Fractals and Visualising Solids | IT

Question 27

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

Question diagram 1
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Solution

To find the shortest path on the surface of a cuboid, we unfold the faces into a flat 2D plane and draw a straight line between the two points.

Step 1 — Understand the cuboid and positions

Let us identify the dimensions of the cuboid and the exact positions of the ant and the laddu. The cuboid has a length of 30 cm. The end faces (where the ant and laddu are) are squares with sides of 12 cm. So, the width of the box is 12 cm and the height of the box is 12 cm.

The laddu is on the back face: It is 6 cm from the left edge of that face. It is 1 cm from the bottom edge of that face.

The ant is on the front face: It is 6 cm from the left edge of that face. It is 1 cm from the top edge of that face.

Diagram 1

Step 2 — Unfold the cuboid

We need to find a path from the back face to the front face. The shortest path will involve crossing one of the four side faces (top, bottom, left, or right) that connect the two end faces. Let us consider unfolding the cuboid over the top face.

Imagine we flatten the back face, the top face, and the front face into a single rectangle. The total "horizontal" distance in this unfolded rectangle will be the length of the cuboid. The total "vertical" distance will be the sum of the distances travelled on each face in that direction.

Let us calculate the horizontal distance: This is the length of the cuboid. Horizontal distance=Length of cuboid\text{Horizontal distance} = \text{Length of cuboid} =30 cm= \mathbf{30 \text{ cm}}

Now, let us calculate the vertical distance for the path over the top face: The path starts at the laddu on the back face. It goes up to the top edge of the back face. Distance from laddu to top edge of back face=Height of face−Laddu’s distance from bottom\text{Distance from laddu to top edge of back face} = \text{Height of face} - \text{Laddu's distance from bottom} =12 cm−1 cm= 12 \text{ cm} - 1 \text{ cm} =11 cm= 11 \text{ cm} Then, the path crosses the top face. The width of the top face is the same as the width of the cuboid. Width of top face=12 cm\text{Width of top face} = \mathbf{12 \text{ cm}} Finally, the path goes down from the top edge of the front face to the ant. Distance from top edge of front face to ant=Ant’s distance from top\text{Distance from top edge of front face to ant} = \text{Ant's distance from top} =1 cm= \mathbf{1 \text{ cm}} The total vertical distance for this path is the sum of these segments. Total vertical distance=11 cm+12 cm+1 cm\text{Total vertical distance} = 11 \text{ cm} + 12 \text{ cm} + 1 \text{ cm} =24 cm= \mathbf{24 \text{ cm}}

Step 3 — Calculate the shortest path length

On the unfolded 2D plane, the shortest path is a straight line. This forms the hypotenuse of a right-angled triangle. We use the Pythagorean theorem, where the two shorter sides are the horizontal and vertical distances we calculated. Let dd be the shortest path length. d2=(Horizontal distance)2+(Vertical distance)2d^2 = (\text{Horizontal distance})^2 + (\text{Vertical distance})^2 d2=(30 cm)2+(24 cm)2d^2 = (30 \text{ cm})^2 + (24 \text{ cm})^2 d2=900 cm2+576 cm2d^2 = 900 \text{ cm}^2 + 576 \text{ cm}^2 d2=1476 cm2d^2 = 1476 \text{ cm}^2 d=1476 cmd = \sqrt{1476} \text{ cm} To simplify the square root: 1476=36×41\sqrt{1476} = \sqrt{36 \times 41} =36×41= \sqrt{36} \times \sqrt{41} =641= 6 \sqrt{41}

We can check other paths (over the bottom face, or either side face). Due to the symmetry of the cuboid and the ant/laddu positions, all these paths will result in the same horizontal distance (30 cm) and vertical distance (24 cm), leading to the same shortest path length.

Answer

The length of the shortest path between the ant and the laddu is 641 cm6\sqrt{41} \text{ cm}.

More questions in IT

Q1

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

By its construction, each step in the sequence has (i) squares of the same size that remain in the figure, and the size of these squares becomes smaller and smaller as the step number increases, and (ii) square holes that are formed by removing square pieces.

Q2

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.]

This fractal is called the Sierpinski Triangle/Gasket.

Q3

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Q4

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.

  1. A solid whose profile has a square outline
  2. A solid whose profile has a circular outline
  3. A solid whose profile has a triangular outline
Q5

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?

Spend some time on this, and if you are finding it difficult to visualise, you may look around and use objects that are around you, or that you will make in the next section. Feel free to consider viewpoints from any direction, including directly above the object.

  1. A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint
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  3. A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint
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  5. A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint

Are there unique solids for each of the conditions, or can you come up with multiple possibilities?

Q6

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 nn sides?

Q7

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 nn-sided polygon?

Q8

What is a net of a cube?

Q9

Visualise how it can be folded to form a cube.

Q10

What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron?

Q11

Are there any other possible nets?

Q12

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

Q13

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

Q14

What is the net of a cylinder?

If the circular faces of a cylinder are unfolded, and if a cut is made along the height of the cylinder, as shown in the figure below, then we get

Q15

What are the sidelengths of the rectangle obtained?

Q16

How will the net of a cone look?

Q17

If the cone is slit open along the line ll and then unrolled, what will we get?

Observe that all the points on the boundary of the base circle are at equal distances from 'O'. So after unrolling the cone, the boundary of the net will be a portion of a circle with centre O.

Q18

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!

Q19

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

Q20

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

Q21

What is the shortest path for the ant to reach the laddu?

Q22

What about in the following case?

Q23

If we think that a certain path is the shortest, how can we be sure that it truly is, among all the infinite possibilities?

Q24

For example, are either of these the shortest path?

Q25

What does this show?

Q26

Have we now completely analysed the problem of finding the shortest path between two points on a cuboid?

Q27

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

Q29

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Q30

Can you now compare the lengths pp and ll?

Q31

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Q32

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Q33

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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.

Q34

What can you say about the projection of an nn-sided regular polygon?

[Hint: Projection of a polygon is composed of the projections of its sides.]

Q35

How would the projections of a cube and a cone look?

Q36

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Q37

Find another object that makes the same projection as that of a given cone.

Q39

Have you played Tetris? There are five basic shapes in Tetris, corresponding to the different ways of arranging four squares.

Q40

Context: In Fig. 4.8, there are five basic shapes in Tetris, corresponding to the different ways of arranging four squares.

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

Q41

Why is this correspondence between directions on isometric paper and axes of the solid so effective for communicating the shape of the solid?

Q42

Can you try drawing the other tetris shapes on isometric paper?

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