NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 12 Quadrilaterals Exercise 12.4

Exercise Set 12.4

Question 1. Justify why the plane cannot be tiled with a regular pentagon. (Hint: Read the first 3 sentences of 'Think and Reflect' in the section on tiling.) (There are many ways to tile the plane using a suitable irregular pentagon. The most recent method was found in 2015.)
Answer:
Angle of a regular pentagon. The angles of a pentagon add to (5 − 2) × 180° = 540° (End-of-Chapter Exercise 21), so each angle of a regular pentagon is \( \frac{540°}{5} \) = 108°.
Condition at a point. In a tiling, the angles around any point add to 360°. At a corner of a pentagon, either only corners meet there, or the point also lies on a side of another pentagon, which then contributes a straight angle of 180°.
Case 1: only corners. We would need 108° × k = 360°, i.e. k = \( \frac{360}{108} \) = \( \frac{10}{3} \), not a whole number. Three corners give 324° (a 36° gap) and four give 432° (overlap).
Case 2: a point on a side. We would need 180° + 108° × k = 360°, i.e. 108° × k = 180°, k = \( \frac{5}{3} \), again not a whole number.

In neither case do the angles make exactly 360°, so regular pentagons always leave gaps or overlap. Hence the plane cannot be tiled with regular pentagons.
(The 2015 irregular pentagon in the textbook has angles 60°, 135°, 105°, 90°, 150°. These can be combined to 360°, e.g. 150° + 105° + 105°, 90° + 135° + 135°, 60° + 150° + 150°, and its side lengths are chosen so that edges match; that is why it tiles.)

Teacher's Note:
1. The whole argument is: 360 is not a whole-number multiple of 108.
2. Do not forget the case where a corner touches the middle of another pentagon's side.

Question 2. Draw a non-convex 4-gon DART. Show how we can tile the plane with copies of DART. Both methods that we discussed earlier will work. Which do you prefer?

Answer:
Method 1 (half-turns). Turn a copy of DART through 180° about the midpoint of one of its sides; the copy fits exactly along that side. Repeat at every free side of every copy.
At each vertex the four angles of DART, including the reflex angle at D, meet once each and add up to 360°. (The reflex angle at D is more than 180°; the other three angles together fill the rest.) So the copies fit around every vertex.
Method 2 (Varignon grid). The midpoints of the sides of DART form a parallelogram (Theorem 9 holds for non-convex 4-gons too; see End-of-Chapter Exercise 13). Draw the grid of copies of this parallelogram and shade it like a chessboard. Place a copy of DART on every shaded cell so that its Varignon parallelogram is that cell (as in Exercise Set 12.3, Q5). The spaces left over are again copies of DART, half-turned, which are then filled in.

Preference: Method 1 is easier by hand: every step is only a half-turn about the midpoint of a side, and the edges match automatically. Method 2 needs the whole grid first, but shows clearly why the pattern repeats. Either choice is fine with a reason.