reto wrote:

Attachment:

T7908.png

Arcs AC, AB and BC are centered at B, C and A respectively, as shown above. If the area of equilateral triangle ABC is √3, what is the area of the shaded region?

A. \(2pi-3\sqrt{3}\)

B. \(\sqrt{3} - pi\)

C. 3pi

D. \(\sqrt{3}+2pi\)

E. 3

1) Choose one vertex and its segment, say B

2) Area of shaded region for B:

(Area of segment) - (Area of triangle)

3) Segment area - find SIDE LENGTH of equilateral triangle, which is the RADIUS of Circle B:

Find side length

Area of equilateral triangle, given: \(\sqrt{3}\)

Formula for that area:\(\frac{s^2\sqrt{3}}{4}\)

\(\frac{s^2\sqrt{3}}{4}\)=\(\sqrt{3}\)

\(s^2\sqrt{3}=4\sqrt{3}\)

\(s^2 =\frac{4\sqrt{3}}{\sqrt{3}}\)

\(s^2 = 4\)

\(s = 2\)

So radius of Circle B = 2

3) SEGMENT AREA: Find Circle B's area. What fraction does segment area equal?

Circle area: \(\pi*r^2 = 4\pi\)

Segment area as fraction? An equilateral triangle has 3 vertices with angle measures of 60°

Fraction:\(\frac{60}{360}=\frac{1}{6}\) of circle's area

Segment area: \((\frac{1}{6})(4\pi)=\frac{2}{3}\pi\)

4) Shaded area * 3

Areas: Shaded = (Segment) - (triangle)

One shaded area\(:\frac{2}{3}\pi -\sqrt{3}\)

There are three such areas, so total shaded area is

\(3(\frac{2}{3}\pi-\sqrt{3})=2\pi-3\sqrt{3}\)

Answer A

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