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TAGS: GRE Instructor Joined: 10 Apr 2015
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If ABCD is a square, and XYZ is an equilateral triangle, [#permalink]
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Question Stats: 83% (01:33) correct 16% (02:22) wrong based on 6 sessions If ABCD is a square, and XYZ is an equilateral triangle, then the area of the square is how many times the area of the triangle?

A) (4√3)/3
B) (8√3)/3
C) 2√6
D) (16√3)/9
E) (16√2)/3

*I'll post a solution in 2 days
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Brent Hanneson – Creator of greenlighttestprep.com Sign up for our free GRE Question of the Day emails Director Joined: 20 Apr 2016
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Re: If ABCD is a square, and XYZ is an equilateral triangle, [#permalink]
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GreenlightTestPrep wrote: If ABCD is a square, and XYZ is an equilateral triangle, then the area of the square is how many times the area of the triangle?

A) (4√3)/3
B) (8√3)/3
C) 2√6
D) (16√3)/9
E) (16√2)/3

*I'll post a solution in 2 days

Here,
Let the side of the equilateral △ = a

and radius of the circle be = r

we know,
area of equilateral △ = $$\frac{\sqrt3}{4} * side^2$$

or area of equilateral △ = $$\frac{\sqrt3}{4} * a^2$$

If equilateral △ is inscribed inside a circle then the side of the equilateral △ =$$\sqrt3 * radius of the circle$$

i.e. $$a = \sqrt3 * r$$(substituting the values)

or $$r = \frac{a}{\sqrt3}$$

Now the side of the square = diameter of the circle = $$2 * radius = 2r = 2 * \frac{a}{\sqrt3}$$

Area of the square = $$(\frac{2a}{\sqrt3})^2$$

Therefore,

$$\frac{Area of the square}{Area of the triangle}= (\frac{2a}{\sqrt3})^2 * \frac{4}{(\sqrt3 * a^2)} = \frac{16}{(3\sqrt3)} = \frac{(16\sqrt3)}{9}$$

i.e. Area of the square = $$\frac{(16\sqrt3)}{9}$$ times the area of the triangle
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GRE Instructor Joined: 10 Apr 2015
Posts: 1546
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Re: If ABCD is a square, and XYZ is an equilateral triangle, [#permalink]
Expert's post
GreenlightTestPrep wrote: If ABCD is a square, and XYZ is an equilateral triangle, then the area of the square is how many times the area of the triangle?

A) (4√3)/3
B) (8√3)/3
C) 2√6
D) (16√3)/9
E) (16√2)/3

If ∆XYZ is EQUILATERAL, then each angle is 60°
So, if we draw a line from the center to a vertex, we'll get two 30° angles.... Now drop a line down like this to create a SPECIAL 30-60-90 right triangle Since the base 30-60-90 right triangle has lengths 1, 2 and √3, let's give the triangle these same measurements... IMPORTANT: This means the circle's radius = 2, which means the circle's DIAMETER = 4
Notice that the circle's diameter = the length of one side of the square
So, each side of the square has length 4, which means the area of the square = (4)(4) = 16

Okay, now let's determine the area of the triangle

Since we know that one side of the special 30-60-90 right triangle has length √3... .... we know that the length of one side of the equilateral triangle = 2√3

This allows us to apply a special area formula for equilateral triangles:
Area of equilateral triangle = (√3)(side²)/4

So, the area of ∆XYZ = (√3)(2√3)²/4
= (√3)(12)/4
= 3√3

The area of the square is how many times the area of the triangle?
Check the answer choices...not there!

Multiply top and bottom by √3 to get: (16√3)/9

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com Sign up for our free GRE Question of the Day emails Re: If ABCD is a square, and XYZ is an equilateral triangle,   [#permalink] 10 Jul 2018, 04:41
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