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Area

by PGMAT » Mon Sep 03, 2012 8:22 am
In the attached figure, is the area of triangular region ABC equal to the area of triangular region DBA ?

(1) (AC)^2=2(AD)^2

(2) ∆ABC is isosceles.

Source: OG13 DS79

[spoiler]OA:C[/spoiler]
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Source: — Data Sufficiency |
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by neelgandham » Mon Sep 03, 2012 10:47 am
In the attached figure, is the area of triangular region ABC equal to the area of triangular region DBA ?
(1) (AC)^2=2(AD)^2
AC = √2 * AD. We don't know the value of BC and hence can't determine the value of AB and DB(relatively). So, statement I is insufficient to answer the question.
(2) ∆ABC is isosceles.
Since we don't know the dimensions of AD or DB, statement II is insufficient to answer the question.
From statement I and statement II
AC = √2 * AD
AC = BC = √2 * AD
Length of side AB = √[(AC^2)+(BC^2)] = √(2*AD^2 + 2*AD^2) = 2*AD
Area of triangle DBA = (1/2)*AD*AB = (1/2)*AD*2*AD = AD*AD
Area of triangle ABC = (1/2) * √2 * AD * √2 * AD = AD*AD
Area of triangle DBA = Area of triangle ABC

IMO C
Anil Gandham
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by mrapaka » Mon Sep 03, 2012 1:33 pm
statement 1: AC^2=2 AD^2 >> no info about the bases of the both triangles. Not sufficient.
statement 2: ABC is an isosceles triangle: => the sides are in the ratio of 1:1:sqrt 2

if AC=BC=X => AB=x* sqrt 2
Area of ABC = 1/2 * AC * BC = 1/2 * x^2
Area of DAB = 1/2 * AB * AD = 1/2 * x* sqrt 2 * AD , but no info about AD. Not sufficient.

Together: AC = sqrt 2 * AD => AD = x/sqrt 2
area od DAB = 1/2 * x/sqrt 2 * x* sqrt 2 = 1/2 * x^2. which is equal to the area of ABC. Sufficient.

So the answer is C
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