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Re: In the figure below, If the area of triangle CDE is 42, what [#permalink]
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Ratio of Areas of 2 Similar triangles = K x K
Where, K is the constant which is equal to the ratio of 3 sides (Assume: A, B and C) in both the triangles
i.e. K = A1/A2 = B1/B2 = C1/C2

Here, We are given that AB = BC = CD = a (assume)

Triangle AGD and CED are Similar
So, K = AD/CD = 3a/a = 3

Therefore,
3 x 3 = Area of AGD / Area of CED
9 = Area of AGD / 42
Area of AGD = 9 x 42 = 378
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Re: In the gure below, If the area of triangle CDE is 42, what [#permalink]
GreenlightTestPrep as per Midsegment Theorem: The segment joining the midpoints of two sides of a triangle is parallel to and half the length of the third side. Also Midsegment divides triangle into two similar ones.

so if AG/BF = 2 , and BF/CE = 2 then AG/CE =4

then ratio of sides of AGD to CDE = 4 : 1 now if two similar triangles have sides in the ratio x/y then their areas are square of their ratios or 16 : 1

hence area of AGD is 16 times area of CDE i.e. 16*42= 672

whats wrong with my reasoning ?
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Re: In the gure below, If the area of triangle CDE is 42, what [#permalink]
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testtaker123 wrote:
GreenlightTestPrep as per Midsegment Theorem: The segment joining the midpoints of two sides of a triangle is parallel to and half the length of the third side. Also Midsegment divides triangle into two similar ones.

so if AG/BF = 2 , and BF/CE = 2 then AG/CE =4

then ratio of sides of AGD to CDE = 4 : 1 now if two similar triangles have sides in the ratio x/y then their areas are square of their ratios or 16 : 1

hence area of AGD is 16 times area of CDE i.e. 16*42= 672

whats wrong with my reasoning ?


There are no midpoints.
We're dividing sides into 3 equal lengths.
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In the gure below, If the area of triangle CDE is 42, what [#permalink]
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This Classic Triangle Similarity Problem has been taken from Gre Math Review.
I always apply ratios in my solution cuz it will magically give u the quickest way to solve any problem (where ratios can be applied)

Here, CD: AD = 1:3 [As each of the sides is equal]
Now according to the Similarity Area Property of Triangle, If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
Area (ADG) : Area (CDE) = AD^2:CD^2
Area (ADG) = 42*9 = 378
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