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Please refer to the figure above, triangle ABC is an isosceles triang

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Please refer to the figure above, triangle ABC is an isosceles triang  [#permalink]

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New post 03 Nov 2018, 10:05
1
5
00:00
A
B
C
D
E

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53% (03:18) correct 47% (02:25) wrong based on 28 sessions

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Please refer to the figure above, triangle ABC is an isosceles triangle with AB = BC and BD is the perpendicular dropped from vertex B to the side AC where D is mid point of AB. Point E is marked on BD such that \(\angle\)DAE = 45 degree and \(\angle\)ECB =15 degree. What is the ratio of the area of triangle BEC to the area of the triangle ABC?


A) \(\frac{(3-√3)}{6}\)
B) \(\frac{(3+√3)}{4}\)
C) \(\frac{(5-√3)}{6}\)
D) \(\frac{(5+√3)}{3√3}\)
E) Cannot be determined

Weekly Quant Quiz #7 Question No 3


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Re: Please refer to the figure above, triangle ABC is an isosceles triang  [#permalink]

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New post 03 Nov 2018, 10:21
BD is the perpendicular bisector, hence AD= CD
moreover angle BAE = angle BCD

triangle AED and triangle EDC are similar and congruent hence angle EAD = 45= angle ECD= 60


hence ABC is a equilateral triangle , hence AB= BC= AC, lets say it is a
AD = sqrt AB*AB - AD/2*AD/2 = sqrt 3/2a

in triangle ADE,
DE =AD = a/2
hence BE = BD-DE= sqrt3 /2 a - a/2

area(triangle BEC)/ area (triangle ABC) = (1/2*CD *BE) /(sqrt 3/4 a^2) = (1/2* a/2*(sqrt3 /2- 1/2) / (sqrt3/4 a^2)= sqrt 3 - 1/sqrt 3/ 2, multiplying sqrt 3 on Numerator and Denominator
gives option A
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Re: Please refer to the figure above, triangle ABC is an isosceles triang  [#permalink]

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New post 03 Nov 2018, 11:00
It cannot be calculated ans E.
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Re: Please refer to the figure above, triangle ABC is an isosceles triang  [#permalink]

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New post 03 Nov 2018, 11:00
Ans is a) its a little complicated but for area of triangle bec we subtract the area of BDC-DEC
so it is (1/2 * AB/2*ABroot3/2 )-1/2*AB/2*AB/2root2
--------------------------------------------------------
root3/2 * ABsquare
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Re: Please refer to the figure above, triangle ABC is an isosceles triang  [#permalink]

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New post 04 Nov 2018, 02:05
ABC is an equilateral triangle.
Assume side = 10
Area of ABC = 25√3

BD = 10√3 / 2 = 5√3
BD = BE + ED
BE = 5√3 - 5 = 5(√3 - 1)

Area of BEA = 1/2 * BE * DC = 25(√3 - 1) / 2

BEA / ABC = 25(√3 - 1) / 2* 25√3

= (√3 - 1)/ 2√3
Multiply num and deno with √3
3-√3 / 6

A
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Re: Please refer to the figure above, triangle ABC is an isosceles triang   [#permalink] 04 Nov 2018, 02:05
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