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In triangle ABC, point X is the midpoint of side AC and [#permalink]
 18 Aug 2009, 18:44
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In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS ? (1) The area of triangular region ABX is 32. (2) The length of one of the altitudes of triangle ABC is 8.
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yezz wrote: In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS ?
(1) The area of triangular region ABX is 32.
(2) The length of one of the altitudes of triangle ABC is 8. SOL: This question makes use of the Midpoint theorem in case of triangles. According to the theorem, the segment joining the midpoints of two sides of a triangle is half the length of the third side and the smaller triangle thus formed is similar to the original triangle. The ratio of sides of the smaller tr to the larger tr = 1/2 => A(smaller tr) : A(Larger tr) = 1:4 From the given info we have: A(CYX) : A(ABC) = 1:4 A(CSR) : A(CYX) = 1:4 => A(CSR) = 1/16 * A(ABC) ST 1: A(ABX) = 1/2 * A(ABC) ....... Since they have the same height and the base of ABX is half the base of ABC Thus from A(ABX), we can calculate A(CSR) => A(ABX)/8 = 4 => SUFFICIENTST 2: We cannot deduce anything from the length of one of the heigths. => NOT SUFFICIENTANS: A
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samrus98 wrote: yezz wrote: In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC, what is the area of triangular region RCS ?
(1) The area of triangular region ABX is 32.
(2) The length of one of the altitudes of triangle ABC is 8. SOL: This question makes use of the Midpoint theorem in case of triangles. According to the theorem, the segment joining the midpoints of two sides of a triangle is half the length of the third side and the smaller triangle thus formed is similar to the original triangle. The ratio of sides of the smaller tr to the larger tr = 1/2 => A(smaller tr) : A(Larger tr) = 1:4 why 1:4? is it because the ratio of sides is 1:2? Though im guess this is the reason but still don't understand the reason behind it From the given info we have: A(CYX) : A(ABC) = 1:4 A(CSR) : A(CYX) = 1:4 => A(CSR) = 1/16 * A(ABC) why 1/16th?ST 1: A(ABX) = 1/2 * A(ABC) ....... Since they have the same height and the base of ABX is half the base of ABC Thus from A(ABX), we can calculate A(CSR) => A(ABX)/ 8 = 4 why divided by 8?=> SUFFICIENTST 2: We cannot deduce anything from the length of one of the heigths. => NOT SUFFICIENTANS: Acan someone please explain the colored text above?? Thanks!!!
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
21 Apr 2012, 10:09
Think of it in terms of a square, if you half both sides of a square, you make a smaller square that is 1/4 the size of the original. The same idea applies to a triangle. A good exercise is to draw out the shapes while solving the problem to visualize.
It is 1/16 because from area of ABC to XYC its a 1:4 ratio and from XYC to RSC is a 1:4 ratio, so going from ABC to RSC is the multiple of the ratios giving us 1:16 ratio.
We divide by 8 at that point because triangle ABX is 1:2 that of ABC, and since CSR is 4 times smaller than ABX we need to multiply the ratios of 1:2 and 1:4 to give us 1:8 ratio.
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
21 Apr 2012, 10:56
aalba005 wrote: Think of it in terms of a square, if you half both sides of a square, you make a smaller square that is 1/4 the size of the original. The same idea applies to a triangle. A good exercise is to draw out the shapes while solving the problem to visualize.
It is 1/16 because from area of ABC to XYC its a 1:4 ratio and from XYC to RSC is a 1:4 ratio, so going from ABC to RSC is the multiple of the ratios giving us 1:16 ratio.
We divide by 8 at that point because triangle ABX is 1:2 that of ABC, and since CSR is 4 times smaller than ABX we need to multiply the ratios of 1:2 and 1:4 to give us 1:8 ratio. Just tried a few example, it seems it is a rule that the area of smaller to area of larger is 1:4 but for the 3rd point about divide by 8, if csr is 4 times smaller than abx, and question gave area of abx, why not just abx/4?
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
21 Apr 2012, 13:49
catty2004 wrote: aalba005 wrote: Think of it in terms of a square, if you half both sides of a square, you make a smaller square that is 1/4 the size of the original. The same idea applies to a triangle. A good exercise is to draw out the shapes while solving the problem to visualize.
