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:
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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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