Constructions And Tilings

Hello, my dear students! Welcome to today's mathematics lesson. I am so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 6, which is all about Constructions and Tilings. This is a wonderful chapter that connects the beauty of geometry with practical applications. Are you ready to explore? Let us begin!

So students, let us start by talking about something very familiar to all of you. Do you remember the 'Eyes' construction that we did in Grade 6? We drew eyes, and we wanted the upper arc and lower arc of each eye to look symmetrical and beautiful. Of course, we could have drawn them freehand, but we wanted to construct them properly so that they look perfect.

We relied on our spatial estimation to determine two centers, let us call them point A and point B, from which we drew the lower arc and upper arc respectively. These arcs define a line, let us call it line XY, that supports the drawing, though it is not part of the final figure. Now, we can start with this supporting line and systematically find the centers A and B. For the eye to be symmetrical, or for the supporting line to be the line of symmetry, the upper and lower arcs should have the same radius. In other words, we must have AX equal to BX.

Now, since AX equals AY and BX equals BY, this actually means that AX equals AY equals BX equals BY. All these four segments are equal to each other. This is a very important observation, students. So, how do we find such points A and B? Let me explain.

From points X and Y, we draw arcs above and below the line XY, using the same radii. The two points at which the arcs meet, above and below XY, give us A and B respectively. We can use this method to construct an eye. Now, let us look at what happens when we join A and B with a line. Where does AB intersect XY, and what is the angle formed between them? We observe that AB passes through the midpoint of XY, and is also perpendicular to it. This is a very important geometric property, students.

A division of a line, or any geometrical object, into two identical parts is called bisection. A line that bisects a given line and is perpendicular to it is called the perpendicular bisector. So, AB is the perpendicular bisector of XY. Now, here is an interesting question. Will the line joining the two points at which the arcs meet, above and below XY, always be the perpendicular bisector of XY? That is, when XY is of any length, and the arcs are drawn using a radius of any length? Let us answer this question through congruence.

Let us consider a line segment XY. We want to find points A and B such that AX equals AY equals BX equals BY. We draw the lines AB, AX, AY, BX, and BY. Let O be the point of intersection between AB and XY. Now, which two triangles should be congruent for AB to be the perpendicular bisector of XY? That is, we need O to be the midpoint of XY and AB to be perpendicular to XY.

If we can show that triangle AOX is congruent to triangle AOY, then OX equals OY, and angle AOX equals angle AOY, because they are corresponding parts of congruent triangles. Since angle AOX and angle AOY together form a straight angle, we have angle AOX plus angle AOY equals 180 degrees. Thus, angle AOX equals angle AOY equals 90 degrees. This establishes that O is the midpoint of XY and AB is perpendicular to XY.

Now, in triangles AOX and AOY, we already know that AX equals AY, and AO is common to both triangles. If we can show that angle XAO equals angle YAO, then by the SAS congruence condition, we can conclude that the triangles are congruent. To show this, we observe that triangle ABX is congruent to triangle ABY. This is so because AX equals AY, BX equals BY, and AB is common to both the triangles. Thus, we have angle XAB equals angle YAB, or angle XAO equals angle YAO, because they are corresponding parts of congruent triangles. Hence, AB is the perpendicular bisector of XY.

Now, students, let us think about eyes of different shapes. How do we get these different shapes? One way is to choose two other points, let us call them C and D, such that CX equals CY equals DX equals DY. An eye of a different shape can be drawn using these points. Now, will C and D lie on the perpendicular bisector AB? The points C and D are at the same distance from both X and Y. We have just seen that joining any two such points gives the perpendicular bisector of XY. Since XY has only one perpendicular bisector, which is the line AB, the points C and D must lie on the line AB.

Let us justify this 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 XY. This is a very important property. So, eyes of different shapes can be drawn by suitably choosing different pairs of points on the perpendicular bisector as centers to construct the upper and lower arcs of the eyes.

Now, students, let us learn how to construct a perpendicular bisector using only an unmarked ruler and a compass. Given a line segment XY, how do we draw its perpendicular bisector? We have seen that joining any two points, one above XY and one below, that are at equal distances from X and Y, gives the perpendicular bisector of XY. This gives us a method to construct the perpendicular bisector.

