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

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

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20 May 2014, 06:34
mbhussain wrote:
2 good bunuel..... how would you rate the difficulty level of this question... simply beyond my understanding

If you are unable to understand a question, try to put all alphabetical statements in algebraic forms. Follow these steps and you may arrive:

The best approach to tackle statement questions in DS is as follows:

Step 1: Convert all the alphabetical statements in algebraic statements
Step 2: Reduce the number of variable to minimum
Step 3: Check how many variables are left. You may probably need that many statements to solve the questions but you might need lesser number of statements to answer the question.

Caution: Don't waste your time in solving the question. You have to analyse the data sufficiency and not solve the question.

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

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24 May 2014, 07:24
Bunuel 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 ?

Look at the diagram below:
Attachment:
Midsegments.png
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.

Hope it's clear.

Hi Bunuel,

I have a question regarding "1. The midsegment is always half the length of the third side. So, $$\frac{AB}{XY}=2$$ and $$\frac{XY}{RS}=2$$ --> $$\frac{AB}{RS}=4$$;"

How do we know that? Don't we only know that their heights are cut in half, how do we know that their base is half as well?

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

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24 May 2014, 09:46
russ9 wrote:
Bunuel 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 ?

Look at the diagram below:
Attachment:
Midsegments.png
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.

Hope it's clear.

Hi Bunuel,

I have a question regarding "1. The midsegment is always half the length of the third side. So, $$\frac{AB}{XY}=2$$ and $$\frac{XY}{RS}=2$$ --> $$\frac{AB}{RS}=4$$;"

How do we know that? Don't we only know that their heights are cut in half, how do we know that their base is half as well?

That's a known property.

Check here: math-triangles-87197.html
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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24 Jun 2014, 12:00
Bunuel wrote:
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....

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
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.

Hope it's clear.

Great question and Fantastic Explanation.
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Re: In triangle ABC, point X is the midpoint of side AC and [#permalink]

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08 Oct 2014, 16:42
argha wrote:
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

Regards

Argha

Hi Argha,
Your solution for (1) states Area(ABX)=1/4 Area(ABC) because the ratio of the bases is 1:2. That would be true, were ABX and ABC similar.
Try this conceptualization for 2 triangles with the constant height but bases in the ratio 1:2.

Area(1)=1= (Bh)/2
h=(2/B)

h=constant=(2/B)
B(2)=2*B(1)
Area(2) =[(2/B)*2B]/2= 2 = 2*(Area1)

So, in the case where height is equal and the triangles aren't similar but share a side, the ratios of base to area are linear.

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

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08 Jan 2015, 01:31
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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 [ 9.81 KiB | Viewed 2182 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 [ 11 KiB | Viewed 2181 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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16 Jan 2017, 20:17
Top Contributor
Attached is a visual that should help.
Attachments

Screen Shot 2017-01-16 at 8.17.06 PM.png [ 173.9 KiB | Viewed 941 times ]

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

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12 Sep 2017, 15:18
Expert's post
Top Contributor
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.

Given: 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

Let's make a few observations before we do anything else.
As others have pointed out before me, the triangles we get by connecting midpoints are similar triangles.
Let's examine a specific case and then generalize...

Consider triangle ABC with these measurements...

The area of triangle ABC = (base)(height)/2 = (16)(12)/2 = 96

Now add in the midpoints X and Y...

This means triangle XCY is similar to triangle ACB

The area of triangle XYC = (base)(height)/2 = (8)(6)/2 = 24
In other words, the area of triangle XYC is 1/4 the area of triangle ABC

Now add midpoint R and S...

The area of triangle RCS = (base)(height)/2 = (4)(3)/2 = 6

In other words, the area of triangle RCS is 1/4 the area of triangle XYC
We can also say that the area of triangle RCS is 1/16 the area of triangle ABC

IMPORTANT: Since connecting midpoints (as we have done above) will always yield similar triangles, the results above (in green) will apply to all triangles.

Now onto the question....

Target question: What is the area of triangular region RCS ?

Statement 1: The area of triangular region ABX is 32
Our task is to determine whether there's a relationship between triangle ABX and triangle RCS
Let's see what triangle ABX looks like on our specific diagram....

If we let AX be the base, then we can see that the area of triangle ABX will be EQUAL to the area of triangle XCB, since both triangles have the same base and the same height.
Since triangles ABX and XCB have the same area, then we can also say that the area of triangle ABX is HALF the area of triangle ABC
So, if the area of triangle ABX is 32, then the area of triangular region ABC is 64
Since we already know that the area of triangle RCS is 1/16 the area of triangle ABC, we can conclude that the area of triangle RSC is 1/16 of 64, which equals 4
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The length of one of the altitudes of triangle ABC is 8
This just tells us one thing about triangle ABC.
Given this information, there's no way to determine the area of triangle ABC, which means there's no way to determine the area of triangle RCS
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

[Reveal] Spoiler:
A

Cheers,
Brent
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Re: In triangle ABC, point X is the midpoint of side AC and   [#permalink] 12 Sep 2017, 15:18

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