Sierpinski Carpet

?In-text Question — Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Carpet.

The first three steps of the Sierpinski Carpet.

Step 0

Draw one square.

Step 1

Divide the square into 9 equal smaller squares.

Remove the central square.

8 squares remain and there is 1 hole.

Step 2

Divide each of the 8 remaining squares into 9 equal squares.

Remove the central square of each.

\(8\times8=64\) squares remain.

Number of holes \(=1+8=9\).

Step 0: 1 square, 0 holes
Step 1: 8 squares, 1 hole
Step 2: 64 squares, 9 holes

?In-text Question — Do you see any pattern in the number of holes and squares that remain at each step?

Let \(R_n\) be the number of squares remaining and \(H_n\) the number of holes at Step \(n\).

Step \(n\)	0	1	2	3
\(R_n\)	1	8	\(8^2=64\)	\(8^3=512\)
\(H_n\)	0	1	\(1+8=9\)	\(1+8+8^2=73\)

Each remaining square gives 8 squares in the next step.

So \(R_n=8^n\).

Each remaining square also gives 1 new hole, and the old holes stay.

So \(H_n=1+8+8^2+\dots+8^{n-1}\).

Using the sum of a geometric series, \(H_n=\dfrac{8^n-1}{8-1}\).

\(H_n=\dfrac{8^n-1}{7}\)

Check: \(H_3=\dfrac{512-1}{7}=73\).

\(R_n=8^n\)
\(H_n=\dfrac{8^n-1}{7}\)

Sierpinski Gasket

?In-text Question — Show that by joining the midpoints of an equilateral triangle, we divide it into 4 identical equilateral triangles. [Hint: Note that the corner triangles are isosceles.]

D, E, F are the midpoints of BC, CA, AB.

Let \(ABC\) be an equilateral triangle of side \(a\).

Let \(D\), \(E\), \(F\) be the midpoints of \(BC\), \(CA\), \(AB\).

Corner triangle AFE

\(AF=\dfrac{a}{2}\) and \(AE=\dfrac{a}{2}\).

So \(\triangle AFE\) is isosceles.

\(\angle A=60^\circ\).

\(\angle AFE=\angle AEF=\dfrac{180^\circ-60^\circ}{2}=60^\circ\).

All three angles are \(60^\circ\), so \(\triangle AFE\) is equilateral.

\(EF=AF=\dfrac{a}{2}\).

Other corner triangles

In the same way, \(\triangle BDF\) and \(\triangle CDE\) are equilateral with side \(\dfrac{a}{2}\).

So \(DF=DE=\dfrac{a}{2}\).

Middle triangle DEF

\(EF=FD=DE=\dfrac{a}{2}\).

So \(\triangle DEF\) is equilateral.

All four triangles

Each of the 4 triangles is equilateral with side \(\dfrac{a}{2}\).

So they are congruent by SSS.

Joining the midpoints divides the triangle into 4 identical equilateral triangles of side \(\dfrac{a}{2}\). Hence proved.

Figure it Out — Set 1

Three questions from page 72 of the textbook.

1Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Triangle.

The first three steps of the Sierpinski Triangle (Gasket).

Step 0

Draw one equilateral triangle.

Step 1

Join the midpoints of the sides.

The triangle breaks into 4 identical equilateral triangles.

Remove the central triangle.

3 triangles remain and there is 1 hole.

Step 2

Repeat the same on each of the 3 remaining triangles.

\(3\times3=9\) triangles remain.

Number of holes \(=1+3=4\).

Step 0: 1 triangle, 0 holes
Step 1: 3 triangles, 1 hole
Step 2: 9 triangles, 4 holes

2Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Let \(T_n\) be the number of triangles remaining and \(H_n\) the number of holes at Step \(n\).

Step \(n\)	0	1	2	3
\(T_n\)	1	3	9	27
\(H_n\)	0	1	\(1+3=4\)	\(1+3+9=13\)

Triangles

Each remaining triangle gives 3 triangles in the next step.

So \(T_{n+1}=3T_n\).

\(T_n=3^n\)

Holes

Each remaining triangle gives 1 new hole, and the old holes stay.

So \(H_{n+1}=H_n+T_n\).

\(H_n=1+3+3^2+\dots+3^{n-1}\)

\(H_n=\dfrac{3^n-1}{3-1}\)

\(H_n=\dfrac{3^n-1}{2}\)

Check: \(H_3=\dfrac{27-1}{2}=13\).

Triangles remaining \(=3^n\)
Holes \(=\dfrac{3^n-1}{2}\)

3Find the area of the region remaining at the nth step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

Sierpinski Carpet

At each step, a square is cut into 9 equal squares.

So each new square has \(\dfrac{1}{9}\) of the area of the earlier square.

Area of each square at Step \(n\) \(=\left(\dfrac{1}{9}\right)^n\)

Number of squares at Step \(n\) \(=8^n\)

Remaining area \(=8^n\times\left(\dfrac{1}{9}\right)^n\)

\(=\left(\dfrac{8}{9}\right)^n\) sq. units

Sierpinski Triangle

At each step, a triangle is cut into 4 equal triangles.

So each new triangle has \(\dfrac{1}{4}\) of the area of the earlier triangle.

Area of each triangle at Step \(n\) \(=\left(\dfrac{1}{4}\right)^n\)

Number of triangles at Step \(n\) \(=3^n\)

Remaining area \(=3^n\times\left(\dfrac{1}{4}\right)^n\)

\(=\left(\dfrac{3}{4}\right)^n\) sq. units

Step \(n\)	0	1	2	3
Carpet	1	\(\dfrac{8}{9}\)	\(\dfrac{64}{81}\)	\(\dfrac{512}{729}\)
Triangle	1	\(\dfrac{3}{4}\)	\(\dfrac{9}{16}\)	\(\dfrac{27}{64}\)

Carpet: \(\left(\dfrac{8}{9}\right)^n\) sq. units
Triangle: \(\left(\dfrac{3}{4}\right)^n\) sq. units

Koch Snowflake

Figure it Out — Set 2

Three questions from page 73 of the textbook.

1Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

The first three steps of the Koch Snowflake.

Step 0

Draw an equilateral triangle.

Step 1

Divide each side into 3 equal parts.

Raise an equilateral triangle on the middle part and remove the middle part.

The triangle becomes a 6-pointed star with 12 sides.

Step 2

Repeat the same on each of the 12 sides.

The shape now has \(12\times4=48\) sides.

Step 0: 3 sides
Step 1: 12 sides
Step 2: 48 sides

2Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.

Each side is replaced by 4 sides: 2 sides of the bump and the 2 remaining outer parts.

So the number of sides gets multiplied by 4 at every step.

Step \(n\)	0	1	2	3
Sides	3	\(3\times4=12\)	\(3\times4^2=48\)	\(3\times4^3=192\)

Number of sides at Step \(n\) \(=3\times4^n\)

Number of sides \(=3\times4^n\)

3Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.

At every step, each side is divided into 3 equal parts.

So the length of each side becomes \(\dfrac{1}{3}\) of the earlier length.

Length of each side at Step \(n\) \(=\left(\dfrac{1}{3}\right)^n\)

Number of sides at Step \(n\) \(=3\times4^n\)

Perimeter \(=3\times4^n\times\left(\dfrac{1}{3}\right)^n\)

\(=3\left(\dfrac{4}{3}\right)^n\) units

Step \(n\)	0	1	2	3
Perimeter	3	4	\(\dfrac{16}{3}\)	\(\dfrac{64}{9}\)

The perimeter keeps increasing, because \(\dfrac{4}{3}>1\).

Perimeter \(=3\left(\dfrac{4}{3}\right)^n\) units

Build it in Your Imagination

?In-text Question — 1. Picture your name, then read off the letters backwards. Make sure to do this by sight, not by sound — really see your name! Now try with your friend’s name.
2. Cut off the four corners of an imaginary square, with each cut going between midpoints of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square?
3. Mark the sides of an equilateral triangle into thirds. Cut off each corner of the triangle, as far as the marks. What shape do you get?
4. Mark the sides of a square into thirds and cut off each of its corners as far as the marks. What shape is left?

1. This is a visualisation exercise, so it has no written answer.

For example, the word MATHS read backwards is SHTAM.

2. Let the square have side 2.

Each corner cut is a right-angled isosceles triangle with legs 1 and hypotenuse \(\sqrt2\).

The shape left over is a smaller square of side \(\sqrt2\), tilted by \(45^\circ\).

Its area is \(2\), which is half of \(4\).

Place the 4 cut corners with their right angles meeting at one point and their hypotenuses on the outside.

They make another square of side \(\sqrt2\).

3. Each cut corner is a small equilateral triangle of side one-third.

The shape left has 6 sides, each of length one-third, and each angle is \(120^\circ\).

It is a regular hexagon.

4. Each cut corner is a right-angled isosceles triangle with legs one-third.

The shape left has 8 sides and each angle is \(135^\circ\).

Its sides are alternately one-third and \(\dfrac{\sqrt2}{3}\), so it is an octagon, but not a regular one.

2. A square (half the area); the 4 corners make another square
3. A regular hexagon
4. An octagon (all angles equal, sides not all equal)

?In-text Question — Can you describe a solid and a viewpoint that would result in each of the following cases? If it helps, you can imagine the solid passing through a wall like Tom did, and leaving a hole of the appropriate shape.
5. A solid whose profile has a square outline
6. A solid whose profile has a circular outline
7. A solid whose profile has a triangular outline

5. A cube, seen from the front.

A cylinder whose height equals its diameter, seen from the side, also gives a square.

6. A sphere, seen from any direction.

A cylinder or a cone, seen from directly above.

7. A cone, seen from the side.

A triangular prism, seen from one of its triangular ends.

A pyramid, seen from the side.

5. Cube (front view)
6. Sphere (any view)
7. Cone (side view)

?In-text Question — As we saw with the elephant, a given solid might have very different profiles from different viewpoints. Can you visualise solids that have the following contrasting profiles?
8. A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint
9. A solid with a circular profile from one viewpoint and a triangular one from another viewpoint
10. A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint
11. A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint
12. A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint

8. A cylinder.

Rectangle from the side, circle from above.

9. A cone.

Triangle from the side, circle from above.

10. A triangular prism lying on a rectangular face.

Rectangle from above, triangle from an end.

11. A cone with its top cut off (a frustum).

Trapezium from the side, circle from above.

12. A pentagonal prism.

Pentagon from an end, rectangle from the side.

8. Cylinder
9. Cone
10. Triangular prism
11. Frustum of a cone
12. Pentagonal prism

?In-text Question — Are there unique solids for each of the conditions, or can you come up with multiple possibilities?

No, the solids are not unique.

Case 10 (rectangle and triangle)

A triangular prism works.

A square pyramid also works: its side view is a triangle and its base seen from below is a square, which is a rectangle.

Case 9 (circle and triangle)

A cone works.

A cone leaning to one side also works, since its top view is still a circle with a point inside and its side view is a triangle.

Case 7 (triangle)

A cone, a pyramid and a triangular prism all give a triangle.

There are many possible solids for each condition.

Making Solids

?In-text Question — If the congruent polygons of a prism have 10 sides, how many faces, edges and vertices does the prism have? What if the polygons have n sides?

Prism with 10-sided polygons

Faces \(=2\) polygons \(+\) 10 side faces \(=12\)

Edges \(=10+10+10=30\)

Vertices \(=10+10=20\)

Prism with n-sided polygons

Faces \(=n+2\)

Edges \(=3n\)

Vertices \(=2n\)

Check with a cube (\(n=4\)): faces \(=6\), edges \(=12\), vertices \(=8\).

