CBSE Class 7 Mathematics Chapter 15: Visualising Solid Shapes NCERT Solutions

To determine which nets can form a cube, we need to visualize how they fold. A cube has 6 square faces. Let's examine each net:

Net (i): This arrangement has 4 squares in a line. If you try to fold this, the two end squares will overlap, and it won't form a closed cube.

Net (ii): This net has 4 squares in a line with one square attached above and one below the second square. This can be folded to form a cube.

Net (iii): This net has 3 squares in a line, with two squares above and one below, arranged to form a closed shape. This can be folded to form a cube.

Net (iv): This net has a 2x3 grid arrangement where one square is missing. This specific arrangement allows for folding into a cube.

Net (v): This net has 4 squares in a line. Similar to net (i), the end squares will overlap, preventing it from forming a cube.

Net (vi): This net has 3 squares in a line, with one square above and two squares below. This arrangement can be folded to form a cube.

Therefore, the nets that can be used to make cubes are (ii), (iii), (iv), and (vi).

We are given that the sum of dots on opposite faces of a die must be 7. Let's analyze each net:

Net (i):

The net shows numbers 3, 2, and 4. Let's assume 3 is the top face. Then the bottom face must be 4 (since 3+4=7). The front face is 2. Its opposite face (back) must be 5 (since 2+5=7). The remaining two faces are the left and right sides. If we place 6 on one side, the opposite side must have 1 (since 6+1=7). The diagram shows 6 and 1 are adjacent, which is possible. So, a possible arrangement is: Top=3, Bottom=4, Front=2, Back=5, Left=6, Right=1.

Net (ii):

The net shows numbers 5, 3, and 4. Let's assume 5 is the top face. Then the bottom face must be 2 (since 5+2=7). The front face is 3. Its opposite face (back) must be 4 (since 3+4=7). The remaining two faces are the left and right sides. If we place 6 on one side, the opposite side must have 1 (since 6+1=7). The diagram shows 6 and 1 are adjacent, which is possible. So, a possible arrangement is: Top=5, Bottom=2, Front=3, Back=4, Left=6, Right=1.

The blanks can be filled as follows:

Net (i): The blank opposite 3 should be 4. The blank opposite 2 should be 5. The blank opposite 6 should be 1.

Net (ii): The blank opposite 5 should be 2. The blank opposite 3 should be 4. The blank opposite 6 should be 1.

No, this cannot be a net for a standard die. A standard die has faces numbered such that opposite faces always sum to 7.

Let's analyze the given net. It appears to be a strip of 5 squares. When folded, the two end squares will become opposite faces, and the three middle squares will form adjacent faces. If we consider the arrangement as given, let's try to identify opposite pairs. If the squares are arranged linearly, the first and the last square would be opposite. In this net, we have 5 squares. Let's assume the squares are numbered sequentially as they appear. If we number them 1 to 5 from left to right, then square 1 and square 5 would be opposite, and square 2 and square 4 would be opposite. The middle square (3) would have its opposite face determined by how the net folds. However, a standard die net usually has a specific structure allowing for clear opposite pairs. A common net structure is 4 squares in a row with one above and one below. If this net is interpreted as a linear strip, and we try to form a cube, we run into issues. For example, if 2 is at one end and 5 at the other, their sum is 7. If 3 is next to 2, and 4 is next to 5, then 3 and 4 might be opposite, summing to 7. However, the number 6 is also present. A cube must have 6 faces. This net only shows 5 distinct numbers. If we assume this is a net for a die, and try to assign numbers to faces, we find contradictions. For instance, if we try to fold this, the two end squares (2 and 5) could be opposite (sum=7). The squares 3 and 4 could be opposite (sum=7). But then where does 6 go? If 6 is adjacent to 3 and 4, it must have an opposite face. This net structure is problematic for forming a standard die. A common issue with such linear nets is that they often result in overlapping faces or gaps when folded into a cube. For a net to be valid for a die, it must be possible to assign numbers 1 through 6 such that opposite faces sum to 7. This net, with the numbers shown, does not allow for such an assignment in a way that forms a closed cube with opposite faces summing to 7.

More specifically, if we consider a common net structure like 4 squares in a row, with one above and one below, this net doesn't fit that. If we try to fold this linear strip, the ends (2 and 5) would be opposite (sum=7). The next pair (3 and 4) would be opposite (sum=7). This leaves the number 6 without an opposite face that sums to 7, or it implies that the net is incomplete or incorrectly formed for a die.

The given incomplete net consists of 3 squares arranged in an 'L' shape. A cube has 6 faces. To complete the net, we need to add 3 more squares such that they can be folded to form a closed cube.

Method 1: Adding squares to form a row

We can attach three more squares to the existing structure. Let's attach two squares adjacent to one of the outer squares of the 'L', extending the shape into a row of 4 squares. Then, we can attach the remaining two squares, one above and one below the second square in the row. This forms a standard cube net.

Diagram 1: Start with the given 3 squares. Add two squares to the right of the top square of the 'L'. Then add one square above the second square from the left, and one square below the second square from the left.

Method 2: Adding squares to form a different shape

Alternatively, we can add the squares in a different configuration. Starting with the given 3 squares, we can add one square adjacent to the top square, and two squares adjacent to the bottom square of the 'L', forming a shape that can also fold into a cube.

Diagram 2: Start with the given 3 squares. Add one square adjacent to the top square. Then add two squares adjacent to the bottom square, extending downwards.

In both completed diagrams, we have added 3 squares to the original 3, making a total of 6 squares, which can be folded to form a cube.