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In triangle ABC, point X is the midpoint of side AC and

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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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In triangle ABC, point X is the midpoint of side AC and [#permalink]

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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.... :oops: :cry:


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 ?

Look at the diagram below:
Attachment:
Midsegments.png
Midsegments.png [ 10 KiB | Viewed 53259 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 \(\frac{x^2}{y^2}\) (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: Triangles [#permalink]

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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
=> SUFFICIENT

ST 2:
We cannot deduce anything from the length of one of the heigths.
=> NOT SUFFICIENT

ANS: A
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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The key here is to recognize the similar triangles. Notice the ratio of the bases between the 3 similar triangles. Notice the ratio of the diagonal "hypotenuse" of these 3 similar triangles.

Notice how that small triangle is similar to the overall big triangle.

A video solution to this triangle inside a triangle question has been provided here:

http://www.gmatpill.com/gmat-practice-t ... stion/3226

Image
Video solutions to similar questions is available for GMAT Pill customers.

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Re: Triangles [#permalink]

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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?
=> SUFFICIENT

ST 2:
We cannot deduce anything from the length of one of the heigths.
=> NOT SUFFICIENT

ANS: A


can 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]

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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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Triangles RCS and ACB are similar since they have an angle in common and since RS || AB other angles are equal too. We are told that CS =1/4 BC, thus area of RCS is 1/16 of the area of ACB. Thus (1) is sufficient.

(2) is not sufficient because we do not know the value of the base (nor can it be derived from the other given info) to arrive at an area.

Answer is (A).

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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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This is a great question, but I don't understand how we can do this in 2 minutes even assuming we knew that ABX = 1/2 * ABC; it took me about 1 minute just to accurately draw the picture. :(

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In triangle ABC, point X is the midpoint of side AC and [#permalink]

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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.


If we observe closesly, in this problem one vertex of the triangle is joined to the midpoint of the other side and this process is done multiple times.

Now, what is so special about the line segment joining the vertex to the midpoint of the other side? more specifically, how will it affect the area of the original triangle? Let's take a look at it.

Consider a triangle ABC of area p square units. Suppose X is the midpoint of side AC. Join B and X as shown below:

Attachment:
median.png
median.png [ 9.81 KiB | Viewed 1772 times ]


What is the area of triangle ABX? Think about it...

The base became half the original size and the height of the vertex from the base remained the same.

Therefore area of the triangle ABX (\(\frac{1}{2} * base* height\)) will be half the area of the original triangle ABC i.e., \(\frac{p}{2}\) square units.

Now what is the area of triangle BXC?

\(p - \frac{p}{2} = \frac{p}{2}\) square units.


So, what did we observe?

A line joining a vertex of a triangle to the midpoint of the opposite side will divide the triangle into two equal parts.

Let us call such a line as a "sword line" of a triangle for easy reference.

Now, with this understanding, let us look at the triangle in the given problem.

Attachment:
OG113DS.png
OG113DS.png [ 11 KiB | Viewed 1771 times ]


If we assume that the area of triangle ABC is \(y\) square units, from our understanding we know that area of triangle BXC is \(\frac{y}{2}\) square units.

Notice that since Y is the midpoint of BC, XY is a "sword line" in the triangle BXC.

Therefore area of triangle XYC = half of area of triangle BXC = \(\frac{y}{4}\) square units.



It is also given that R is the midpoint of XC. Therefore, in triangle XYC, YR is a "sword line".

Therefore area of triangle YRC = half of area of triangle XYC = \(\frac{y}{8}\) square units.



Finally, it is given that S is the midpoint of YC. Therefore, RS is a "sword line" in triangle YRC.

Therefore area of triangle RCS = half of area of triangle YRC = \(\frac{y}{16}\) square units.



Now statement 1 says that area of \(triangle ABX = 32\)

therefore, according to our nomenclature, \(\frac{y}{2} = 32\)

\(y = 64\)

\(\frac{y}{16} = 4\)

Therefore area of \(triangle YRC = 4\) square units.

Therefore statement 1 is sufficient.



Now we cannot determine anything with statement 2 unless we are given the vertex from which (or the side to which) the altitude is drawn.

