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The figure above consists of triangle ADF and rectangle AB
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08 Oct 2018, 12:36
08 Oct 2018, 12:36
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Question Stats:
39% (03:17) correct
60% (02:43) wrong based on 68 sessions
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The figure above consists of triangle ADF and rectangle ABCD. BE = CG = 1.5 and EG = 2. If the area of triangle EFG = x, what is the area of rectangle ABCD?
A. \(5x\)
B. \(\frac{5x}{2}\)
C. \(15x\)
D. \(\frac{15x}{2}\)
E. \(\frac{25x}{2}\)
Re: The figure above consists of triangle ADF and rectangle AB
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11 Apr 2020, 10:43
11 Apr 2020, 10:43
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BC = 1.5 + 2 + 1.5 = 5 = AD
Since ABCD is a rectangle, BC is parallel to AD. So triangle EFG is similar to triangle AFD.
\(Sides Ratio = \frac{EG}{AD} = \frac{2}{5}\)
Ratio of altitudes will be 2/5 too
Area of EFG = x = (1/2)*Altitude * 2
Altitude of triangle EFG = x
So altitude of triangle AFD \(= \frac{5x}{2}\)
\(CD = \frac{5x}{2} - x = \frac{3x}{2}\)
Area of rectangle ABCD \(= AD * CD = 5 * \frac{3x}{2} = \frac{15x}{2}\)
Since ABCD is a rectangle, BC is parallel to AD. So triangle EFG is similar to triangle AFD.
\(Sides Ratio = \frac{EG}{AD} = \frac{2}{5}\)
Ratio of altitudes will be 2/5 too
Area of EFG = x = (1/2)*Altitude * 2
Altitude of triangle EFG = x
So altitude of triangle AFD \(= \frac{5x}{2}\)
\(CD = \frac{5x}{2} - x = \frac{3x}{2}\)
Area of rectangle ABCD \(= AD * CD = 5 * \frac{3x}{2} = \frac{15x}{2}\)
Re: The figure above consists of triangle ADF and rectangle AB
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08 Sep 2025, 09:24
08 Sep 2025, 09:24
Carcass it took me 3:17
if this sort of question were to occur in section of quants then is this time good enough ?
if this sort of question were to occur in section of quants then is this time good enough ?