10 sides: 12 faces, 30 edges, 20 vertices
n sides: \(n+2\) faces, \(3n\) edges, \(2n\) vertices

?In-text Question — If the base of a pyramid has 10 sides, how many faces, edges and vertices does the pyramid have? What if the base is an n-sided polygon?

Pyramid with a 10-sided base

Faces \(=1\) base \(+\) 10 triangles \(=11\)

Edges \(=10\) (base) \(+\) 10 (slant) \(=20\)

Vertices \(=10+1=11\)

Pyramid with an n-sided base

Faces \(=n+1\)

Edges \(=2n\)

Vertices \(=n+1\)

Check with a square pyramid (\(n=4\)): faces \(=5\), edges \(=8\), vertices \(=5\).

10 sides: 11 faces, 20 edges, 11 vertices
n sides: \(n+1\) faces, \(2n\) edges, \(n+1\) vertices

In-text Question — Visualise how it can be folded to form a cube.

Fig. 4.1

Fig. 4.1 with its six squares named A to F.

The column of four squares is A, B, C, D (from the top), E is the square on the left of C and F is the square on the right of C.

Keep C flat on the table. It is the bottom face.

Fold B up along its edge with C. It becomes the back wall.

Fold D up along its edge with C. It becomes the front wall.

Fold E up along its edge with C. It becomes the left wall.

Fold F up along its edge with C. It becomes the right wall.

Fold A over along its edge with B. It lands on top of the cube, opposite C.

Opposite faces: C and A, B and D, E and F.

C is the bottom, A the top, B and D the back and front, E and F the sides. The net folds into a cube.

Figure it Out — Set 3

Three questions from pages 80–81 of the textbook.

Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try.

The six shapes of the question.

The six shapes redrawn. Green: a net of a cube. Red: not a net.

(i) Fold the row of 4 squares into a band around the cube.

The two squares above the end of the row land on the same face.

So one face is covered twice and another is left empty.

Not a net.

(ii) Two rows of 3 squares, joined at one square.

Folds into a cube.

It is a net.

(iii) A staircase of 3 pairs of squares.

Folds into a cube.

It is a net.

(iv) A row of 4 squares with one square above and one below.

The row makes the band and the two extra squares become the top and bottom.

It is a net.

(v) A row of 4 squares with 2 squares hanging below the same square.

The first hanging square becomes the bottom.

The second one then lands on the face opposite to it, which is already covered by the band.

Not a net.

(vi) A row of 4 squares with one square above and one below, in different columns.

The band is made by the row, and the top and bottom are covered once each.

It is a net.

(i) No
(ii) Yes
(iii) Yes
(iv) Yes
(v) No
(vi) Yes

Try This — A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. For example, the following nets are all considered the same —
Find all the 11 nets of a cube.

These nets are all considered the same.

The 11 nets of a cube, grouped by the number of squares in each row.

How the nets are found

Every net has 6 squares joined edge to edge.

Sort them by the number of squares in each row.

Six nets with rows 1-4-1

A row of 4 squares, with one square above and one below.

The square above can be over any of the 4 squares, and so can the one below.

This gives 6 different nets after removing rotations and flips.

Three nets with rows 2-3-1

One net with rows 2-2-2

One net with rows 3-3

Total \(=6+3+1+1=11\).

The 11 nets are shown in the figure: 6 of type 1-4-1, 3 of type 2-3-1, 1 of type 2-2-2 and 1 of type 3-3.

3Try This — Draw a net of a cuboid having sidelengths:
(i) 5 cm, 3 cm, and 1 cm
(ii) 6 cm, 3 cm, and 2 cm

A cuboid has 6 faces, in 3 pairs of equal rectangles.

(i) 5 cm, 3 cm, 1 cm

Faces: \(5\times3\) (2 faces), \(5\times1\) (2 faces), \(3\times1\) (2 faces).

A net of the 5 cm by 3 cm by 1 cm cuboid. Lengths are in cm.

(ii) 6 cm, 3 cm, 2 cm

Faces: \(6\times3\) (2 faces), \(6\times2\) (2 faces), \(3\times2\) (2 faces).

A net of the 6 cm by 3 cm by 2 cm cuboid. Lengths are in cm.

Steps to draw

Draw the \(a\times b\) rectangle in the middle.

Draw the two \(a\times c\) rectangles above and below it.

Draw the two \(c\times b\) rectangles on the left and right.

Attach the last \(a\times b\) rectangle below the lower \(a\times c\) rectangle.

Nets are drawn above. Any net that folds into the cuboid with these 3 pairs of faces is acceptable.

In-text Question — What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron?

The four arrangements of the question.

The four arrangements of equilateral triangles from the textbook, redrawn. Green: a net. Red: not a net.

A regular tetrahedron has 4 faces, all equilateral triangles.

A net of it is 4 equilateral triangles joined edge to edge that fold into the tetrahedron.

(i) A big triangle made of 4 small ones.

The 3 outer triangles fold up over the middle one.

It is a net.

(ii) A strip of 3 triangles with a fourth triangle below the first.

Two of the triangles land on the same face and one face is left empty.

Not a net.

(iii) A strip of 4 triangles.

It folds into the tetrahedron.

It is a net.

(iv) A strip of 4 triangles with another triangle below the first.

This has 5 triangles, but a tetrahedron has only 4 faces.

Not a net.

(i) and (iii) are nets of a regular tetrahedron.

MTMath Talk — Draw a net with appropriate measurements that can be folded into a regular tetrahedron. Verify if it works by making an actual cutout.

All 4 faces must be equilateral triangles of the same side.

Take the side as 5 cm.

Draw 4 equilateral triangles of side 5 cm joined edge to edge, either as a big triangle of side 10 cm cut into 4, or as a strip of 4.

The two nets of a regular tetrahedron. Each small triangle has side 5 cm.

Cut out and fold along the inner lines.

The 3 corners meet at one point and the tetrahedron closes.

Four equilateral triangles of side 5 cm, joined as shown, fold into a regular tetrahedron.

