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Triangle area

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by selango » Thu May 27, 2010 7:16 am
In the below figure, if the area of triangle on the right is twice the area of triangle on the left,then in terms of s, S is

A)(sqrt2) * S/2

B)(sqrt3) * S/2

C)(sqrt2) * S

D)(sqrt3)*S

E) 2S

Any better approach to solve this problem quickly?
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by gmatmachoman » Thu May 27, 2010 7:30 am
Let ABC be triangle having "s" as side

DEF be triangle having S as side

ABC is similar to DEF

So A2/A1= (S/s)^2

From stem A2= 2 * A1

2=(S/s)^2

S = s. sqrt(2)

Pick C
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by selango » Thu May 27, 2010 7:40 am
gmatmachoman wrote:Let ABC be triangle having "s" as side

DEF be triangle having S as side

ABC is similar to DEF

So A2/A1= (S/s)^2

From stem A2= 2 * A1

2=(S/s)^2

S = s. sqrt(2)


gmatmachoman,

Can you explain this "A2/A1= (S/s)^2"


Pick C
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by indiantiger » Thu May 27, 2010 7:40 am
Question is based on similarity ratio :

The two given triangles are similar based on AAA

two get the ratio of the areas we square the similarity ratio
=> s^2/S^2 = 1/2

=>s/S = sqrt(1/2)
=> sqrt(2)*s = S (answer)

Does choice (C) have small s C)(sqrt2) * S
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by thephoenix » Thu May 27, 2010 8:48 am
Can you explain this "A2/A1= (S/s)^2"
its property of similar triangles
if two triangles are similar then square of the ratio of the two corresponding sites is equal to ratio of respective areas
Many of the great achievements of the world were accomplished by tired and discouraged men who kept on working
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by selango » Thu May 27, 2010 8:58 am
Got it..thanks...
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by Patrick_GMATFix » Thu May 27, 2010 10:02 am
Because no restriction is given to the degree measure of angles, and because all the info we have is about ratios (area1 is twice area2) we can plug in any pair of triangles and we'll get the same result. We have to make sure that the triangles we use are similar (they have the same angles)

The easiest triangles to use would be two 45-45-90 triangles. The area of these triangle is(1/2)(bh) where base and height are the same. So the area can be re-written as (1/2)(s)(s) for the small triangle and (1/2)(S)(S) for the big.

Let the area of the big triangle be 8, and the area of the small triangle be 4.

Big triangle: 8 = (1/2)(S)(S) --> S=4
Small triangle: 4 = (1/2)(s)(s) --> s=2*root(2)

Now we can answer the question: S = s*root(2) because [2*root(2)]*root(2) = 4.

The answer is C.


Alternatively, in general if you have two geometric shapes with proportional measurements, if the ratio of length to length is p/r, then the ratio of area to area will be (p/r)^2 and the ratio of volume to volume will be (p/r)^3.

For example if the ratio fo side of square A to side of square B is 1/2, then the rato of areas will be (1/2)^2 and the ratio of volumes (of the cubes) will be (1/2)^3.

In this question, we can use this property to solve in 10 seconds.

The ratio of area to area of the triangles is (2/1). This means that (p/r)^2 =(2/1) where p/r is the ratio of length to length. Take the square root of both sides and you'll get (p/r)=root(2)/1. In other words, the longer side is root(2)*the shorter side. S=root(2)*s. The answer is C.

If you have trouble understanding these explanations, you can find a step by step video solution at GMATPrep question 1049
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