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2 years ago

Which of the following is the number if sides a polygon can have to form a regular tessallation?

Mathematics

1 answer:

Arisa [49]2 years ago

6 0

None of the options is true.

Three Different, Regular tessellation is formed 1. if we join regular polygons of side 3, called equilateral triangle,2. Regular polygon of side 4, called Square, and 3. Regular polygon of side,6 called hexagon,

to form a pattern.

<h3>What is the polygon?</h3>

A polygon is said to be regular if all interior angles and all sides are equal.

The regular polygon of side 3, when joined together can be named: 3.3.3.

The regular polygon of side 4, when joined together can be named 4.4.4.

The regular polygon of side 6, when joined together can be named 6.6.6.

So, a regular tessellation is formed, from a regular polygon having,3,4, and 6 Sides.

None of the sides of the polygon, which is given in the option, can be used to form a regular tessellation.

None of the options is true.

To learn more about the polygon visit:

brainly.com/question/1592456

#SPJ1

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Read 2 more answers

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$