It is 1/16 because from area of ABC to XYC its a 1:4 ratio and from XYC to RSC is a 1:4 ratio, so going from ABC to RSC is the multiple of the ratios giving us 1:16 ratio.
We divide by 8 at that point because triangle ABX is 1:2 that of ABC, and since CSR is 4 times smaller than ABX we need to multiply the ratios of 1:2 and 1:4 to give us 1:8 ratio. Just tried a few example, it seems it is a rule that the area of smaller to area of larger is 1:4 but for the 3rd point about divide by 8, if csr is 4 times smaller than abx, and question gave area of abx, why not just abx/4? Because CSR is 8 times smaller than ABX not 4. CRS has 1/2 the base of ABX but also 1/4 the height of ABX (or other way round depending on how you drew it). It is not 1/2 the base and 1/2 the height of ABX.
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
22 Apr 2012, 18:11
aalba005 wrote: catty2004 wrote: aalba005 wrote: Think of it in terms of a square, if you half both sides of a square, you make a smaller square that is 1/4 the size of the original. The same idea applies to a triangle. A good exercise is to draw out the shapes while solving the problem to visualize.
It is 1/16 because from area of ABC to XYC its a 1:4 ratio and from XYC to RSC is a 1:4 ratio, so going from ABC to RSC is the multiple of the ratios giving us 1:16 ratio.
We divide by 8 at that point because triangle ABX is 1:2 that of ABC, and since CSR is 4 times smaller than ABX we need to multiply the ratios of 1:2 and 1:4 to give us 1:8 ratio. Just tried a few example, it seems it is a rule that the area of smaller to area of larger is 1:4 but for the 3rd point about divide by 8, if csr is 4 times smaller than abx, and question gave area of abx, why not just abx/4? Because CSR is 8 times smaller than ABX not 4. CRS has 1/2 the base of ABX but also 1/4 the height of ABX (or other way round depending on how you drew it). It is not 1/2 the base and 1/2 the height of ABX. please dont be frustrated with me......im completely and utterly lost in all these 1:16, 1:4, 1:2......now 1:8....  
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
22 Apr 2012, 22:43
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catty2004 wrote: please dont be frustrated with me......im completely and utterly lost in all these 1:16, 1:4, 1:2......now 1:8.... Look at the diagram below: Attachment: 
Midsegments.png [ 10 KiB | Viewed 5132 times ]
Notice that XY is the midsegment of triangel ABC and RS is the midsemgent of triangle XYC (midsegment is a line segment joining the midpoints of two sides of a triangle). Several important properties: 1. The midsegment is always half the length of the third side. So, \frac{AB}{XY}=2 and \frac{XY}{RS}=2 --> \frac{AB}{RS}=4; 2. The midsegment always divides a triangle into two similar triangles. So, ABC is similar to XYC and XYC is similar to RSC --> ABC is similar to RSC, and according to above the ratio of their sides is 4:1; 3. If two similar triangles have sides in the ratio \frac{x}{y}, then their areas are in the ratio \frac{x}{y} (or in another way in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \frac{AREA}{area}=\frac{S^2}{s^2}.). So, since ABC is similar to RSC and the ratio of their sides is 4:1 then \frac{area_{ABC}}{area_{RSC}}=4^2=16, so the area of ABC is 16 times as large as the area of RSC; 4. Each median divides the triangle into two smaller triangles which have the same area. So, since X is the midpoint of AC then BX is the median of ABC and the area of ABX is half of the area of ABC. From the above we have that the area of ABX is 16/2=8 times as large as the area of RSC. So, to find the area of RSC we need to find the area of ABX. For more check Triangles chapter of Math Book: math-triangles-87197.html(1) The area of triangular region ABX is 32 --> the area of RSC=32/8=4. Sufficient. (2) The length of one of the altitudes of triangle ABC is 8. Only altitude is no use to get the area. Not sufficient. Answer: A. Hope it's clear.
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
23 Apr 2012, 10:06
Great explanation and diagram. Thanks!
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
29 Sep 2012, 13:23
2 good bunuel..... how would you rate the difficulty level of this question... simply beyond my understanding
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]
30 Sep 2012, 04:45
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Re: In triangle ABC, point X is the midpoint of side AC and
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30 Sep 2012, 04:45
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