azamaka wrote:

The figure shown is a regular hexagon with center h. the shaded area is a parallelogram that shares three vertices with the hexagon and its fourth vertex is the center of the hexagon. If the length of one side of the hexagon is 8 centimeters, what is the area of the unshaded region?

A) \(16\sqrt{3}cm^2\)

B) \(96 cm^2\)

C) \(64\sqrt{3}cm^2\)

D) \(96\sqrt{3}cm^2\)

E) \(256cm^2\)

Attachment:

hexshade.jpg [ 34.8 KiB | Viewed 742 times ]
This prompt's wording and wordiness could confuse. The figure is better.

Unshaded region area = (hexagon area) - (parallelogram area)

Dividing shapes into triangles is an easy way to find area, especially when the polygons are regular (equal angles, equal sides).

A regular polygon can be divided into congruent isosceles triangles.

A regular hexagon is the only regular polygon that can be divided into congruent

equilateral triangles.

Connect the vertices. There are 6 equilateral triangles.

An equilateral triangle has three 60° angles.

Six sides of a hexagon divide 360° at center into six 60° angles at center

The other two angles of each triangle also = 60°; both bisect a vertex of 120° = 60° each

Given: each side of hexagon = 8 cm

From above: the six triangles are equilateral. Each side of each triangle = 8 cm

Find the area of one of these triangles.

Multiply by 6 for the area of the hexagon.

Multiply by 2 for the area of the parallelogram, which = two of those triangles

Area of equilateral triangle* is a handy thing to know, s = side:

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

\(A =\frac{8^2\sqrt{3}}{4}\)

\(A =16\sqrt{3}\)Area of hexagon (six equilateral triangles):

\(6 * 16\sqrt{3}= 96\sqrt{3}\) sq cm

Area of parallelogram (two equilateral triangles)

\(2 *16\sqrt{3} = 32\sqrt{3}\) sq cm

Area of unshaded region:

(Hexagon area) - (parallelogram area) =

\((96\sqrt{3}- 32\sqrt{3}) = 64\sqrt{3}\) sq cm

Answer C

*If you don't remember the formula for the area of an equilateral triangle, draw one. Drop an altitude, which is a perpendicular bisector of the opposite side and of the vertex.

That altitude creates two congruent right 30-60-90 triangles with side lengths that correspond to 30-60-90, in ratio \(x : x\sqrt{3} : 2x\)

Side lengths? Side opposite the 90° angle = \(2x = 8\). Side opposite 30° angle is half of that, i.e., \(x, x = 4\). Side opposite 60° angle = height of triangle = \(x\sqrt{3}\) or \(4\sqrt{3}\)

Area\(=\frac{b*h}{2} = (8*4\sqrt{3})*\frac{1}{2} = 16\sqrt{3}\) square cm
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