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

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.

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

I'd say it's 700 level question.

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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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).

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:

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


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

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!

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...

B is a vertex of triangle ABC, it's not a midpoint of any side.

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.

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:



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.



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.

Attached is a visual that should help.

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

Answer: A

Cheers,
Brent

Bunuel could you please refer to some similar questions? Thanks.

found this topic and understood much better, thanks everyone

Excellent work!!!

Dear Sir,

I had a question. Does it matter if this is a right triangle, isosceles triangle or equilateral? The area does not change right?