Here are the steps. First, taking some fixed radius, from X and then from Y, we construct two sufficiently long arcs above XY. Name the point where the arcs meet as A. Second, using the same radius, from X and then from Y, we construct two sufficiently long arcs below XY. Name the point where the arcs meet as B. Third, AB is the required perpendicular bisector. Thus, the perpendicular bisector can be constructed using the simplest geometric tools, an unmarked ruler and a compass. We will use only these two tools for all the other geometric constructions in this chapter, unless there is a need for drawing lines of specific lengths in standard units.

This method of constructing the perpendicular bisector is not only geometrically exact but also a practical way to construct it accurately. This method to find the midpoint of a line segment is more accurate than measuring the length using a marked scale. Students, you must try this construction yourself to understand it better.

Now, let us learn how to construct a 90 degree angle at a given point. Can we extend the method of constructing the perpendicular bisector to construct a 90 degree angle at any point on a line? Draw a line and mark a point O on it. We need to construct a 90 degree angle at point O. How do we do this? We need to find a segment of this line for which O is the midpoint. So, we extend the line on either side of O. Using a compass, we mark two points X and Y at equal distance from O, so that O is the midpoint of XY. The perpendicular bisector of XY will pass through O and is perpendicular to the given line. In this case, do we need to draw two pairs of intersecting arcs to get the perpendicular bisector of XY? No, we do not. We already have one point, O, lying on the perpendicular bisector. So we only need to draw one pair of arcs to find another point on the perpendicular bisector.

Now, students, I want to tell you something very interesting. Ancient mathematicians from different civilizations, including India, knew exact procedures to construct perpendiculars and perpendicular bisectors. In India, the earliest known texts containing these methods are the Sulba-Sutras. These are geometric texts of the Vedic period dealing with the construction of fire altars for rituals. The Sulba-Sutras are part of one of the six Vedangas, a term that literally means limbs of the Vedas. The Sulbas contain the methods that we developed earlier to construct a perpendicular and the perpendicular bisector.

All the construction methods in the Sulba-Sutras make use of a different kind of compass from what you would have used, a rope. A rope can be used to draw circles or arcs. It can also be stretched to form a straight line. In addition, the Sulba-Sutras also contain other methods to construct perpendicular lines. Here is an interesting construction of the perpendicular bisector using a rope, from the Katyayana-Sulbasutra.

Let XY be the given line segment, drawn on the ground, for which we need to construct a perpendicular bisector. Fix a small pole or peg vertically into the ground at each point X and Y. Take a sufficiently long rope. Make two loops at its ends. Without taking into account the parts of the rope that has gone into the loops, fold the rope into half to find and mark its midpoint. Fasten the two loops at the ends of the rope to the poles at X and Y. Pull the midpoint of the rope above XY, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as A. Now, similarly pull the midpoint of the rope below XY, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as B. AB is the required perpendicular bisector. This is amazing, isn't it, students? Our ancestors used ropes to do geometry thousands of years ago!

Now, let us learn about angle bisection. How do we construct the figure that you see in your textbook? The supporting lines for this figure look like this. What is the angle between two adjacent lines? We need the angle between every pair of adjacent lines to be equal. Since 360 degrees is equally divided into 8 parts, every angle is 45 degrees. How do we construct a 45 degree angle using only a ruler and a compass? We know how to construct a 90 degree angle. If we can divide it into two equal parts, or bisect it, then we get a 45 degree angle.

We will now develop a general method to bisect any angle. Consider an angle XOY. We can bisect it if we can draw two congruent triangles, triangle OBC and triangle OAC, as shown in the figure. Then angle BOC equals angle AOC. How do we construct these congruent triangles, given the angle? If A and B are marked such that OA equals OB, and if C is chosen such that BC equals AC, then by the SSS congruence condition, triangle OBC is congruent to triangle OAC. So we can bisect an angle as follows.

