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

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Math Expert
Joined: 02 Sep 2009
Posts: 59725
A wire is cut into two pieces, one of length a and the other of length  [#permalink]

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20 Mar 2019, 04:53
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Difficulty:

35% (medium)

Question Stats:

75% (02:49) correct 25% (02:18) wrong based on 8 sessions

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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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20 Mar 2019, 07:17
1
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
Math Expert
Joined: 02 Aug 2009
Posts: 8325
Re: A wire is cut into two pieces, one of length a and the other of length  [#permalink]

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20 Mar 2019, 07:23
1
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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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
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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# A wire is cut into two pieces, one of length a and the other of length

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