6. Constructions and Tilings

1. Is it necessary to have the same radius for the arcs above and below XY?

No. The point above is equidistant from X and Y. The point below is also equidistant from X and Y (with another radius). Both points lie on the perpendicular bisector. Two points are enough to draw the line.

2. Can we construct both pairs of arcs on the same side of XY?

Yes, but we must use two different radii. Both meeting points are equidistant from X and Y, so both lie on the perpendicular bisector. Joining them and extending the line gives the perpendicular bisector. (If the two points are very close, the line may be less accurate.)

3. While drawing one pair of arcs, must the radii be the same?

Yes. If the radii from X and Y are different, the meeting point is not at equal distances from X and Y. Then it does not lie on the perpendicular bisector.

4. Recreate the 4-petal design.

Draw a segment and construct its perpendicular bisector. Now we have two perpendicular lines meeting at O.

Mark four equal segments from O along the four arms (use a compass). Draw an “eye” (petal) on each segment, as we did for the eye construction. Trace the petals with a colour pencil.

1. Justify why AB in Fig. 6.4 is the perpendicular bisector.

A is the midpoint of the rope, so AX = AY (half the rope each). Similarly, BX = BY. Both halves of the rope have the same length, so AX = AY = BX = BY.

A and B are equidistant from X and Y. So, both lie on the perpendicular bisector of XY. Hence, AB is the perpendicular bisector.

2. Different methods to construct a 90° angle at a point using a rope.

(i) Use the rope as a compass to mark X and Y on the line with OX = OY. Then use the rope method of Fig. 6.4 on XY. The line through O is perpendicular.

(ii) Make a rope with 12 equal parts. Stretch it into a triangle with sides 3, 4 and 5 parts, with O at the corner between 3 and 4. The angle at O is 90°. (This method is also found in the Śulba-Sūtras.)

1. Construct at least 4 different angles and draw their bisectors.

Do it yourself. For each angle, mark OA = OB on the arms. From A and B, cut arcs of the same radius to meet at C. OC is the bisector.

2. Construct the 8-petalled figure (Fig. 6.5).

Construct two perpendicular lines through O. Bisect each of the four 90° angles. Now there are 8 lines, 45° apart.

Mark equal lengths along all 8 lines with a compass. Draw an “eye” (petal) on each of these segments.

3. If arcs are drawn on the other side, will OC still be the angle bisector?

Yes. OA = OB and CA = CB. So, both O and C are equidistant from A and B. Both lie on the perpendicular bisector of AB.

Also, ∆OAC ≅ ∆OBC (SSS), so the line OC makes equal angles with OA and OB. Hence, OC bisects ∠AOB. (C must not coincide with O.)

4. What other angles can be constructed? Can you construct 65.5°?

From 90°: 45°, 22.5°, 11.25°, … From 60°: 30°, 15°, 7.5°, … We can also add them: 75° = 60° + 15°, 105° = 90° + 15°, 135° = 90° + 45°, 67.5° = 45° + 22.5°.

No, 65.5° cannot be made this way. It cannot be got by adding or subtracting the angles that come from bisecting 90° and 60°.

5. A method to construct the angle bisector using a rope.

Using a fixed length of rope from O, mark A and B on the two arms (OA = OB).

Tie the two ends of another rope at A and B. Pull its midpoint tight away from O and mark it C. Then AC = BC.

Stretch a rope from O through C. OC is the angle bisector (∆OAC ≅ ∆OBC by SSS).

6. Construct the petals of maximum size inside a square.

Find the midpoint of each side using perpendicular bisectors.

With each midpoint as centre and radius = half the side, draw a semicircle inside the square. Each semicircle passes through two corners and the centre of the square.

The four semicircles make the four petals. The petals are of the maximum size, because each arc just touches the sides of the square at the corners.

1. Construct 4 different angles and copy them.

Do it yourself using the steps above.

2. Construct Fig. 6.6.

See the steps given above.

1. Construct 4 pairs of parallel lines.

Do it yourself. Draw a line m and a transversal l through A. Copy the angle at A to a point B on l. The new line is parallel to m.

2. Construct the 8-pointed star (ABCD…).

Draw 8 lines through O, 45° apart (90° angles and their bisectors).

Mark A, B, C, D, E, F, G and H on these lines at the same distance from O.

Through A, draw a line parallel to OH. Through H, draw a line parallel to OA. They meet at S. So, OASH is a rhombus.

Repeat for every pair of neighbouring points to get T, U, V, W, X, Y and Z. Join O to each outer point. Shade alternate triangles.

1 and 2. Make pointed arches and your own arch designs.

Do it yourself. A larger radius gives a taller, sharper arch. A smaller radius gives a flatter arch.

1. Construct the figures (a) to (e).

(a) Inflexed arc: draw two equal parallel vertical lines. On top, draw two S-shaped curves that meet at a point on the perpendicular bisector. Each S-curve is made of two arcs — one bending out, one bending in.

(b) Flower: draw a circle and step its radius 6 times on it to get 6 points. With these points as centres, draw equal circles (petals) that just touch each other. Draw the centre circle over them and erase the arcs inside it. Only a compass is needed.

(c) Draw a circle. Step its radius 6 times on it. Join the points to get the hexagon inside the circle.

(d) Construct a regular hexagon. With each vertex as centre, draw a circle of radius half the side. The 6 circles touch each other.

(e) Construct a regular hexagon. Divide it into 24 small equilateral triangles (mark the midpoints of the sides and draw lines parallel to the sides). In each small triangle, find its centre and join it to the three vertices.

2. Optical illusion.

We see a white triangle in the middle, but it is not drawn. The cut-outs in the black circles and the V-shapes make our brain fill in the missing edges.

To recreate: construct an equilateral triangle lightly. Draw black circles at its vertices. Draw an upside-down equilateral triangle as V-shapes. Erase the support lines.

3. Construct the star inside a hexagon.

Angles: each angle of the hexagon is 120°. Each point of the star is 60°. So, the two small angles beside it are 30° each.

Construct a regular hexagon. Join alternate vertices to form two equilateral triangles. Keep only the outer edges of the star.

4. Construct a perpendicular to line l through a point P outside it.

With P as centre, draw an arc cutting l at X and Y. Now PX = PY, so P is on the perpendicular bisector of XY.

From X and Y, draw arcs of equal radius on the other side of l. They meet at Q. Join PQ. PQ is perpendicular to l.

1. Can the region be tiled with the L-shaped tile?

Yes. The region has 12 squares, so 4 L-tiles are needed. One way:

2. Can the 8 × 8 grid with two opposite corners removed be tiled with 2 × 1 tiles?

No. Colour it like a chessboard. The two removed corners have the same colour.

So, 62 squares remain: 32 of one colour and 30 of the other. Each tile covers one square of each colour. So, a tiling is not possible.