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In the figure above, ABCD is a square with sides equal to 1, AFC is
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Bunuel
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In the figure above, ABCD is a square with sides equal to 1, AFC is [#permalink] New post  15 May 2018, 01:27
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In the figure above, ABCD is a square with sides equal to 1, AFC is an arc of a circle centered at B, and AEC is an arc of a circle centered at D. What is the area of rhombus AECF?


A. \(2 - \sqrt{2}\)

B. \(\sqrt{2} - 1\)

C. \(\sqrt{2}(2 - \sqrt{2})\)

D. \(\sqrt{2}\)

E. \(1 + \sqrt{2}\)

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In the figure above, ABCD is a square with sides equal to 1, AFC is [#permalink] New post  15 May 2018, 03:24
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Solution



Given:
• ABCD is a square with side = 1
• AFC is an arc, centered at B
• AEC is an arc, centered at D

To find:
• Area of the rhombus AECF

Approach and Working:
• To know the area of a rhombus, we need to find the length of its diagonals, as area of a rhombus = \(\frac{1}{2}\) * product of the diagonals
• For square ABCD with side 1-unit, diagonal AC = \(\sqrt{2}\)
    o AC is also the longer diagonal of AECF
• Another diagonal EF = BD – BE – DF
    o Now, DE = 1, as the radius of the semicircle is equal to the side of the square
    o Therefore, BE = BD – DE = \(\sqrt{2} – 1\)
    o Similarly, DF = \(\sqrt{2} – 1\)
    o Hence, EF = \(\sqrt{2} – 2 * (\sqrt{2} – 1) = 2 – \sqrt{2}\)
    o Area of the rhombus AECF = \(\frac{1}{2} * AC * EF = \frac{1}{2} * \sqrt{2} * 2 – \sqrt{2} = \sqrt{2} – 1\)


Hence, the correct answer is option B.
Answer: B
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Re: In the figure above, ABCD is a square with sides equal to 1, AFC is [#permalink] New post  23 Mar 2022, 03:39
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Re: In the figure above, ABCD is a square with sides equal to 1, AFC is [#permalink]
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