Suppose we have to cover the XY-plane with id | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) regular polygon with $n$ sides can tile the plane if and only if its interior angle is a divisor of $360^{\circ}$.The interior angle of a regular

Vedclass · Accepted Answer

exactly three of the four shapes

Vedclass · Answer

(B) regular polygon with $n$ sides can tile the plane if and only if its interior angle is a divisor of $360^{\circ}$.The interior angle of a regular $n$-gon is given by $\frac{(n-2) 	imes 180^{\circ}}{n}$.$(1)$ For an equilateral triangle $(n=3)$: Interior angle = $\frac{(3-2) 	imes 180^{\circ}}{3} = 60^{\circ}$. Since $360^{\circ} / 60^{\circ} = 6$,it can tile the plane.$(2)$ For a square $(n=4)$: Interior angle = $\frac{(4-2) 	imes 180^{\circ}}{4} = 90^{\circ}$. Since $360^{\circ} / 90^{\circ} = 4$,it can tile the plane.$(3)$ For a regular pentagon $(n=5)$: Interior angle = $\frac{(5-2) 	imes 180^{\circ}}{5} = 108^{\circ}$. Since $360^{\circ} / 108^{\circ} = 3.33$ (not an integer),it cannot tile the plane.$(4)$ For a regular hexagon $(n=6)$: Interior angle = $\frac{(6-2) 	imes 180^{\circ}}{6} = 120^{\circ}$. Since $360^{\circ} / 120^{\circ} = 3$,it can tile the plane.Thus,the shapes that can tile the plane are the equilateral triangle,the square,and the regular hexagon. There are exactly three such shapes.

Suppose we have to cover the $XY$-plane with identical tiles such that no two tiles overlap and no gap is left between the tiles. Suppose that we can choose tiles of the following shapes: equilateral triangle,square,regular pentagon,regular hexagon. Then,the tiling can be done with tiles of

Similar Questions

If a point $P$ on the line $3x + 5y = 15$ is equidistant from the coordinate axes,then $P$ lies

Let $A(2, -3)$ and $B(-2, 1)$ be vertices of a triangle $ABC$. If the centroid of this triangle moves on the line $2x + 3y = 1$,then the locus of the vertex $C$ is the line

The point whose abscissa is equal to its ordinate and which is equidistant from the points $(1, 0)$ and $(0, 3)$ is

Let the algebraic sum of the perpendicular distances from the points $A(2, 0)$,$B(0, 2)$,and $C(1, 1)$ to a variable line be zero. Then,all such lines:

The equation of the locus of the point of intersection of the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module