Here are the steps for angle bisection. First, mark points A and B such that OA equals OB. Second, choosing any sufficiently long radius, cut arcs from A and B, keeping the radius the same. Mark the point of intersection as C. Third, OC bisects angle AOB. So, a 45 degree angle can be constructed by first constructing a 90 degree angle and then bisecting it. This is very useful, students. Remember this method, as we will use it many times.

Now, let us learn about repeating units and repeating angles. Look at the figure in your textbook. In this figure, there is a single unit repeating itself. To construct this figure, we need to make exact copies of this unit in two different orientations. In order to make exact copies, all the units must have the same arm lengths and the same angle between the arms. We can ensure equal arm lengths using a compass, but how do we ensure equal angles? Let us develop a method to create an exact copy of a given angle.

Draw an angle. Create a copy of this angle using only a ruler and compass. Here is one simple approach. First, draw an arc from point A. This gives us three points that form the isosceles triangle ABC. Second, draw an arc of the same radius from point X. Third, measure BC using a compass. Transfer this length on the arc from Z to get YZ equals BC. Fourth, by the SSS congruence condition, triangle ABC is congruent to triangle XYZ. So angle A equals angle X. This procedure to copy an angle finds an important application in constructing parallel lines using only a ruler and compass.

Now, let us learn how to construct a line parallel to the given line. Recall that in the construction using a ruler and a set square, we constructed equal corresponding angles to get parallel lines. How do we implement this idea using a ruler and a compass? Suppose there is a line m to which we need to draw a parallel line. We construct a line l that intersects m. Line l will serve as a transversal to line m and to the line parallel to m that we are going to construct. Let us choose a point B on l through which we are going to draw the parallel line. This parallel line must make the same corresponding angle as shown in the figure. This can be done by copying the angle between m and l. The steps involve constructing arcs of equal radius from A and B, then transferring the length to get the parallel line.

Now, students, let us learn about arch designs. Have you seen these beautiful arches in buildings like the Diwan-i-Aam in the Red Fort or in Central Park in New York City? How did they make these arches? The first step is to be able to draw them on a plane surface such as paper or stone. Let us 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 equals CD, and angle BAD equals angle CDA. How would you construct these support lines? We construct equal angles at A and D. Mark B and C such that AB equals CD. Use these support lines to construct an arch. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing.

Some arches look pointed, like the ones in the Red Fort. How do we construct this shape? What supporting lines will you use to draw this arch? Remember 'Wavy Wave' from the Grade 6 Textbook? The supporting lines are just two line segments of equal length. If their midpoints are marked, will you be able to construct a pointed arch? Yes, you will!

Now, let us learn about regular hexagons. Recall that a regular polygon has equal sides and equal angles. A regular polygon with 3 sides is an equilateral triangle, and a regular polygon with 4 sides is a square. We have constructed these figures earlier. How do we construct a regular pentagon, which is a 5-sided figure, and a regular hexagon, which is a 6-sided figure? To begin with, try to construct a pentagon and hexagon with equal side lengths. To construct a regular pentagon, we first need to have a better understanding of triangles and pentagons. We will discuss this in later years. However, constructing a regular hexagon is within our reach!

Can we break a regular hexagon into smaller pieces that can be constructed? Let us think about the relationship between regular hexagons and equilateral triangles. What happens when we join the opposite points of a regular hexagon? Since a regular hexagon has equal sides and angles, can we expect a figure like this? Will all the triangles in the figure be equilateral triangles?

To answer these questions, we will reverse our approach. Can six congruent equilateral triangles be placed together as shown in the figure? If yes, will it result in a regular hexagon? If six congruent equilateral triangles can indeed be placed as shown, then the sides of the resulting hexagon are equal, and their angles are 60 plus 60 equals 120 degrees. So what we really need to examine is whether six congruent equilateral triangles can fit this way without overlapping and without leaving any gaps around the centre.

We have defined a degree by taking the complete angle around a point to be 360 degrees. So all the angles around the centre should add up to 360 degrees. Consider this figure. Will the 70 degree angle fit into the gap? What is the gap angle AOI? We have 40 degrees plus 60 degrees plus 50 degrees plus 30 degrees plus 40 degrees plus 90 degrees plus gap angle equals 360 degrees. Use this to determine whether the 70 degree angle fits the gap.

