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ps - square in square

by ccassel » Tue Apr 19, 2011 2:44 pm
I solved this correctly by plugging in good numbers but how would you solve it without plugging in? It took me too much time.

The triangels in a figure (a small triangle in a large triangle not necessarily touching sides) are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region (area between the small triangle and large triangle) in terms of K?

A. 3/4K
B. 2/3K
C. 1/2K
D. 1/3K
E. 1/4K

Answer is A
Last edited by ccassel on Wed Apr 20, 2011 8:07 am, edited 1 time in total.
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by therealtomrose » Tue Apr 19, 2011 6:00 pm
More information is needed. Please describe the shaded region. Where are these questions coming from BTW? I can't figure out the source. They're not OG, and they're not MGMAT.
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by manpsingh87 » Tue Apr 19, 2011 9:04 pm
ccassel wrote:I solved this correctly by plugging in good numbers but how would you solve it without plugging in? It took me too much time.

The triangels in a figure (a small triangle in a large triangle not necessarily touching sides) are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?

A. 3/4K
B. 2/3K
C. 1/2K
D. 1/3K
E. 1/4K

Answer is A
i think by saying shaded region he means the region bounded between the small triangle and the large triangle..!!!
let p be side of larger triangle, q be side of smaller triangle.
area of a larger triangular region=1/2*(Sqrt(3)/2)*p^2=k;
area of a smaller traingular region=1/2*(sqrt(3)/2)*q^2;
also p/q=2/1; q=p/2;
submitting this information we have are of smaller traingular region=1/4{1/2*(Sqrt(3)/2)*p^2;=k/4

therefore area of shaded region=k-k/4=3k/4;
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by force5 » Wed Apr 20, 2011 1:25 am
ccassel what is the shaded region. It cant be assumed.
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by GMATGuruNY » Wed Apr 20, 2011 3:45 am
ccassel wrote:I solved this correctly by plugging in good numbers but how would you solve it without plugging in? It took me too much time.

The triangels in a figure (a small triangle in a large triangle not necessarily touching sides) are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region in terms of K?

A. 3/4K
B. 2/3K
C. 1/2K
D. 1/3K
E. 1/4K

Answer is A
Plugging in is a very efficient way to solve this problem if you know the formula for the area of an equilateral triangle: A = (b^2)/4 * √3.

In the larger triangle, let B = 2.
In the smaller triangle, let b = 1.
Since the area of each triangle will include 1/4 * √3, and we're looking only for the difference between the two areas, we can ignore the 1/4 * √3 and determine only the values of B^2 and b^2:
Larger triangle = B^2 = 2^2 = 4. Thus, K = 4.
Smaller triangle = b^2 = 1^1 = 1.
Thus, the area outside the smaller triangle = 4-1 = 3. This is our target.

Now we plug K=4 into the answers to see which yields our target of 3.

Only answer choice A works:
(3/4)K = (3/4)*4 = 3.

The correct answer is A.
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by m.abdulk » Wed Apr 20, 2011 7:25 am
guessing the shaded region to be the region between the 2 traingles,

the approach i followed is

Let the Larger triangle be ABC and smaller one be XYZ

since both are equilateral => both are similar triangles

hence AB/XY=BC/YZ=AC/XZ

and we know from the question that ratio is 2/1

so areas will be in the ratio of (2/1)^2(i.e, squared) =>4/1

so the area between the 2triangles shuld be 3/4K and hence A.

Can someone tell is this the right way or any easier way of doing this?
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by ccassel » Wed Apr 20, 2011 8:10 am
Sorry for the missing piece of information. The shaded region is the area between the small triangle and large triangle - ie. inside the large, and outside the small. I have edited the queston.

Cheers,
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by Stuart@KaplanGMAT » Wed Apr 20, 2011 11:01 am
ccassel wrote:I solved this correctly by plugging in good numbers but how would you solve it without plugging in? It took me too much time.

The triangels in a figure (a small triangle in a large triangle not necessarily touching sides) are equilateral and the ratio of the length of a side of the larger triangle to the length of a side of the smaller triangle is 2/1. If the area of the larger triangular region is K, what is the area of the shaded region (area between the small triangle and large triangle) in terms of K?

A. 3/4K
B. 2/3K
C. 1/2K
D. 1/3K
E. 1/4K

Answer is A
Mitch has provided a great example of picking numbers, so let's try something different: applying basic principles and avoiding math.

First, we need to understand the question: we have a smaller equilateral triangle inside a larger one and the triangles are similar - in other words, the proportions are identical. The area of the shaded region is simply the area of the big minus the area of the small.

We're given the relationship between linear measurements of the triangles: the smaller one has sides 1/2 the length of the larger one.

We're asked about the relationship of two-dimensional measurements of the triangles: we want the area of the big - the area of the small.

Well, since area is a two-dimensional measurement, we square the original ratio to get the new ratio. Since the linear ratio of small:large = 1:2, the square ratio of small:large is 1:4.

Accordingly, the small triangle must have 1/4 the area of the large triangle.

So, if the area of the large triangle is K, the area of the small triangle is (1/4)K.

K - (1/4)K = (3/4)K... choose (A).
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