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A wire is cut into two pieces, one of length a and the other of length

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A wire is cut into two pieces, one of length a and the other of length  [#permalink]

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New post 20 Mar 2019, 04:53
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A wire is cut into two pieces, one of length a and the other of length b. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length b is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is a/b ?

(A) 1

(B) \(\frac{\sqrt{6}}{2}\)

(C) \(\sqrt{3}\)

(D) 2

(E) \(\frac{3\sqrt{2}}{2}\)

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Re: A wire is cut into two pieces, one of length a and the other of length  [#permalink]

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New post 20 Mar 2019, 07:17
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Bunuel wrote:
A wire is cut into two pieces, one of length a and the other of length b. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length b is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is a/b ?

(A) 1

(B) \(\frac{\sqrt{6}}{2}\)

(C) \(\sqrt{3}\)

(D) 2

(E) \(\frac{3\sqrt{2}}{2}\)


Since a and b forms equilateral triangle and regular hexagon respectively, therefore a and b are the perimeters.

So side of triangle = \(\frac {a}{3}\)
and
side of hexagon = \(\frac {b}{6}\)

Area of equilateral triangle = \(\frac {\sqrt{3}}{4}*(side)^2 = \frac {\sqrt{3}}{4}*(\frac {a}{3})^2 = \frac {\sqrt{3}}{4}*\frac {a^2}{9}\)
and

Area of regular hexagon = \(\frac {3\sqrt{3}}{2}*(side)^2 = \frac {3\sqrt{3}}{2}*(\frac {b}{6})^2 = \frac {3\sqrt{3}}{2}*\frac {b^2}{36}\)

Since both the area are equal, equating both the equations we get,

\(\frac {a}{b} = \frac {\sqrt{6}}{2}\)

Hence the answer is B
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Re: A wire is cut into two pieces, one of length a and the other of length  [#permalink]

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New post 20 Mar 2019, 07:23
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Bunuel wrote:
A wire is cut into two pieces, one of length a and the other of length b. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length b is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is a/b ?

(A) 1

(B) \(\frac{\sqrt{6}}{2}\)

(C) \(\sqrt{3}\)

(D) 2

(E) \(\frac{3\sqrt{2}}{2}\)



You hav eto know the area of the equilateral triangle and a regular hexagon

(I) Equilateral triangle..
Perimeter is a, so side = \(\frac{a}{3}\), and area = \(\frac{\sqrt{3}*side^2}{4}\)= \(\frac{\sqrt{3}*(a/3)^2}{4}\)

(II) Regular Hexagon..
Perimeter is b, so side = \(\frac{b}{r}\), so the regular hexagon consists of 6* equilateral triangle of side b/6,
and area = 6*\(\frac{\sqrt{3}*side^2}{4}\)= 6* \(\frac{\sqrt{3}*(b/6)^2}{4}\)

Areas are equal so \(\frac{\sqrt{3}*(a/3)^2}{4}\) = 6* \(\frac{\sqrt{3}*(b/6)^2}{4}\)....... \({a^2/9}\) = 6* \({b^2/36}\)..
\(\frac{a}{b}=\frac{\sqrt{6}}{2}\)

B
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A wire is cut into two pieces, one of length a and the other of length  [#permalink]

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New post 20 Mar 2019, 07:32
Bunuel wrote:
A wire is cut into two pieces, one of length a and the other of length b. The piece of length $a$ is bent to form an equilateral triangle, and the piece of length b is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is a/b ?

(A) 1

(B) \(\frac{\sqrt{6}}{2}\)

(C) \(\sqrt{3}\)

(D) 2

(E) \(\frac{3\sqrt{2}}{2}\)


area of equilateral trgl ; √3 * s2/ 4
hexagon has 6 equilaterla triangles = 6 * √3 * s2/ 4 ; 3√3*s2/2
so ratio
side of triangle a/3 and side of hexagon b/6
√3 * s2/ 4/ 3√3*s2/2
\(\frac{\sqrt{6}}{2}\)
IMO B
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A wire is cut into two pieces, one of length a and the other of length   [#permalink] 20 Mar 2019, 07:32
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