Thus, if there are angles that add up to 360 degrees, their vertices can be joined together at a single point such that the angles do not overlap and they completely cover the region around the point. Since each angle in an equilateral triangle is 60 degrees, six such angles add up to 360 degrees. Therefore, six congruent triangles can be arranged as shown in the figure. In the figure, can you explain why AOD, BOE, and COF are straight lines? This is because the angles around the centre add up to 180 degrees along each straight line.

Now, let us construct a regular hexagon with a side length of 4 centimeters using a ruler and a compass. We can construct a regular hexagon more directly if we can construct a 120 degree angle using a ruler and compass. How do we do it? This can be done if we can construct a 60 degree angle. If we construct this, we also get the 120 degree angle.

How do we construct a 60 degree angle? We get a 60 degree angle if we construct an equilateral triangle! We can use the following steps for this. Suppose we need a 60 degree angle at point A on a line segment AX. Step one, construct an arc with centre A and any radius. Step two, with the same radius, cut another arc from B that meets the first arc. Let C be the point at which the arcs meet. We have angle CAX equals 60 degrees. Why is angle CAX equals 60 degrees? Is there an equilateral triangle here? Yes, triangle ABC is equilateral because AB equals AC equals BC, so all its angles are 60 degrees.

We can use these ideas to construct a regular hexagon. Construct a regular hexagon of side length 5 centimeters. How do we do this? We construct 60 degree angles at each vertex, and use the same side length for all sides. This will give us a regular hexagon.

Now, let us learn about constructing 30 degree and 15 degree angles. How will you construct 30 degree and 15 degree angles? For 30 degrees, we construct a 60 degree angle and then bisect it. For 15 degrees, we bisect the 30 degree angle. This is as simple as that!

Now, let us learn about the 6-pointed star. Construct the 6-pointed star shown in your textbook. Note that it has rotational symmetry. Do you see a hexagon here? Are the six triangles forming the 6 points of the star, triangles AGH, BHI, CIJ, DJK, ELK, and FLG, equilateral? Why? We need to find the angles in these triangles to answer this question. The hint is that the star is made by constructing equilateral triangles on the sides of a hexagon.

Now, students, let us move on to the second part of this chapter, which is about tilings. Tilings are everywhere around us, in floor patterns, wall designs, and even in nature. Let us learn about this interesting topic.

Tangrams are puzzles that originated in China. They make use of 7 pieces obtained by dividing a square as shown in your textbook. For the problems ahead, we need these 7 tangram pieces. These are provided at the end of the book. Or, by looking at the figure, you could make cardboard cutouts of the pieces. We can form interesting pieces by rearranging the tangram pieces. Here is an arrow made from tangram pieces.

Now, what is tiling? Covering a region using a set of shapes, without gaps or overlaps, is called tiling. Consider a rectangular grid made of unit squares. We call this a 4 by 6 grid, since it has 4 rows and 6 columns. Can a 4 by 6 grid be tiled using multiple copies of 2 by 1 tiles? We are allowed to rotate a 2 by 1 tile and use it. Here is one way to tile a 4 by 6 grid with 2 by 1 tiles. Obviously, this is not the only tiling possible.

Can a 4 by 7 grid be tiled using 2 by 1 tiles? What about a 5 by 7 grid? To see that there is no way to tile a 5 by 7 grid using 2 by 1 tiles, observe that this grid has 35 unit squares. Each tile covers exactly 2 unit squares. Since 35 is odd, we cannot cover all the squares without leaving one out. So, it is not possible.

Now, let us think about when an m by n grid is tileable with 2 by 1 tiles. Is an m by n grid tileable with 2 by 1 tiles if both m and n are even? If yes, come up with a general strategy to tile it. One general strategy for this case is to cover each column with vertical tiles. This is possible because the number of rows is even. So, yes, it is always possible when both dimensions are even.

Is an m by n grid tileable with 2 by 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. Yes, it is possible. We can fill the even number of rows completely with vertical tiles, and for the remaining one row, we can use horizontal tiles. Since the number of columns is odd, we will have one column left, which cannot be filled. Wait, let us think again. Actually, if one dimension is even and the other is odd, we can tile it by using horizontal tiles in pairs of rows. Since the number of rows is even, we can pair them up and fill each pair with horizontal tiles. So, yes, it is possible.

