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gmat prep - similar triangles.

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gmat prep - similar triangles.

by g.shankaran » Tue Jun 14, 2011 7:22 am
Triangle ABC and Triangle CDF has angle a = angle c , angle d = angle b and angle c = angle f. the area of triangle cdf is twice the area of triangle ABC. if the base of triangle ABC is s,and the base of triangle CDF is S, then in terms of s, S =

1. sqrt(2)/2 s
2. sqrt(3)/2 s
3. sqrt(2)s
4. sqrt(3)s
5. 2s
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by Frankenstein » Tue Jun 14, 2011 7:59 am
Hi,
If a,b,c are the sides of the triangle,
Area of triangle is, a = (1/2)b*c*sin(angle A)
Similar triangle will be having same angles with sides p*a,p*b,p*c
Area is A = (1/2)p*b*p*c*sin(angle A).
Given that A =2a =>p^2 = 2 =>p=sqrt(2).
So, S=p*s = ssqrt(2)

Hence, 3
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by gmaths » Tue Jun 14, 2011 8:03 am
first of all these two triangles are similar as all 3 angles of ABC are equal to corresponding angles of CDF....therefore, the ratio of all the sides will be equal(lets say that ratio to be K)...
then the property says that the ratio of corresponding areas must be equal to K^2.....here we are given the ratio of the areas of the two triangles to be equal to 2:1....and so the ratio of their bases must be sq root of 2 :1....and hence S:s::sqrt(2).....or S=sqrt(2)s.....hence option 3.
if it is to be , it is up to me...
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by newgmattest » Tue Jun 14, 2011 9:30 pm
Hi GMAT Experts,

Please help to explain the logic to solve this.

Thanks!
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by winniethepooh » Tue Jun 14, 2011 10:22 pm
Frankenstine, I don't understand what you write!
What is p doing there?
& what is trigonometry included for?
No other method?
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by Frankenstein » Tue Jun 14, 2011 10:44 pm
winniethepooh wrote:Frankenstine, I don't understand what you write!
What is p doing there?
& what is trigonometry included for?
No other method?
Okay.. I will try to make it clear.
if a,b,c are the sides of a triangle. Then similar triangle will have sides in the same ratio as a:b:c
So, let the sides of similar triangle be p*a,p*b,p*c. So I write them as pa,pb,pc
if 's' is the semis perimeter of the triangle, ie. s = (a+b+c)/2 [Please don't confuse this's' with the base length 's' in the question]
then area of triangle is given by sqrt[s*(s-a)*(s-b)*(s-c)]
so, area of 2nd triangle will be sqrt[ps*(ps-pa)*(ps-pb)*(ps-pc)]= sqrt[p^4*s*(s-a)*(s-b)*(s-c)]=p2*sqrt[s*(s-a)*(s-b)*(s-c)]
From the given data p^2 = 2.So, p=sqrt(2).
So, S=p*s = sqrt(2)s
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by GMATGuruNY » Wed Jun 15, 2011 3:38 am
g.shankaran wrote:Image
We can plug in any two triangles that are similar.
Let each triangle be a 45-45-90 triangle with sides proportioned x: x : x: x√2:
Image
In the smaller triangle, let s=1.
Area of smaller triangle = (1/2)*1*1 = 1/2.
Since the larger triangle is twice the size, area of the larger triangle = 2*(1/2) = 1.
Thus, given the area of the larger triangle:
(1/2)*S*S = 1.
S^2 = 2.
S = √2. This is our target.

Now we plug s=1 into the answers to see which yields our target of √2.

Only answer choice C works:
√2s = √2*1 = √2.

The correct answer is C.
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by winniethepooh » Wed Jun 15, 2011 9:50 am
Thanks Mitch.
This clarification was indeed needed. But, I would like to know what really pushed you to assume that this question is based on an Isosceles triangle? Or is it that you have assumed this in the absence of information ?
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by GMATGuruNY » Wed Jun 15, 2011 1:15 pm
winniethepooh wrote:Thanks Mitch.
This clarification was indeed needed. But, I would like to know what really pushed you to assume that this question is based on an Isosceles triangle? Or is it that you have assumed this in the absence of information ?
I made no assumptions; the triangles do not have to be isosceles. I chose to make them isosceles right triangles in order to make the math easier. In a 45-45-90 triangle, the base = the height, so the area is easy to calculate.

When plugging in, choose values that:
-- satisfy the conditions given in the problem
-- make the math easy

In the problem above, isosceles right triangles accomplish both of these goals.
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by winniethepooh » Wed Jun 15, 2011 4:07 pm
I love your way of solving problems!!
Thanks a ton!!
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