Therefore statement 2 is not sufficient.


Since we arrive at a unique answer using Statement 1 alone, option A is the correct answer for this Data Sufficiency question.

Foot Note:

The sword line we used in this problem is widely referred to as the median of a triangle.

The meaning of the word ‘median’ is clear from its name itself. The word "median" comes from the Latin root medius, which means ‘in between’. The English word ‘middle’ too comes from the same root. So, a median is the line that joins a vertex to the mid point of the opposite side of a triangle. (Obviously, every triangle will have 3 medians - one from each vertex).

However, we do not need to know its name to understand how it works within the scope of the GMAT. This is the reason why I focused on first illustrating how that line works rather than telling its name. :)



Hope this helps. :)



- Krishna.
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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New post 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]

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New post 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]

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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.... :oops: :cry:

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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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Great explanation and diagram. Thanks!

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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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2 good bunuel..... how would you rate the difficulty level of this question... simply beyond my understanding

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Re: GMAT Geometry DS Question [#permalink]

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New post 20 Sep 2013, 21:14
IMO A,

Soln:

(1)....Area of triangle ABX=1/4 of triangle ABC(X is the mid point of AC)

Therefore area of triangle ABC=4*32

Area of triangle RCS=1/16* area of ABC=8

Hence sufficient

(2)...Length of an altitiude is not sufficient to calculate the area of triangle ABC. Hence insufficient



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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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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.

X is the midpoint of AC meaning AX = XC. If ABX = 32 so regardless of the slope of BC and as long as the line from B to X will create two triangles equal in area. (i.e. the area of ABX = CBX) We know there is a segment drawn from midpoint X to Y and midpoint R to S. Because these increasingly small triangles are built with midpoints, we know that their ratios are proportionate to one another and that they are similar. For example, the ratio of the length of SC to YC is 1:2. The ratio of the areas of similar triangles can be found by taking the ratio of lengths (i.e. 1:2) and squaring it. Therefore, the area of RSC to XYC = 1:4.

Ratio of area XYC:ABC = 1:4 (because XYC was created from the midpoints of two of ABC's legs) the ratio of XYC:RSC = 1:4 The ratio of ABC:RSC = 1/16. If we know the area of ABX we can find the area of CBX - they are the same. Sufficient.

(2) The length of one of the altitudes of triangle ABC is 8.
Altitude but it doesn't tell us anything. Altitude doesn't give us midpoints (unless noted) so we can't even determine a single definite area for the triangle. Insufficient

A

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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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New post 04 Apr 2014, 10:25
Though explanations given so far are great, we really don't need to know the heights of ABC and RSC to solve this question.

The question is testing properties of similar triangles here. There 3 ways to tell if the triangles are similar:

1. AAA (angle angle angle)
All three pairs of corresponding angles are the same.
(We actually need two angles really:)

2. SSS in same proportion (side side side)
All three pairs of corresponding sides are in the same proportion

3. SAS (side angle side)
Two pairs of sides in the same proportion and the included angle equal.

It is this 3rd property that is quite handy here. Two triangles ABC and RSC are similar since RC/AC=SC/BC=1/4 and they include the same angle THETA (as shown in the figure https://drive.google.com/file/d/0B3it2i ... sp=sharing)

1) Tells us that Area(ABX) =32. Since height of triangle ABC is same that of triangle ABX and the base is twice, area (ABC) = 32*2= 64. Since ABC and RSC are similar and their sides are in the ration of 1:4 their ares will be in the ration of 1:16. Sufficient to solve.
2) Tells us one of the altitudes is 8. Since it does not tell which one, insufficient!

Hope it makes sense!
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Last edited by NoHalfMeasures on 31 May 2014, 01:54, edited 1 time in total.

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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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New post 04 May 2014, 03:10
I only have one question: why isnt the diagram made in such a way that point B is the midpoint of AC. I know the question doesnt say it...but then it could go either way, right? B could be made the mid point, or couldnt be...

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Re: In triangle ABC, point X is the midpoint of side AC and   [#permalink] 04 May 2014, 03:10

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