MTMath Talk — Draw a net with appropriate measurements that can be folded into a square pyramid. Verify if it works by making an actual cutout.

A square pyramid has 1 square base and 4 triangular faces.

Take the base as a square of side 4 cm.

Take each triangular face as an equilateral triangle of side 4 cm, so that it fits each side of the square.

Draw the square and join one triangle to each of its 4 sides.

A net of a square pyramid. The square has side 4 cm and each triangle has all sides 4 cm.

Cut out and fold the 4 triangles upwards.

Their top corners meet at the apex.

A square of side 4 cm with an equilateral triangle of side 4 cm on each side folds into a square pyramid.

In-text Question — What are the sidelengths of the rectangle obtained?

The cylinder is cut along its height and opened out.

Nets of a cylinder (left) and a cone (right).

Cut the curved surface of the cylinder along its height and open it out.

It becomes a rectangle.

One side of the rectangle is the height \(h\) of the cylinder.

The other side is the distance round the circular base.

That is the circumference \(2\pi r\).

Sides of the rectangle: \(h\) and \(2\pi r\)

Math Talk — What surface do you construct by using the above net, in which O is not the centre of the boundary circle? Make a physical model to help you answer this question!

A net in which O is not the centre of the boundary circle.

In the net of a cone, \(O\) is the centre of the circular arc.

The cone and its net, with \(O\) as the centre of the arc.

Every point of the boundary is at the same distance from \(O\), so all slant lines are equal.

If \(O\) is not the centre, the distances from \(O\) to the boundary are different (the dashed lines in the question figure).

When the two straight edges are joined, the surface still comes to a point at \(O\), but the slant lengths vary from place to place.

So the model leans to one side: the point is not directly above the centre of the opening.

It is a slanting (oblique) cone-like surface, not a right circular cone.

A slanting (oblique) cone-like surface, not a right circular cone.

MTMath Talk — Draw a net with appropriate measurements that can be folded into a triangular prism. Verify that it works by making an actual cutout.

A triangular prism has 2 triangular faces and 3 rectangular faces.

Take the triangle as equilateral with side 3 cm and the prism length as 5 cm.

Draw 3 rectangles of \(3\text{ cm}\times5\text{ cm}\) side by side.

Draw an equilateral triangle of side 3 cm on the top edge of one rectangle.

Draw another on the bottom edge of the same rectangle.

A net of a triangular prism: three 3 cm by 5 cm rectangles and two equilateral triangles of side 3 cm.

Cut out and fold the rectangles into a triangular tube.

Fold the two triangles to close the ends.

Three \(3\times5\) rectangles in a row with a triangle of side 3 cm on top and below one of them.

Math Talk — Taking all the triangles in the net to be equilateral, make a cutout of the net and fold it to form an octahedron.

A net of the octahedron.

An octahedron has 8 faces, all equilateral triangles, with 4 triangles meeting at each of its 6 vertices.

The net with its triangles numbered 1 to 8.

Draw the net with 8 equilateral triangles of the same side, say 4 cm, and cut it out.

Triangles 1 to 6 form a zig-zag strip. Curl the strip into a ring by joining the slanting edge of 1 to the slanting edge of 6.

The ring has a triangular opening at the top and another at the bottom.

Triangle 7 is joined to triangle 4 along its top edge. Fold it down to close the top opening.

Triangle 8 is joined to triangle 3 along its lower edge. Fold it up to close the bottom opening.

The result is two square pyramids joined at their square bases.

Triangles 1 to 6 make a ring; triangles 7 and 8 close the top and the bottom. This folds into an octahedron.

?In-text Question — Net of a sphere? Experiment and see if you can make a paper cutout that can perfectly wrap around a ball without leaving any wrinkles, gaps or overlaps.

Try wrapping a ball with a flat paper.

Paper can be rolled into a cylinder or a cone without stretching.

A sphere is curved in every direction, so flat paper can cover it only with wrinkles, gaps or overlaps.

So a sphere has no net.

No. A sphere cannot be covered exactly by a flat cutout.

Math Talk — What is the shortest path for the ant to reach the laddu?

The laddu is at the centre of the top face. The ant is at the centre of the side face.

A path on the surface can be changed into a path on the net of the same length, so find a straight line on a net.

The laddu is on the top face and the ant is on the end face. These two faces share an edge.

Unfold the end face next to the top face, so that the two faces lie in one plane.

The top face with the end face unfolded beside it.

The laddu and the ant are both at the centres of their faces, so they lie on one straight line that crosses the shared edge at right angles.

Let the top face be \(L\) long in this direction, and let the end face be \(h\) high.

The straight line has length \(\dfrac L2+\dfrac h2\).

On the cuboid this is the path that goes straight along the middle of the top to the edge and then straight down the middle of the end face.

Let the width of the top face be \(w\). Another net, with the end face attached to the front face, gives the straight line of length \(\dfrac12\sqrt{(L+w)^2+(w+h)^2}\).

The first path is not longer than this one when \(Lh\le w(L+w+h)\). This holds, for example, for a cube.

So in such cases the path along the middle of the top and down the middle of the end face is the shorter of the two.

The book gives no measurements for this cuboid, so the length is left in terms of \(L\), \(w\) and \(h\). For a very thin cuboid the second unfolding can give the shorter path, so compare the two lengths for the cuboid you have.

Go straight along the middle of the top face to the edge, then straight down the middle of the end face: length \(\dfrac{L+h}{2}\).

Shortest Paths on a Cuboid

In-text Question — Find the shortest path between the ant and the laddu in the following case:

The cuboid, the laddu and the ant.

The cuboid is 8 cm long and 4 cm by 4 cm at the ends.

The laddu is on the bottom edge of the long face, 2 cm from the end.

The ant is at the centre of the end face.

Unfold the end face next to the long face, as the textbook does in the second net.

The unfolding in which the straight line stays inside the net.

