The figure above shows Line L, Circle 1 with center at C1, and Circle

Question stem:- Is the area of ΔABC1 less than the area of ΔDEC2 ?
Or, Is \(\frac{1}{2}*AB*h<\frac{1}{2}*DE*h\) ? (points C1 and C2 are equidistant from line L)
Or, Is AB<DE ?

St1:- \((C_{1}A=C_{1}B)< (C_{2}D=C_{2}E)\)
As points C1 and C2 are equidistant from line L; \(C_{1}H_{1}=C_{2}H_{2}\)
So AB<DE (3rd side of the triangle) (Figure enclosed)
Sufficient

St2:- AB<DE
Sufficient.

Ans. (D)

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Can anyone please explain why D is the right choice. I am not sure why the second statement is true. Even though the length of the chord AB is smaller than the second circle, can the part where the two points intersect the line can differ, hence, can the radius of Circle with center C1 be bigger?

Hi shyamhp

Please refer the two triangles under discussion in the two circles, height of the two triangles are equal.(Given in question)

So, while calculating area(1/2 base*height), the only parameter which is the key for comparing the areas of two triangles is the measure of base.

Hence, the triangle with larger measure of base would have larger area and vice-versa.
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how do you know that the height of the two triangles is equal? is it from the statement 'points C1 and C2 are equidistant from line L'?

Thanks

Hi Skn1608,

Exactly.
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Hi,

Can someone please explain why statement 1 is sufficient?

My rationale is that the legs of the triangle with radius of 8 will stretch to a greater extent than the legs of the triangle, which has radius of let's say 6.

Thanks!

Posted from my mobile device

Might be a stupid question but regarding to (1)

How do I know that A-B is the radius? Couldn't it be the case that it looks like a radius but isn't exactly the radius, hence we need a statement that says "AB is the radius"?

I am confused because I've seen a similar DS question a few days ago and it was only sufficient with statement number (2) that said that a given chord is indeed the diameter / radius.
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This question has contradicting answer statements.
For statement 1. if the both the circles have equal heights then radius of circle 1 being less than radius of circle 2 would translate into length of chord AB> length of chord DE as per the Pythagoras theorem.
Statement 2 is clearly sufficient, but it says DE>AB which is the opposite of statement 1.

Can someone please advise if what i have deduced is incorrect?

Check the diagram in this post: https://gmatclub.com/forum/the-figure-a ... l#p2106705 It SHOWS that the radius of Circle 1 being less than the radius of Circle 2, results into AB < DE, not AB > DE.
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The way I thought about it was that first we know that the area of a triangle is 1/2 base * height. So any info we have about these measurements might help.

For (1) given that the radius of circle 1 is less than the radius of circle 2, this means that the distance from C1 to A and to B is shorter than the distance from C2 to D and E. Meaning that the base for the calculation of the area of ABC is smaller therefore the area is smaller.

For (2) I followed this same logic since it is given that the base of triangle ABC is shorter than CDE.

(1) As the heights of the circles are equal the circle with a larger radius will produce a larger area for its triangle. Because applying the Pythagorean formula we will get a larger base for the circle \(C_2\). Sufficient.

(2) Larger cord will give a larger base for the triangle and as the heights are equal so the circle \(C_2\) will have a larger area for its triangle. Sufficient.

The answer is \(D\)

Solution:

In the triangles ABC1and DC2E, let the height be "h" in each of the triangles.

This height is "h" for both as the distance b/w C1 and AB = h = distance b/w C2 and DE.

The area = 1/2 (AB)*h for triangle ABC1 and 1/2 DE*h for triangle DC2E

So, we are being asked if 1/2 (AB)*h < 1/2 DE*h?

=> Is AB <DE ?

​St(1):-The radius of Circle 1 is less than the radius of Circle 2.

The height being same we have 1/2(C1A)*h and 1/2 (C2D)*h as areas of the triangles.

C1A <C2D is given and thus answer to the question stem is yes.(Sufficient)

St(2):- AB<DE

This is exactly asked as a question. Thus (Sufficient) (option d)

Devmitra Sen(GMAT Quant Expert)


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No Bullcrap approach,

Let the radius be r1 & r2 ,

For the 1st circle by Pythagorean theorem-
(r1)^2 = (base1)^2 + (height)^2
or, (height)^2 = (r1)^2 - (base1)^2

from Statement (1) -

As the height is constant and as we increase r1(say to r2) base(say base2) of the 2nd triangle in circle 2 will also increase.
or, r1 < r2 => (base1) < (base2)

As the height is constant, the triangle with bigger base will have larger area => statement 1 is Sufficient

CrackverbalGMAT
How can you assume that the heights are the same?

Hey woohoo921,
Thanks for posting your query.


You’re missing a very important piece of information given in the question. This is common in questions with too much information but should not happen if you wish not to lose points to avoidable mistakes. 😊


Always read very carefully! For this question, let me help you:


READING THE QUESTION AGAIN:
Below is the question stem of this DS question. I have bolded the part that I want you to focus on. Read and then try to infer what the bolded part could mean. Let’s go!

“The figure above shows Line L, Circle 1 with center at \(C_1\), and Circle 2 with center at \(C_2\). Line L intersects Circle 1 at points A and B, Line L intersects Circle 2 at points D and E, and points \(C_1\) and \(C_2\) are equidistant from line L. Is the area of Δ\(ABC_1\) less than the area of Δ\(DEC_2\)?”


EXPLANATION:
I hope you tried inferring on your own. Let me elaborate now -

• The bolded statement in the question is what you missed completely. We are given that points \(C_1\) and \(C_2\) are equidistant from line L. To understand the meaning of this, you first need to understand how the distance between a point and a line is calculated.
• Here’s the definition – the distance between a point and a line is equal to the length of the perpendicular drawn from the point to the line.
  
  • In our question, this means that if we draw \(C_1M\) perpendicular to AB (M lying on AB), then \(C_1M\) is the distance between \(C_1\) and line L.
  • Similarly, that if we draw \(C_2N\) perpendicular to DE (N lying on DE), then \(C_2N\) is the distance between \(C_2\) and line L.
• Per our given information (the bolded part), we have \(C_1M = C_2N\). -----------(1)

Now, note that \(C_1M\) and \(C_2N\) are precisely the heights of triangles \(ABC_1\) and \(DEC_2\)! And hence, from (1), we get that the heights of these two triangles are equal.


And that’s it!


TAKEAWAYS

1. Read a question piece by piece and keep translating English to Math as you go.
2. Keep making inferences as you gather more and more information while reading questions or solving them.

Hope this helps!


Best,
Shweta
Quant Product Creator, e-GMAT
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