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03 Nov 2018, 10:05
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (03:16) correct
58%
(02:57)
wrong
based on 45
sessions
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Time
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Not Attempted Yet
Attachment:
5.jpg [ 9.71 KiB | Viewed 3544 times ]
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
03 Nov 2018, 10:21
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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
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
SUNILAA
Joined: 09 Apr 2018
Last visit: 21 Nov 2021
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Schools: IIMA PGPX"20
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03 Nov 2018, 11:00
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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
so it is (1/2 * AB/2*ABroot3/2 )-1/2*AB/2*AB/2root2
--------------------------------------------------------
root3/2 * ABsquare