Horizontal distance from the laddu to the ant \(=2+2=4\) cm.

Vertical distance \(=2\) cm (the ant is 2 cm above the bottom).

Shortest path \(=\sqrt{4^2+2^2}\)

\(=\sqrt{16+4}\)

\(=\sqrt{20}\)

\(=2\sqrt5\) cm

\(\approx4.47\) cm

Shortest path \(=2\sqrt5\) cm \(\approx4.47\) cm

Try This — What is the length of the shortest path between the ant and the laddu?

The box with the laddu (stuck to the back of the box) and the ant.

The box is 30 cm long with square ends of 12 cm.

The laddu is on one end, 6 cm from the side and 1 cm above the bottom.

The ant is on the other end, 6 cm from the side and 1 cm below the top.

Try the different unfoldings and find each straight-line distance.

Via the bottom face only

Distance \(=1+30+11=42\) cm

Via one side face only

Horizontal \(=6+30+6=42\) cm, vertical \(=10\) cm

Distance \(=\sqrt{42^2+10^2}=\sqrt{1864}\approx43.2\) cm

Via the bottom face and a side face

Horizontal \(=1+30+6=37\) cm, vertical \(=6+11=17\) cm

Distance \(=\sqrt{37^2+17^2}=\sqrt{1658}\approx40.7\) cm

Via the bottom, a side face and the top

The unfolding through the bottom, side and top faces.

Horizontal \(=1+30+1=32\) cm

Vertical \(=6+12+6=24\) cm

\(d^2=24^2+32^2\)

\(d^2=576+1024=1600\)

\(d=\sqrt{1600}=40\) cm

Compare: \(40<40.7<42<43.2\).

Shortest path \(=40\) cm

In-text Question — What happens to the length of a line in its projection? Let us consider projections in Fig. 4.3. For example let l be the actual length of the line and p be the length of its projection. Draw AE ⊥ BC. AECD is a rectangle (why?). So, AE = DC = p. Also, ∠AEB = 90°. Can you now compare the lengths p and l? When is the length of the projected line equal to its actual length?

Fig. 4.3

Why is AECD a rectangle?

\(AD\) and \(BC\) are both perpendicular to the plane, so \(AD\parallel EC\).

\(BC\) is perpendicular to the plane, so \(BC\perp DC\).

Also \(AE\perp BC\), so \(AE\parallel DC\).

So \(AECD\) is a parallelogram.

\(\angle ADC=90^\circ\), because \(AD\) is perpendicular to the plane.

A parallelogram with one right angle is a rectangle.

Comparing p and l

In the right triangle \(AEB\), \(AB=l\) is the hypotenuse and \(AE=p\) is a leg.

The hypotenuse is the longest side.

So \(p\le l\).

When is p = l?

\(p=l\) when \(BE=0\), that is when \(A\) and \(B\) are at the same distance from the plane.

This happens when the line is parallel to the plane.

\(p\le l\)
\(p=l\) when the line is parallel to the plane of projection.

MTMath Talk — What do you think are the different possible projections of a square that we get based on its orientation?

Facing the plane

The projection is a square of the same size.

Tilted about a line parallel to a side

One pair of sides shrinks.

The projection is a rectangle.

Tilted in any other way

Parallel sides stay parallel.

The projection is a parallelogram.

Edge-on

The projection is a line segment.

A square, a rectangle, a parallelogram, or a line segment.

MTMath Talk — What do you think is the projection of a parallelogram under different orientations? Can this ever be a quadrilateral that is not a parallelogram? As a starting point, you could think about the projection of a pair of parallel lines.

The projection of a pair of parallel lines is a pair of parallel lines.

A parallelogram has two pairs of parallel sides.

So in its projection both pairs of sides stay parallel.

Hence the projection is again a parallelogram.

If the parallelogram is edge-on, the projection is a line segment.

So the projection can never be a quadrilateral that is not a parallelogram.

Always a parallelogram (or a line segment when edge-on). Never a non-parallelogram quadrilateral.

MTMath Talk — What can you say about the projection of an n-sided regular polygon? [Hint: Projection of a polygon is composed of the projections of its sides.]

The projection of each side is a line segment.

So the projection is a closed figure with \(n\) sides.

Facing the plane, it is a regular \(n\)-sided polygon of the same size.

When tilted, it is a squashed \(n\)-sided polygon.

Parallel sides stay parallel, so if \(n\) is even, opposite sides stay parallel and equal.

Edge-on, it is a line segment.

An n-sided polygon (a line segment when the polygon is edge-on).

In-text Question — How would the projections of a cube and a cone look?

Fig. 4.4

Fig. 4.5

Cube

Facing a face: a square.

Tilted about an edge direction: a rectangle.

In other positions: a six-sided figure (hexagon).

Cone

Along its axis: a circle with the apex at the centre.

From the side: a triangle.

In between: a triangle joined to part of an ellipse.

Cube: square, rectangle or hexagon
Cone: circle, triangle or triangle with a curved base

In-text Question — See Figures 4.2 – 4.5. In each case, see if you can visualise another object that gives the same projection.

Fig. 4.2

Fig. 4.3

Fig. 4.4

Fig. 4.5

Line (Fig. 4.2)

A longer, tilted line can have the same projection.

Cube (Fig. 4.4)

A cuboid with the same front face has the same square projection.

Cone (Fig. 4.5)

A pyramid or a triangular prism gives the same triangle.

Fig. 4.6

Different lines and different cuboids give the same projections.

Many different objects can give the same projection.

In-text Question — Find another object that makes the same projection as that of a given cone.

The given cone and its projection (Fig. 4.5).

Seen from the side, a cone gives a triangle.

A square pyramid, or a triangular prism standing on its edge, also gives a triangle.

Seen from above, a cone gives a circle.

A cylinder or a sphere also gives a circle.

A pyramid (triangle) or a cylinder (circle).

Figure it Out — Set 4

Three questions from pages 92–93 of the textbook.

