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11 Jun 2019, 01:44
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
65% (01:29) correct
35%
(01:41)
wrong
based on 77
sessions
History
Date
Time
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Not Attempted Yet
In the figure attached, the length of line segment CD is twice the length of line segment BC. The ratio of the area of ΔABC to the area of ΔABD is:
A. \(1:\sqrt{2}\)
B. \(1:\sqrt{3}\)
C. 1:2
D. 1:3
E. 1:4
Attachment:
216155.17.g500054img01.gif [ 1.06 KiB | Viewed 3072 times ]
11 Jun 2019, 02:24
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Bunuel
Area of right angled triangle = 1/2*leg1*leg2; Leg1 is same for both triangles (y) and Leg 2 is X for ABC and ABD=3X(X+2X); So area is 1:3 IMO Option D
11 Jun 2019, 06:25
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Bunuel
Let BC = x --> CD = 2x
BD = BC + CD = x + 2x = 3x
Let AB = y
--> Area of ΔABC = 1/2*AB*BC = 1/2*y*x = xy/2
--> Area of ΔABD = 1/2*AB*BD = 1/2*y*3x = 3xy/2
--> ratio of the area of ΔABC to the area of ΔABD = (xy/2)/(3xy/2) = 1:3
Option D
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