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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