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# Two isosceles Triangles have equal vertical angles and their

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Two isosceles Triangles have equal vertical angles and their [#permalink]

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15 Nov 2012, 07:39
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Question Stats:

80% (00:52) correct 20% (01:14) wrong based on 126 sessions

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Two isosceles Triangles have equal vertical angles and their areas are in the ratio 16:25. Find the ratio of their corresponding heights .

(A) 4/5

(B) 5/4

(C) 3/2

(D) 5/7

(E) 2/3
[Reveal] Spoiler: OA

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Re: Two isosceles Triangles have equal vertical angles and their [#permalink]

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15 Nov 2012, 07:59
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vomhorizon wrote:
Two isosceles Triangles have equal vertical angles and their areas are in the ratio 16:25. Find the ratio of their corresponding heights .

(A) 4/5

(B) 5/4

(C) 3/2

(D) 5/7

(E) 2/3

We are basically given that the triangles are similar.

In two similar triangles, the ratio of their areas is the square of the ratio of their sides and also, the square of the ratio of their corresponding heights.

Therefore, area/AREA=height^2/HEIGHT^2=16/25 --> height/HEIGHT=4/5.

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Re: Two isosceles Triangles have equal vertical angles and their [#permalink]

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15 Nov 2012, 08:23
Is there any other way to solve the problem in case you didn't know the formula of isosceles triangle = 1/2 (leg)^2

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Re: Two isosceles Triangles have equal vertical angles and their [#permalink]

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15 Nov 2012, 08:31
nelz007 wrote:
Is there any other way to solve the problem in case you didn't know the formula of isosceles triangle = 1/2 (leg)^2

This question trumped me the first time i attempted it, because SIMILARITY and properties of SIMILAR triangles are easy to forget, but they are very important, as i have seen quite a few questions where knowledge of similarity is important, especially when we have one big triangle being divided into two etc etc (MGMAT questions, in their advanced book) ..

In addition to the explanation given by Bunuel, this link is worth bookmarking :

http://www.mathopenref.com/similartriangles.html
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Re: Two isosceles Triangles have equal vertical angles and their [#permalink]

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16 Nov 2012, 18:30
nelz007 wrote:
Is there any other way to solve the problem in case you didn't know the formula of isosceles triangle = 1/2 (leg)^2

You can use the property of two similar triangles : If two similar triangles have corresponding side lengths in ratio a:b, then their areas will be in ratio a$$2$$ : b$$2$$

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Re: Two isosceles Triangles have equal vertical angles and their [#permalink]

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19 Aug 2013, 07:11
I have similar solution:

Area1: 16=1/2*a*h => a=32/h
Area2: 25=1/2*A*H => A=50/H

Find: h/H

Similar triangles so: h/H=a/A

Solve:

h/H=a/A

h/H=(32/h)/(50/H)

h/H=(32/h)*(H/50)

h/H=(32/50)*(H/h)

h/H=(16/25)*(H/h)

h^2/H^2=16/25

h/H=4/5

Answ: A.

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Re: Two isosceles Triangles have equal vertical angles and their [#permalink]

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25 Sep 2014, 11:19
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Re: Two isosceles Triangles have equal vertical angles and their   [#permalink] 25 Sep 2014, 11:19
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