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# Is the area of equilateral triangle A greater than the area of square?

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Intern
Joined: 06 Oct 2012
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Is the area of equilateral triangle A greater than the area of square?  [#permalink]

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29 Jan 2016, 12:22
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Is the area of equilateral triangle A greater than the area of square B ?

(1) The product of the length of a side of the square and the length of a side of the triangle is 120.

(2) The ratio of the length of a side of the square to the length of a side of the triangle is 5 to 6.
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Joined: 05 Mar 2015
Posts: 980
Re: Is the area of equilateral triangle A greater than the area of square?  [#permalink]

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29 Jul 2016, 20:11
dheerajkaushik90 wrote:
Is the area of equilateral triangle A greater than the area of square B ?

(1) The product of the length of a side of the square and the length of a side of the triangle is 120.

(2) The ratio of the length of a side of the square to the length of a side of the triangle is 5 to 6.

Pls explain the solution.

(1) let the side of sq. be 2 and side of triangle be 60
then triangle>square
let the side of sq. be 60 and side of triangle be 2
then square>triangle.
Not suff...
(2) Side of square=5/6*Side of triangle------>0.8(side of triangle)
area of triangle=sq.root3/4*((side of triangle)^2----->0.4*((side of triangle)^2
and area of square= (0.8(side of triangle))^2=0.64*((side of triangle)^2
then square area> triangle area
Suff...

Ans B
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Joined: 02 Jul 2017
Posts: 294
GMAT 1: 730 Q50 V38
Re: Is the area of equilateral triangle A greater than the area of square?  [#permalink]

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14 Sep 2017, 02:39
Find : if area of equilateral triangle A > the area of square B
Let side of equilateral triangle A = a
let side of square B = b

Area of square = $$b^2$$
Area of triangle = $$\frac{\sqrt[]3}{4} a^2$$

(1) The product of the length of a side of the square and the length of a side of the triangle is 120.

=> a*b=120
As we don't know value of a and b , either b > a or a >b.

let a= 4 => b= 30
Area of square = $$b^2 = 900$$
Area of triangle = $$\frac{\sqrt[]3}{4} a^2 =\frac{\sqrt[]3}{4} * 16 = 4\sqrt[]3$$
=> Area of square > area of triangle

let a= 40 => b= 3
Area of square = $$3^2 = 9$$
Area of triangle =$$\frac{\sqrt[]3}{4} a^2 = \frac{\sqrt[]3}{4} * 1600 = 400\sqrt[]3$$
=> Area of square < area of triangle
Not sufficient

(2) The ratio of the length of a side of the square to the length of a side of the triangle is 5 to 6.
=>$$\frac{b}{a}=\frac{5}{6}$$
=>$$b= \frac{5}{6} a$$
=>$$b^2 = \frac{25}{36} a^2$$

Area of triangle = $$\frac{\sqrt[]3}{4} a^2 = \frac{\sqrt[]3}{4} * \frac{36}{25} b^2 = \frac{9\sqrt[]3}{25} b^2 = \frac{9\sqrt[]3}{25} * Area of Square$$
even if we consider$$\sqrt[]3 = 2$$
Area of triangle > Area of Square
Sufficient

Re: Is the area of equilateral triangle A greater than the area of square? &nbs [#permalink] 14 Sep 2017, 02:39
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