Math Talk — Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths?

Front view, top view and side view of the three lines of Fig. 4.6.

Let the front view, top view and side view lengths of a line be \(F\), \(T\) and \(S\).

Measure each length on the printed figure with a ruler (in mm).

Line	\(F\)	\(T\)	\(S\)	\(F^2+S^2\)	\(T^2\)
1st	20.4	20.4	0.0	416	416
2nd	20.4	21.4	3.9	431	458
3rd	20.4	22.2	9.5	506	493

The front view is the same length \(F\) in all three lines.

In the 1st line the side view is only a point, so \(S=0\) and \(T=F\).

In the 2nd and 3rd lines the side view grows, and the top view becomes longer than the front view.

In every line \(F^2+S^2\) is close to \(T^2\). The small differences come from measuring.

So the top view is the hypotenuse of a right triangle whose other two sides are the front view and the side view.

The lengths are measured by ruler on the printed figure, so they depend on the size of the print. Check the relation with your own measurements.

\(T^2=F^2+S^2\)

2Math Talk — Find the front view, top view and side view of each of the following solids, fixing its orientation with respect to the vertical, horizontal and side planes: cube, cuboid, parallelepiped, cylinder, cone, prism, and pyramid. If needed, see the next problem for clues.

The orientation chosen for each solid is given in the first column.

Solid	Front view	Top view	Side view
Cube	Square	Square	Square
Cuboid (resting on a face)	Rectangle	Rectangle	Rectangle
Parallelepiped (base horizontal, top slid sideways)	Parallelogram	Rectangle	Rectangle
Cylinder (standing)	Rectangle	Circle	Rectangle
Cone (standing, apex up)	Triangle	Circle with the apex at its centre	Triangle
Triangular prism (lying on a rectangular face, triangles at the ends)	Rectangle	Rectangle	Triangle
Square pyramid (standing on its base)	Triangle	Square with its diagonals	Triangle

See the table for the three views of each solid.

Match each of the following objects with its projections.

The eight objects in the question	Colour in the book
Jug	purple
Funnel	red
Hammer	green
Car	yellow
Slide with a box	pink
Chair	blue
Ceiling fan	yellow-green
Pot with a lid	pink

Each object is to be matched with one of the eight rows of projections (front view, top view, side view) printed in the book.

Each row of the projections table shows the front view, top view and side view of one object.

Match them by shape.

Object	Row of projections (from the top)	What the views show
Jug (purple)	8th row	Spout in the top view, handle in the side view
Funnel (red)	6th row	Circle in the top view, funnel shape from the front and side
Hammer (green)	7th row	Small head in the top view, hammer from the side
Car (yellow)	1st row	Car from the front and side, a car outline from above
Slide with a box (pink)	3rd row	Long rectangle with a square in the top view, ramp in the side view
Chair (blue)	4th row	Chair from the front and side, square seat from above
Ceiling fan (yellow-green)	5th row	Three blades from above, fan from the front and side
Pot with lid (pink)	2nd row	Circular lid from above, pot from the front and side

Jug: 8th row
Funnel: 6th row
Hammer: 7th row
Car: 1st row
Slide with a box: 3rd row
Chair: 4th row
Ceiling fan: 5th row
Pot with lid: 2nd row

TTTry This — Place an object in front of a plane, such as a wall of your room. Shine a torch light on the object in a direction perpendicular to the wall. What do you see? Observe what happens to the size of the shadow as you vary the distance between your torch and your object. Why does this happen?

What you see

The shadow on the wall has nearly the same shape as the projection of the object.

It may be scaled up, stretched or slightly distorted.

Moving the torch

When the torch is close to the object, the shadow is large.

When the torch is moved away, the shadow becomes smaller.

Why

A torch sends light out in a spreading cone from a point.

The nearer the torch, the more the rays spread, so the shadow is bigger.

As the torch moves far away, the rays become almost parallel and perpendicular to the wall.

Then the shadow matches the projection of the object.

The Sun is such a faraway light source.

The shadow shrinks towards the true projection as the torch moves farther away, because the light rays become nearly parallel.

Figure it Out — Set 5

Six questions from pages 95–97 of the textbook.

Draw the top view, front view and the side view of each of the following combinations of identical cubes.

The six combinations of identical cubes (numbered left to right, top to bottom).

Number the solids 1 to 6 from left to right, top to bottom.

In each figure, the squares in a view are the faces of the cubes seen from that direction.

Top view: front of the solid at the bottom of the drawing.

Side view: seen from the right, so the front of the solid is on the left of the drawing.

Solid 1 (3 cubes)

Solid 1 with its front, top and side views.

Solid 2 (4 cubes)

Solid 2 with its front, top and side views.

Solid 3 (4 cubes)

Solid 3 with its front, top and side views.

Solid 4 (14 cubes)

Solid 4 with its front, top and side views.

Solid 5 (4 cubes)

Solid 5 with its front, top and side views.

Solid 6 (13 cubes)

Solid 6 with its front, top and side views.

The positions of the cubes are read from the textbook figures, so the views follow the cubes that can be seen in each figure.

The three views of each solid are drawn above.

Math Talk — Imagine eight identical cubes, glued together along faces to form the letter ‘C’.
(i) This looks like a ‘C’ from the front. What does it look like from the side? From the top?
(ii) Glue additional cubes to make a shape that looks like ‘C’ from the front and ‘A’ from the top.
(iii) Now, can you glue even more cubes to make it look like ‘C’ from the front, ‘A’ from the top, and ‘F’ from the side?
(iv) Can you think of other letter combinations to make with a single combination of cubes in this manner?

The letters used in the question: C from the front, A from the top and F from the side.

(i) The letter C is 3 cubes wide and 4 cubes tall, and one cube thick.

The 8-cube letter C and its views.

From the side: a vertical bar of 4 cubes (1 wide, 4 tall).

From the top: a horizontal bar of 3 cubes (3 wide, 1 deep).

