What kinds of regular polygons can be used for regular tesselations

Mathematics

2 answers:

Andreyy893 years ago

8 0

Answers are four, six and three sided

Ahat [919]3 years ago

5 0

To tessellate a surface using a regular polygon, the interior angle must be a sub-multiple (i.e. factor) of 360 degrees to cover completely the surface.

For a regular three-sided polygon, the interior angle is (180-360/3)=60 °
Since 6*60=360, so a regular three-sided polygon (equilateral triangle) tessellates.
For a regular four-sided polygon, the interior angle is (180-360/4)=90 °
Since 4*90=360, so a regular four-sided polygon (square) tessellates.
For a regular five-sided polygon, the interior angle is (180-360/5)=108 °
Since 360/108=3.33... (not an integer), so a regular five-sided polygon (pentagon) does NOT tessellate.
For a regular six-sided polygon, the interior angle is (180-360/6)=120 °
Since 3*120=360, so a regular six-sided polygon (hexagon) tessellates.

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What is the equation of a line parallel to y=1/2×+6 that passes through (-4,1)

IgorC [24]

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

$y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so we're really looking for the equation of a line whose slope is 1/2 and passes through (-4 , 1)

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{ \cfrac{1}{2}}(x-\stackrel{x_1}{(-4)}) \\\\\\ y-1=\cfrac{1}{2}(x+4)\implies y-1=\cfrac{1}{2}x+2\implies y=\cfrac{1}{2}x+3$