide the triangle in 2 halves equal in area
2) If a line connects midpoints of any two sides, then the smaller triangle thus formed will be proportional to the area of the original triangle in area. ie area of larger : area of smaller triangle = 4:1

Now connect XB. area AXB = area BXC = 16

connect XY, (applying rule 1 again for BXC) area BYX = area CYX = 8

area rsc : area cyx = 1:4

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Re: DS Q109, OG 12 [#permalink]

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New post 01 Jul 2009, 12:27
Thanks a lot .. i understand it clearly now

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Re: DS Q109, OG 12   [#permalink] 01 Jul 2009, 12:27
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Q109. In triangle ABC, pt. X is the midpt. of side AC and

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