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A piece of wire is bent so as to form the boundary of a squa
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06 Apr 2020, 05:56
06 Apr 2020, 05:56
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A piece of wire is bent so as to form the boundary of a square with area A. If the wire is then bent into the shape of an equilateral triangle, what will be the area of the triangle thus bounded in terms of A?
A. \(\frac{\sqrt{3}}{64}A^2\)
B. \(\frac{\sqrt{3}}{4}A\)
C. \(\frac{9}{16}A\)
D. \(\frac{3}{4}A\)
E. \(\frac{4\sqrt{3}}{9}A\)
Kudos for a correct solution.
A. \(\frac{\sqrt{3}}{64}A^2\)
B. \(\frac{\sqrt{3}}{4}A\)
C. \(\frac{9}{16}A\)
D. \(\frac{3}{4}A\)
E. \(\frac{4\sqrt{3}}{9}A\)
Kudos for a correct solution.
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Joined: 10 Apr 2015
Posts: 6218
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Re: A piece of wire is bent so as to form the boundary of a squa
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06 Apr 2020, 05:59
06 Apr 2020, 05:59
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Carcass wrote:
A piece of wire is bent so as to form the boundary of a square with area A. If the wire is then bent into the shape of an equilateral triangle, what will be the area of the triangle thus bounded in terms of A?
A. \(\frac{\sqrt{3}}{64}A^2\)
B. \(\frac{\sqrt{3}}{4}A\)
C. \(\frac{9}{16}A\)
D. \(\frac{3}{4}A\)
E. \(\frac{4\sqrt{3}}{9}A\)
Kudos for a correct solution.
A. \(\frac{\sqrt{3}}{64}A^2\)
B. \(\frac{\sqrt{3}}{4}A\)
C. \(\frac{9}{16}A\)
D. \(\frac{3}{4}A\)
E. \(\frac{4\sqrt{3}}{9}A\)
Kudos for a correct solution.
Let's say 4x = the length of the wire
So when we create a square, each side has length x, which means the area of the square = (x)(x) = x²
This means A = x²
Now take the same length of wire (length = 4x) and create an equilateral triangle
This means each side will have length 4x/3
Nice formula: Area of equilateral triangle = (√3/4)(side length)²
= (√3/4)(4x/3)²
= (√3/4)(16x²/9)
= (4√3/9)x²
= (4√3/9)A
Answer: E
Cheers,
Brent
Re: A piece of wire is bent so as to form the boundary of a squa
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07 Oct 2023, 12:47
07 Oct 2023, 12:47
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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