(ii) The front view C and the top view A are both measured along the same left-to-right direction, so the two letters must be equally wide.

As printed, C is 3 cubes wide but A is 4 cubes wide.

So no single solid gives exactly this C from the front and this A from the top.

(iii) The depth of the top view A is the width of the side view F.

As printed, the top view A is 4 cubes deep, but the side view F is only 3 cubes wide.

So no single solid gives all three letters as printed.

Book error: as printed, the letters do not fit together. C is 3 cubes wide, A is 4 wide and 4 deep, and F is 3 wide, so (ii) and (iii) cannot be built exactly. The count is from the letter pictures in the textbook; recount them in your copy.

(iv) For two letters of the same width, fill every position \((x,y,z)\) whose front cell \((x,z)\) and top cell \((x,y)\) are both part of the letters.

Then check that no cell of either letter is left empty.

Example: ‘C’ from the front and ‘C’ from the top.

This 24-cube solid gives C from the front and C from the top.

(i) Side: a bar of 4 cubes; top: a bar of 3 cubes
(ii) Not possible with the letters as printed
(iii) Not possible with the letters as printed
(iv) Yes, e.g. C from the front and C from the top

Which solid corresponds to the given top view, front view, and side view?

The given front view, top view and side view.

Read the heights from the three views.

Front view (2 columns)

Left column is 3 squares high.

Right column is 2 squares high.

Side view (3 columns, front on the left)

Front column is 1 high.

Middle column is 1 high.

Back column is 3 high.

Top view

The back row has 2 squares, the middle row has 2 squares, and the front row has only the left square.

So the solid has a back wall, 3 high on the left and 2 high on the right, in front of which is a single layer.

The front layer is 1 high with a cube missing at the front right.

The solid that gives the given front, top and side views.

Check the options against these heights.

Options (i), (iv), (v), (vii): no step in the back wall (it is level, or the tall part is a single column).

Option (vi): a notch is cut into the top of the back wall.

Option (iii): there is a step, but its front layer has two separate cubes with a gap between them, so its front and top views are wider.

The only option with a back wall 3 high on the left and 2 high on the right, and a single front cube on the left, is (ii).

Solid (ii)

Using identical cubes, make a solid that gives the following projections.

The three sets of projections of the question.

Each row of three views in the question is one solid.

Read the heights in each view and place cubes so that all three views are matched.

Views (i), (ii), (iii) (9 cubes)

Top view: back row 2 squares, middle row 3, front row 2 with a gap in the middle.

Front view: heights 1, 2, 2.

Side view: heights 1, 1, 2 (front to back).

The solid for views (i), (ii), (iii), with its three views.

Views (iv), (v), (vi) (7 cubes)

Top view: back and middle rows 2 squares, front row the left square only.

Front view: heights 2, 2.

Side view: heights 1, 1, 2 (front to back).

The solid for views (iv), (v), (vi), with its three views.

Views (vii), (viii), (ix) (8 cubes)

Top view: back and middle rows 2 squares, front row the left square only.

Front view: heights 2, 3.

Side view: heights 1, 1, 3 (front to back).

The solid for views (vii), (viii), (ix), with its three views.

The three solids are drawn above, with 9, 7 and 8 cubes.

Find the number of cubes in this stack of identical cubes.

The stack of identical cubes.

The stack is made in layers, each smaller than the one below it.

Top layer

1 cube.

Second layer

\(1+2=3\) cubes (rows of 1 and 2).

Third layer

\(1+2+3=6\) cubes.

Bottom layer

\(1+2+3+4=10\) cubes (the bottom row visible in front has 4 cubes).

Total \(=1+3+6+10\)

\(=20\)

The cubes behind the front ones cannot be seen in the picture. The count assumes that the stack has no gaps, so each layer is a full triangle.

20 cubes

6Math Talk — What are the different shapes the projection of a cube can make under different orientations?

Looking straight at a face

The projection is a square.

Tilted about an edge direction

The projection is a rectangle.

Tilted in any other way

Opposite faces project to equal parallel figures.

The outline is a six-sided figure (hexagon).

Looking along a body diagonal

The outline is a regular hexagon (the isometric projection).

A square, a rectangle or a hexagon.

Math Talk — Construct a model of a cube and use your hands to keep it balanced on one corner vertex. Can you try to understand why all the projected edges have equal length?

The isometric projection of a cube.

Balance the cube on a corner so that the diagonal from that corner to the opposite corner is vertical.

The 3 edges at the top corner make equal angles with this vertical line.

Edges with equal angles to the vertical have equal projections on the floor.

Every other edge of the cube is a copy of one of these 3 edges, turned about the vertical line.

So all 12 edges project to the same length.

This is why the outline is a regular hexagon.

All edges make the same angle with the vertical diagonal, so their projections are equal.

Activity — Imagine these are cubes, not squares. Draw each of these on your isometric paper (you can find it at the end of the book).

The five Tetris shapes (Fig. 4.8).

The five shapes drawn as cubes, each in the same orientation as in the question.

Draw edge by edge.

Move up the page for height, along \(/\) for the depth axis and along \(\backslash\) for the length axis.

Count the units along each axis before drawing each line.

Draw faint lines first and darken only the lines that can be seen.

The five shapes are drawn above.

In-text Question — How would you draw a 2 × 2 × 2 cube?

A 1 × 1 × 1 cube drawn on the isometric grid.

A 1 × 1 × 1 cube and a 2 × 2 × 2 cube on an isometric grid.

Draw the \(1\times1\times1\) cube first: 3 faces, with 9 edges visible.

For the \(2\times2\times2\) cube, go 2 units along each of the 3 axes instead of 1.

Draw the 3 visible faces, each a rhombus of \(2\times2\) units.

Add the grid lines inside each face to show the 8 small cubes.

Draw the same figure with every edge 2 units long.

Figure it Out — Set 6

Four questions from pages 100–102 of the textbook.

