Question 20

How do we find such $A$ and $B$?

From $X$ and $Y$, draw arcs above and below $XY$, with the same radii. The two points at which the arcs meet, above and below $XY$, give us $A$ and $B$, respectively.

Use this to construct an eye.

In Fig. 6.1, join $A$ and $B$ with a line. Where does $AB$ intersect $XY$, and what is the angle formed between them?

Will the line joining the two points at which the arcs meet, above and below $XY$, always be the perpendicular bisector of $XY$, i.e., when $XY$ is of any length, and the arcs are drawn using a radius of any length?

Which two triangles should be congruent for $AB$ to be the perpendicular bisector of $XY$ (that is, $O$ is the midpoint of $XY$ and $AB$ is perpendicular to $XY$)?

How do we get these different shapes? Try!

Will $C$ and $D$ lie on the perpendicular bisector $\text{AB}$?

Justify the following statement using the facts that we have established.

Any point that has the same distance from $X$ and $Y$ lies on the perpendicular bisector of $\text{XY}$.

Given a line segment $XY$, how do we draw its perpendicular bisector using only an unmarked ruler and a compass?

Can we extend the method of constructing the perpendicular bisector to construct a $90^\circ$ angle at any point on a line? Draw a line and mark a point $O$ on it. Construct a $90^\circ$ angle at point $O$.

Find a segment of this line for which $O$ is the midpoint.

How do we construct this figure?

What is the angle between two adjacent lines?

How do we construct a $45^\circ$ angle using only a ruler and a compass?

Construct the following figure.

Draw an angle. Create a copy of this angle using only a ruler and compass.

How do we implement this idea using a ruler and a compass?

How did they make these arches?

Construct this arch shape on a piece of paper.

Let us think about the support lines this figure will need.

For symmetry, we should have $AB = CD$, and $\angle BAD = \angle CDA$. How would you construct these support lines?

Use these support lines to construct an arch. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing.

If their midpoints are marked, will you be able to construct a pointed arch?

How do we construct a regular pentagon (5-sided figure) and a regular hexagon (6-sided figure)? To begin with, try to construct a pentagon and hexagon with equal sidelengths.

Can we break a regular hexagon into smaller pieces that can be constructed?

Can six congruent equilateral triangles be placed together as in Fig. 6.12? If yes, will it result in a regular hexagon?

Consider this figure. Will the $70^\circ$ angle fit into the gap? What is the gap angle $\angle AOI$?

We have, $40^\circ + 60^\circ + 50^\circ + 30^\circ + 40^\circ + 90^\circ + \text{gap angle} = 360^\circ$

Use this to determine whether the $70^\circ$ angle fits the gap.

In Fig. 6.12 can you explain why $AOD$, $BOE$ and $COF$ are straight lines?

Construct a regular hexagon with a sidelength $4\text{ cm}$ using a ruler and a compass.

Context: We can construct a regular hexagon more directly if we can construct a $120^\circ$ angle using a ruler and a compass.

Q. How do we do it?

Why is $\angle \text{CAX} = 60^\circ$? Is there an equilateral triangle here?

Construct a regular hexagon of sidelength $5\text{ cm}$.

How will you construct $30^\circ$ and $15^\circ$ angles?

Construct the following 6-pointed star. Note that it has a rotational symmetry.

Are the six triangles forming the 6 points of the star — $\Delta \text{AGH}$, $\Delta \text{BHI}$, $\Delta \text{CIJ}$, $\Delta \text{DJK}$, $\Delta \text{ELK}$, $\Delta \text{FLG}$ — equilateral? Why?

[Hint: Find the angles.]

Can a $4 \times 6$ grid be tiled using multiple copies of $2 \times 1$ tiles? We are allowed to rotate a $2 \times 1$ tile and use it.

Can a $4 \times 7$ grid be tiled using $2 \times 1$ tiles?

What about a $5 \times 7$ grid?

Complete the justification.

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

Is an $m \times n$ grid tileable with $2 \times 1$ 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.

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

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

Is the following region tileable with $2 \times 1$ tiles?

Context: We are considering whether a given region can be tiled using $2 \times 1$ tiles.

Q. 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 $5 \times 3$ grid, makes it non-tileable?

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?

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