Is an m by n grid tileable with 2 by 1 tiles if both m and n are odd? Give reasons. No, it is not possible because the total number of unit squares would be odd, and each tile covers 2 squares, so we cannot cover all squares.

Now, here is a 5 by 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2 by 1 tiles? Sometimes yes, sometimes no, depending on which square is removed. Let us explore this.

Is the following region tileable with 2 by 1 tiles? 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 by 3 grid, makes it non-tileable?

There is an interesting way to look at these questions. For any tiling problem of this kind, we can create a similar problem with the unit squares coloured black and white so that black squares have only white neighbours and white squares have only black neighbours. In the black-and-white region, the problem is to tile the region with the 2 by 1 black-and-white tiles so that each black square of a tile sits on a black square of the grid, and each white square sits on a white square. 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? It can be seen that the answer to both the questions is yes.

Is the black-and-white region in the figure tileable? Any region tiled with black-and-white tiles must have an equal number of black squares and white squares. Since the black-and-white region has 8 white squares and 6 black squares, it can never be tiled with these tiles! This is a very clever method, students. By introducing colours and making the problem more complicated, it becomes easier to tackle. What a creative way of looking at the problem!

Now, let us learn about tiling the entire plane. So far we have seen how to tile a given region. What about tiling the entire plane? Can you think of a shape whose copies can tile the entire plane? Clearly, squares can. Are there other regular polygons that can tile the plane? What about equilateral triangles? Yes, equilateral triangles can tile the plane. Tiling with equilateral triangles shows the possibility of tiling with another regular polygon. A plane can be tiled using regular hexagons as well. A plane can also be tiled using more than one shape, and using non-regular polygons. People such as the great Dutch artist Escher, whose works explored mathematical themes such as tiling, have come up with creative ways of tiling a plane with animal shapes. Mathematicians are still exploring various ways of tiling the plane!

Have you seen tilings in daily life? They are often used in buildings and in designs. Tilings are found in nature too. The front face of bee hives and some wasp nests are tiled using hexagonal cells. These cells are used by the insects to keep their eggs, larvae and pupae safe, as well as to store food. Because the region is tiled, no space is wasted. Scientists still wonder how bees and wasps are able to make hexagonal cells. Next time you see any tiling, pay closer attention to it! Tilings are still one of the most exciting and active areas of research in geometry.

Now, students, let us summarize what we have learned in this chapter.

First, we learned about the perpendicular bisector. A division of a line segment, or any geometrical quantity, into two identical parts is called bisection. Any point that is of equal distance from the two endpoints of a given line segment lies on its perpendicular bisector. This property can be used to construct the perpendicular bisector using a ruler and compass.

Second, we learned that the method of constructing the perpendicular bisector can be modified to draw a 90 degree angle at any point on a line using only a ruler and compass.

Third, we learned that an angle can be bisected and copied using the congruence properties of triangles. We learned the steps to bisect an angle and to copy an angle.

Fourth, we learned that a 60 degree angle can be constructed using a ruler and compass by constructing an equilateral triangle. From this, we can construct 30 degree, 15 degree, 45 degree, 90 degree, and 120 degree angles.

Fifth, we learned about regular hexagons and their relationship with equilateral triangles. Six equilateral triangles can be arranged to form a regular hexagon.

Sixth, we learned about tilings. Covering a region using a set of shapes, without gaps or overlaps, is called tiling. We learned about tiling with 2 by 1 tiles, and we learned the clever black and white colouring argument to determine when a region is not tileable.

Seventh, we learned about tiling the entire plane with regular polygons like triangles, squares, and hexagons, and also with non-regular shapes and even with animal shapes, as done by the artist Escher.

This brings us to the end of our lesson, students. I hope you have understood all the concepts clearly. Remember, geometry is all around us, and constructions and tilings are used in many real-life applications. Keep exploring, keep questioning, and keep learning. Thank you for your attention, and see you in the next lesson!