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01 Aug 2014, 06:05
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walker
Fig
Either the question is wrong or this diagram. I don't see ACF as a right angle.
01 Aug 2014, 06:47
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kamranjkhan
walker
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Either the question is wrong or this diagram. I don't see ACF as a right angle.
Angle AFC is a right angle, not ACF.
01 Feb 2023, 08:47
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I have a simple solution to this Q. We are required to find ratio of area of triangle BCE to triangle ABD.
Let a be side of square ADEF and x be side of equilateral triangle ABC
Consider triangle AFC: \(x^2 = a^2 + FC^2\) ....eqn 1
Consider triangle BCE: \(x^2 = BC^2 + CE^2 = 2CE^2\)
or \(CE^2 = \frac{x^2}{2}\)...eqn 2
(Note: As equilateral triangle is inscribed in the square, by symmetry, we can figure out that BC = CE )
Now, \(x^2 = 2CE^2 = (a - FC)^2 = 2(a^2 + FC^2 -2a*FC)\) ....eqn 3
Substitute \(x^2\) from eqn 1 in eqn 3
\(a*FC = \frac{x^2}{4}\)
Now the ratio of area of triangle BCE to triangle ABD is
\(\frac{CE^2}{a*FC} = (x^2/2)/(x^2/4) = 2\)
Ans : C
Let a be side of square ADEF and x be side of equilateral triangle ABC
Consider triangle AFC: \(x^2 = a^2 + FC^2\) ....eqn 1
Consider triangle BCE: \(x^2 = BC^2 + CE^2 = 2CE^2\)
or \(CE^2 = \frac{x^2}{2}\)...eqn 2
(Note: As equilateral triangle is inscribed in the square, by symmetry, we can figure out that BC = CE )
Now, \(x^2 = 2CE^2 = (a - FC)^2 = 2(a^2 + FC^2 -2a*FC)\) ....eqn 3
Substitute \(x^2\) from eqn 1 in eqn 3
\(a*FC = \frac{x^2}{4}\)
Now the ratio of area of triangle BCE to triangle ABD is
\(\frac{CE^2}{a*FC} = (x^2/2)/(x^2/4) = 2\)
Ans : C
