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20 Mar 2019, 04:53
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
65% (03:05) correct
35%
(02:23)
wrong
based on 31
sessions
History
Date
Time
Result
Not Attempted Yet
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) \(\frac{3\sqrt{2}}{2}\)
(A) 1
(B) \(\frac{\sqrt{6}}{2}\)
(C) \(\sqrt{3}\)
(D) 2
(E) \(\frac{3\sqrt{2}}{2}\)
20 Mar 2019, 07:17
Kudos
Bookmarks
Bunuel
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
20 Mar 2019, 07:23
Kudos
Bookmarks
Bunuel
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