Math Talk — In addition to the 5 ways shown in Fig. 4.8, are there any additional ways of gluing four cubes together along faces? Can you visualise and draw these as well?

The 5 ways of Fig. 4.8.

The 5 ways in Fig.

4.8 are flat: all four cubes lie in one layer.

Four cubes can also be glued so that they do not lie in one layer.

Three more ways of gluing four cubes.

Tripod

3 cubes are glued to the 3 faces of one cube that meet at a corner.

Left screw and right screw

Two cubes in a row, then a turn, then a cube on top.

They are mirror images of each other, and one cannot be turned into the other.

So there are 3 additional ways.

Total \(=5+3=8\) ways.

Yes. There are 3 more ways (a tripod, a left screw and a right screw), so 8 in all.

Draw the following figures on the isometric grid. [Hint: It may be useful to determine whether the edge to be currently drawn — say, along the height — goes from down to up or up to down. Accordingly, draw the line segment on the grid either in the direction of the height axis or opposite to it.]

The three figures to be drawn on the isometric grid.

Draw each figure edge by edge, with 1 unit equal to one edge of the grid.

First figure (L-shaped)

An L made of a vertical bar 3 units high and a foot 3 units long.

The L-shaped figure.

Start at the bottom front corner.

Go 3 units along the depth axis, then 3 units up the height axis.

Come back down the other face, closing each face of the bar with a line along the length axis of 1 unit.

Second figure (T-shaped)

A base 3 units long and a post 1 unit wide and 1 unit high in the middle.

The T-shaped figure.

Draw the base first: 3 units along the length axis, 1 unit up the height axis, 1 unit along the depth axis.

Then raise the post of 1 unit on the middle square.

Third figure (staircase)

3 steps, each 1 unit high and 1 unit deep.

The staircase.

Go up 1 unit, back 1 unit, up 1 unit, back 1 unit, up 1 unit.

Then complete the sides along the length axis.

The number of units in each figure is read from the textbook drawing, so check the lengths against your book.

Figures are drawn above on the isometric grid.

Math Talk — Is there anything strange about the path of this ball? Recreate it on the isometric grid. [Hint: Consider a portion of this figure that is physically realisable and identify the 3 primary directions.]

The path of the ball.

What is strange

Follow the arrows from the red ball. The ball moves along the front row, then climbs up two steps at the right.

It then walks along the top row and along the left row, and arrives back where it started.

The ball has gone up two steps and never come down, yet it is back at its starting height.

On a real solid this cannot happen, so the figure is an impossible figure, like an endless staircase.

Recreating it

Take a part of the figure that can really exist. Leave out the left row, for example.

Identify the 3 primary directions: the height (vertical lines of the grid), and the length and depth (the two slanting directions).

Draw the front row first: 5 cubes along the length direction.

At the right end, draw the two steps going up: a column of 2 cubes, then a column of 3 cubes along the depth direction.

From the top of the 3-cube column, draw the row going back along the length direction at the height of the top.

Now draw the left row of 4 cubes along the depth direction at the height of the first row, so that it joins the front row at the corner where the ball starts.

The figure recreated on the isometric grid.

The left row is drawn so that it looks joined to the top row. A real model cannot have both, because the top row is higher than the front row.

The ball goes up step after step but returns to its start, which is impossible in a real solid.

Math Talk — Observe this triangle.
(i) Would it be possible to build a model out of actual cubes? What are the front, top, and side profiles of this impossible triangle?
(ii) Recreate this on an isometric grid.
(iii) Why does the illusion work?

The triangle of the question.

(i) No, a real model cannot be built from actual cubes.

The triangle is made of 3 straight arms of 5 cubes, one along each of the 3 primary directions: height, length and depth.

Take the vertical arm, the arm along the length direction at its top, and the arm along the depth direction at its bottom.

The top arm and the bottom arm are in different places in space, and they do not touch.

Each corner looks joined only because the picture shows one point where a near point and a far point lie on the same line of sight.

The profiles of the arms, seen from the front, the top and the side, are drawn below.

The front, top and side views of the arms.

From the front: a vertical bar of 5 cubes with a bar of 4 cubes at its top.

From the top: a bar of 5 cubes in the front row with a bar of 4 cubes going back from its left end.

From the side: a vertical bar of 5 cubes with a bar of 4 cubes at its bottom.

None of the three views shows a closed triangle.

(ii) Recreate the triangle on the isometric grid.

Draw one arm of 5 cubes along the height axis.

Draw the second arm of 4 more cubes along the length axis, starting at the top of the first arm.

Draw the third arm of 4 more cubes along the depth axis, starting at the bottom of the first arm.

Draw the arms so that they appear to join at each corner.

The impossible triangle recreated on the isometric grid.

(iii) The illusion works because a flat picture loses depth.

In an isometric drawing, a step along the length axis, then a step along the depth axis in the opposite direction, then a step up the height axis, bring you back to the same spot on the paper.

So the end of the top arm and the end of the bottom arm can be drawn at the same place, although they are far apart in space.

Our brain reads each corner on its own as a proper 3D corner, so the 3 corners each look correct, though together they cannot exist.

(i) No
(ii) Draw 3 arms along the 3 axes
(iii) The picture loses depth, so each corner looks right but the whole cannot exist.

Continue with Ganita Prakash, Part II

Move on to Chapter 5: Tales by Dots and Lines, or head back to the Ganita Prakash hub for the complete chapter list.

Chapter 5: Tales by Dots and Lines Ganita Prakash Hub

Part II, Chapter 4Exploring Some Geometric Themes

Sierpinski Carpet

Sierpinski Gasket

Figure it Out — Set 1

Koch Snowflake

Figure it Out — Set 2

Build it in Your Imagination

Making Solids

Figure it Out — Set 3

Shortest Paths on a Cuboid

Figure it Out — Set 4

Figure it Out — Set 5

Figure it Out — Set 6

Continue with Ganita Prakash, Part II