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01 Feb 2018, 00:07
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
80% (01:16) correct
20%
(01:49)
wrong
based on 79
sessions
History
Date
Time
Result
Not Attempted Yet
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?
(A) 1/2*(x + y)
(B) x + y
(C) xy
(D) 1/2
(E) 1
Attachment:
2018-02-01_1104.png [ 14.24 KiB | Viewed 3164 times ]
01 Feb 2018, 02:48
Kudos
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Bunuel
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?
(A) 1/2*(x + y)
(B) x + y
(C) xy
(D) 1/2
(E) 1
Attachment:
2018-02-01_1104.png
Area of triangle ADC = \(\frac{1}{2}*AD*x\)
Area of triangle ABC = Area of triangle ADB - Area of triangle ADC
Area of triangle ABC = \(\frac{1}{2}*AD*(x+y) - \frac{1}{2}*AD*x = \frac{1}{2}*AD*(2x-x)\) (as x=y) = Area of triangle ADC
Therefore, Ratio of area of (triangle ABC)/area of (triangle ADC) = 1(Option E)
01 Feb 2018, 04:17
Kudos
Bookmarks
Bunuel
In the figure above, x = y. What is the ratio area of (triangle ABC)/(triangle ADC) ?
(A) 1/2*(x + y)
(B) x + y
(C) xy
(D) 1/2
(E) 1
Attachment:
2018-02-01_1104.png
Area of the triangle = (1/2)*Base*Height
For triangle ABC, Area of the triangle = (1/2)*y*AD
For triangle ADC, Area of the triangle = (1/2)*x*AD
Now since Area of ABC/Area of ADC = (1/2)*y*AD/(1/2)*x*AD = y/x
But since, x=y hence
Now since Area of ABC/Area of ADC = y/y = 1
Answer: option E
