Question 1

Can a 4 × 7 grid be tiled using 2 × 1 tiles?
What about a 5 × 7 grid?

Accepted Answer

A 4 × 7 grid has 28 unit squares. Since each 2 × 1 tile covers 2 squares, and 28 is even, it is possible to tile the 4 × 7 grid completely with 2 × 1 tiles. For example, by placing vertical tiles in each column.

A 5 × 7 grid has 35 unit squares, which is odd. Since each tile covers 2 squares, it is impossible to cover an odd number of squares completely without gaps or overlaps. Hence, a 5 × 7 grid cannot be tiled using 2 × 1 tiles. — Since each tile covers 2 squares, the total number of squares must be even for tiling to be possible. 4 × 7 = 28 (even), so tiling is possible. 5 × 7 = 35 (odd), so tiling is impossible.

Question 2

Complete the justification.
Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even? If yes, come up with a general strategy to tile it.

Accepted Answer

If both m and n are even, then the total number of unit squares m × n is even. One general strategy is to cover each column with vertical 2 × 1 tiles. Since the number of rows m is even, each column can be fully covered by vertical tiles without gaps or overlaps.

Thus, the entire m × n grid can be tiled by placing vertical tiles column-wise. — Because m is even, each column has an even number of squares, allowing vertical tiles to cover the column completely. Repeating this for all columns covers the entire grid.

Question 3

Is an m × n grid tileable with 2 × 1 tiles, if one of m and n is even and the other is odd? If yes, come up with a general strategy to tile it.

Accepted Answer

If one of m or n is even and the other is odd, then the total number of squares m × n is even (since even × odd = even). Therefore, tiling with 2 × 1 tiles is possible.

A general strategy is to cover the grid by placing tiles along the dimension which is even. For example, if m is even and n is odd, place vertical tiles column-wise; if n is even and m is odd, place horizontal tiles row-wise. — Because the total number of squares is even, and the dimension with even length allows complete coverage by tiles oriented along that dimension, the entire grid can be tiled.

Question 4

Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd? Give reasons.

Accepted Answer

If both m and n are odd, then m × n is odd (odd × odd = odd). Since each 2 × 1 tile covers 2 squares, it is impossible to cover an odd number of squares completely without gaps or overlaps.

Therefore, an m × n grid with both dimensions odd cannot be tiled using 2 × 1 tiles. — Tiling requires the total number of squares to be even because each tile covers 2 squares. An odd total number of squares makes tiling impossible.

Question 5

Here is a 5 × 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2 × 1 tiles?

Accepted Answer

The 5 × 3 grid has 15 squares. Removing one square leaves 14 squares, which is even. However, tiling depends not only on the number of squares but also on the arrangement.

By coloring the grid in a checkerboard pattern (black and white squares alternating), each 2 × 1 tile covers one black and one white square. The original 5 × 3 grid has 8 black and 7 white squares. Removing one square changes the count.

If the removed square is black, the grid has 7 black and 7 white squares, making tiling possible. If the removed square is white, the counts are unequal, making tiling impossible.

Therefore, depending on which square is removed, the grid may or may not be tileable. — Tiling with 2 × 1 tiles requires equal numbers of black and white squares in the checkerboard coloring. Removing a square changes this balance.

Question 6

Is the following region tileable with 2 × 1 tiles?

Accepted Answer

To determine tileability, color the region in a checkerboard pattern. Each 2 × 1 tile covers one black and one white square.

Count the number of black and white squares. If they are equal, tiling is possible; if not, tiling is impossible.

For the given region (Fig. 6.13), the numbers of black and white squares are unequal, so it is not tileable with 2 × 1 tiles. — Checkerboard coloring helps identify tileability by ensuring each tile covers one black and one white square. Unequal counts mean tiling is impossible.

Question 7

What about this one? Were you able to tile this? How can we be sure that this is not tileable? Can you find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?

Accepted Answer

The given region (Fig. 6.13) is not tileable because the number of black and white squares in its checkerboard coloring is unequal.

By removing certain unit squares from a 5 × 3 grid, the balance between black and white squares is disturbed, making tiling impossible.

Another unit square that can be removed to make the 5 × 3 grid non-tileable is any square whose removal causes unequal numbers of black and white squares. For example, removing a white square when black and white counts were equal before removal. — Checkerboard coloring and counting black and white squares provide a method to prove non-tileability. Unequal counts imply no perfect tiling with 2 × 1 tiles.

Question 8

If the plain grid is tileable, is the black-and-white-grid tileable?
If the black-and-white grid is tileable, is the plain grid tileable?

