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The corners of an equilateral triangle are rounded to form semi-circle

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The corners of an equilateral triangle are rounded to form semi-circle  [#permalink]

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New post 14 Sep 2018, 02:10
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The corners of an equilateral triangle are rounded to form semi-circles of radius 1. If the length of a side of the triangle before rounding is 10, what is the area of the resulting figure?


A. \(25\sqrt{3} + 3\)

B. \(25\sqrt{3} + \frac{3\pi}{2}\)

C. \(25\sqrt{3}\)

D. \(22\sqrt{3} + 3\)

E. \(22\sqrt{3} + \frac{3\pi}{2}\)

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The corners of an equilateral triangle are rounded to form semi-circle  [#permalink]

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New post Updated on: 28 Sep 2018, 05:57
Bunuel wrote:
The corners of an equilateral triangle are rounded to form semi-circles of radius 1. If the length of a side of the triangle before rounding is 10, what is the area of the resulting figure?


A. \(25\sqrt{3} + 3\)

B. \(25\sqrt{3} + \frac{3\pi}{2}\)

C. \(25\sqrt{3}\)

D. \(22\sqrt{3} + 3\)

E. \(22\sqrt{3} + \frac{3\pi}{2}\)


Area of the resulting figure = Area of Bigger equilateral - (Area cut out)

Area cut out from one corner = (1/3)*(Area of Equilateral triangle with side 2√3 - Area of Circle with radius 1) .

[The inscribed circle with radius 1 will give us the side of equilateral as 2√3 using 30-60-90 property)

Area cut out from one corner \(= (1/3)*[(√3/4)* 2√3)^2 - π1^2) = √3-(π/3)\) .
Total Area cut out \(= 3*[√3-(π/3)] = 3√3 - π\)

Area of the resulting figure = \((√3/4)* 10)^2 - (3√3 - π) = 22√3 +π\)

I think answer must be option E but highlighted part is still a bit confusing. Taking semicircle out from corners will involve different calculations.
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Originally posted by GMATinsight on 14 Sep 2018, 02:30.
Last edited by GMATinsight on 28 Sep 2018, 05:57, edited 1 time in total.
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Re: The corners of an equilateral triangle are rounded to form semi-circle  [#permalink]

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New post 14 Sep 2018, 20:42
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got the answer correct but took 3 mins first find the area of remaining figure = root3/4 10*10 - 3*root3/4*2*2( this is the area of equi triangle formed by making a semi circle and then add area of semi circle to remaining fig (pi*1^2/2 )*3
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Re: The corners of an equilateral triangle are rounded to form semi-circle  [#permalink]

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New post 28 Sep 2018, 04:37
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Any better way?
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The corners of an equilateral triangle are rounded to form semi-circle  [#permalink]

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New post 30 Sep 2018, 21:08
Try to work out the sides of the big triangles and small triangle (pyto theorem) first.

(a) Area of the big equilateral triangle = (10/2) * 5 (3^(1/2)) = 25(3^(1/2))
(b) Area of 1 semi-circle = π (1^2) /2 = π/2
(c) Area of 1 small equilateral triangle= (3^(1/2))
(d) Area of 3 cut-out parts= (c)- (b) = 3*[(3) - (1) ]= 3[(3^(1/2)) - (π/2)] = 3(3^(1/2)) - 3π/2
Answer = (d) - (a) = 25(3^(1/2)) - [3(3^(1/2)) - 3π/2] = 22(3^(1/2))+ (3π/2)------ (E)
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The corners of an equilateral triangle are rounded to form semi-circle &nbs [#permalink] 30 Sep 2018, 21:08
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The corners of an equilateral triangle are rounded to form semi-circle

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