NCERT Solutions · Class 9 Mathematics Quadrilaterals

Each angle of a regular pentagon is \[\frac{(5-2)\times 180^\circ}{5} = 108^\circ \] At a vertex of a tiling the angles add to \(\displaystyle 360^\circ\). If \(\displaystyle k\) pentagons meet there, \[108^\circ \times k = 360^\circ \;\Rightarrow\; k = \frac{10}{3} \] which is not a whole number. \[3 \times 108^\circ = 324^\circ \ (\text{gap } 36^\circ), \qquad 4 \times 108^\circ = 432^\circ > 360^\circ \ (\text{overlap}) \] A vertex on the side of another pentagon needs \[108^\circ \times k = 180^\circ \;\Rightarrow\; k = \frac{5}{3} \] which also fails.Answer: No whole number of \(\displaystyle 108^\circ\) angles adds to \(\displaystyle 360^\circ\) (or \(\displaystyle 180^\circ\)), so regular pentagons cannot tile the plane.

Take a DART with its dent at \(\displaystyle D\): \[D(2,2),\quad A(0,0),\quad R(0,6),\quad T(4,2) \] \[\angle D+\angle A+\angle R+\angle T = 225^\circ+45^\circ+45^\circ+45^\circ = 360^\circ \] Midpoints of \(\displaystyle DA, AR, RT, TD\): \[M_1(1,1),\quad M_2(0,3),\quad M_3(2,4),\quad M_4(3,2) \] Method 1. Rotate DART by \(\displaystyle 180^\circ\) about the midpoint of a side; the copy shares that side. About \(\displaystyle M_1\): \[P\mapsto 2M_1-P:\quad D\mapsto A,\ \ A\mapsto D,\ \ R\mapsto(2,-4),\ \ T\mapsto(-2,0) \] Repeat at every side of every copy.Method 2. Midpoint Theorem in \(\displaystyle \triangle DAR\) and \(\displaystyle \triangle ART\): \[M_1M_2\parallel DR,\ \ M_1M_2=\tfrac12 DR;\qquad M_2M_3\parallel AT,\ \ M_2M_3=\tfrac12 AT \] So \(\displaystyle M_1M_2M_3M_4\) is the Varignon parallelogram. Draw its grid; on every second cell of every second row fit a same-way-up DART. Each gap is a half-turned DART.Answer: DART \(\displaystyle D(2,2),\,A(0,0),\,R(0,6),\,T(4,2)\) tiles the plane by either method; I prefer Method $\displaystyle 1$ (one rule, no grid), but either is valid.