Accepted Answer

Yes, if the plain grid is tileable, then the black-and-white grid is also tileable because the tiles cover one black and one white square each.

Similarly, if the black-and-white grid is tileable, then the plain grid is tileable because the coloring does not affect the ability to tile the region with 2 × 1 tiles.

Thus, tileability of the plain grid and the black-and-white grid are equivalent. — The coloring is a tool to analyze tileability. Since each tile covers one black and one white square, tileability in one corresponds to tileability in the other.

Question 9

Is the black-and-white region in Fig. 6.14 tileable?

Accepted Answer

No, the black-and-white region in Fig. 6.14 is not tileable with 2 × 1 black-and-white tiles because it has 8 white squares and 6 black squares.

Since each tile covers one black and one white square, the number of black and white squares must be equal for tiling to be possible.

Here, the counts are unequal, so tiling is impossible. — Tileability requires equal numbers of black and white squares in the checkerboard coloring. Unequal counts prevent perfect tiling.

Question 10

Use this idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?

Accepted Answer

By coloring the 5 × 3 grid in a checkerboard pattern, removing a square that causes the number of black and white squares to become unequal will make the grid non-tileable.

For example, removing a white square when the counts were equal before removal will create an imbalance, making tiling impossible.

Thus, any unit square whose removal causes unequal black and white counts will make the grid non-tileable. — Checkerboard coloring helps identify which squares are critical for maintaining tileability. Removing a square that breaks the balance prevents tiling.

Question 11

Are the following tilings possible?
1.
2.

Accepted Answer

1. To determine if the first tiling is possible, check if the region can be covered completely by 2 × 1 tiles without gaps or overlaps. Use checkerboard coloring to verify if the number of black and white squares is equal.

2. Similarly, for the second tiling, use the same method.

If the counts are equal and the arrangement allows placing tiles, the tiling is possible; otherwise, it is not. — Checkerboard coloring and counting black and white squares is a standard method to check tileability with 2 × 1 tiles.

Question 12

Can you think of a shape whose copies can tile the entire plane?

Accepted Answer

Squares can tile the entire plane by placing copies side by side without gaps or overlaps.

Other shapes that can tile the plane include equilateral triangles and regular hexagons.

Thus, shapes like squares, equilateral triangles, and regular hexagons can tile the entire plane. — These shapes fit together perfectly due to their angles and side lengths, allowing infinite repetition to cover the plane.

Question 13

What are the two basic tools used in geometric constructions, and what are their primary functions?

Accepted Answer

Compass for drawing arcs, ruler for drawing straight lines without measurement markings — The two basic tools used in geometric constructions are the compass and the ruler (straightedge). The compass is used to draw arcs and circles, while the ruler is used to draw straight lines but without any measurement markings to ensure constructions are done geometrically, not by measurement.

Question 14

Why is it important to use only a compass and an unmarked ruler for geometric constructions instead of measuring tools like a protractor or a marked ruler?

Accepted Answer

Geometric constructions rely on the exactness of shapes and angles created through logical steps rather than measurements. Using only a compass and an unmarked ruler ensures that the constructions are precise and based on geometric principles, not on approximations from measurements. This method allows for creating figures that satisfy specific conditions exactly, such as perpendicular bisectors and angle bisectors. — Geometric constructions are designed to produce exact figures by following specific steps using only a compass and an unmarked ruler. Measuring tools like protractors or marked rulers introduce approximations and errors. The compass and unmarked ruler ensure that the properties of the figures, like equal lengths or right angles, are maintained through congruence and geometric reasoning rather than measurement.

Question 15

Describe the step-by-step procedure to construct the perpendicular bisector of a given line segment AB using a compass and an unmarked ruler.

Accepted Answer

To construct the perpendicular bisector of line segment AB: 
1. Place the compass pointer at point A and draw arcs above and below the line segment with a radius more than half of AB.
2. Without changing the compass width, draw similar arcs from point B, intersecting the previous arcs above and below AB.
3. Mark the points of intersection of the arcs as points C and D.
4. Using the ruler, draw a straight line through points C and D. This line is the perpendicular bisector of AB, dividing it into two equal parts at a right angle. — The construction works because points C and D are equidistant from A and B. Thus, the line joining C and D is perpendicular to AB and bisects it. This method does not require measuring the length or angle but uses the properties of congruent triangles formed by the arcs.

Constructions

Constructions — Study Notes

Introduction to Geometric Constructions

Constructing a Perpendicular Bisector of a Line Segment

Constructing a Perpendicular to a Line from a Point on the Line

Practice Questions — Constructions

All 7 Chapters in Ganita Prakash-II