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GMAtprep question

Expert replies
by mlane25269 » Sun Jun 22, 2008 3:58 pm
How do I solve?
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Source: — Problem Solving |

by Nycgrl » Sun Jun 22, 2008 6:36 pm
Area of bigger trangle A = 2 a(area of smaller triangle)

A/a= 2

all the three angles are equal therefore both the triangles are similar

A/a =S/s=2 =>S=2s
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ans

by smashinonions » Sun Jun 22, 2008 6:53 pm
https://www.mathopenref.com/similartrianglesareas.html


for similar triangles,

side = Sqrt (area)

let s = a (area)
and S= A (area)

given,


A = 2a


also,

s / S = sqrt(a)/ sqrt(2a)

or S = Sqrt(2a) * s/ sqrt(a)

= Sqrt (2a) Sqrt (a) * s/ Sqrt (a) * sqrt (a) [simplification]

= sqrt (2) * s (ANS)
------------------------------------------------------
Five years ago: GMAT - 740

Five years later: 1st Attempt - 620
GOING for 2nd attempt: Target - 740
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by AleksandrM » Mon Jun 23, 2008 9:49 am
smashonions,

Is this some sort of a rule?

for similar triangles,

side = Sqrt (area)


Can you elaborate. Thanks.
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by egybs » Mon Jun 23, 2008 11:58 am
The rule is:

Where:
A1: area of similar triangle 1
A2: area of other similar triangle 2
L1: length of any side (the base) or height on triangle 1
L2: length of related side or height on triangle 2


A1/A2 = X

L1 = L2 * sqrt(X)

If you want the proof, let me know and I'll work out it for you.
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by smashinonions » Mon Jun 23, 2008 3:12 pm
the ratio of the sides of two similar triangles is equal to the sqrt of the ratio of their areas is the rule

which is if triangles A1 and A2 have sides S1 and S2 and areas Ar1 and Ar2 then,

S1 / S2 = Sqrt ( Ar1 / Ar2)

pls visit https://www.mathopenref.com/similartrianglesareas.html for more details :)
------------------------------------------------------
Five years ago: GMAT - 740

Five years later: 1st Attempt - 620
GOING for 2nd attempt: Target - 740
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by AleksandrM » Tue Jun 24, 2008 9:04 am
Kool, thank you guys.
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by rosh26 » Thu Jun 26, 2008 10:07 am
Is there another way to solve the problem?

I'm just not understanding the solution above.

Thanks in advance.
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by mlane25269 » Thu Jun 26, 2008 10:14 am
I agree with rosh26!! I'm having trouble understanding the concept. Is this based on a formula because I haven't seen this concept covered in the official GMAT guide or any of the Kaplan resources I've looked at. Would this fall under geometry and how likely is something like this to appear on the real GMAT?
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by AleksandrM » Thu Jun 26, 2008 3:04 pm
If there are two similar triangles, one with Area = 8 the other one with Area = 2 the bigger angle has side S the smaller side s.

Say that side S is the base and = 4, which means that height is also 4.

Now, what is the measure of side s.

S/s = sqroot(Area/area)

4/s =sqroot(8/2)

s = 2.

I believe that this is how it's done, though I just learned this approach myself.
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by rosh26 » Sat Jun 28, 2008 9:16 pm
but the answer is sqrt 2 (s) not 2s...
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by AleksandrM » Sun Jun 29, 2008 6:45 am
That's because that is as far as you can simplify it.
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