CBSE Class 8 Maths Exemplar Chapter 11: Mensuration NCERT Solutions

When a cube of side 5 cm is sliced into 1 cm cubes, the total number of smaller cubes is $5 \times 5 \times 5 = 125$.

The cubes with exactly one face painted are those located in the center of each face of the original cube. These cubes are not on the edges or corners.

For a cube of side 'n' cm cut into 1 cm cubes, the number of cubes with exactly one face painted is given by the formula $6 \times (n-2)^2$, where 'n' is the side length of the larger cube.

In this case, $n=5$. So, the number of cubes with exactly one face painted is:

$$6\; \backslash times\; (5-2)^2\; =\; 6\; \backslash times\; (3)^2\; =\; 6\; \backslash times\; 9\; =\; 54$$

Therefore, 54 small cubes will have exactly one face painted.

The original cube has a side length of 4 cm.

The surface area of the original cube is calculated using the formula $6 \times side^2$.

Surface Area (Original) = $6\; \backslash times\; (4\; \backslash text\{\; cm\})^2\; =\; 6\; \backslash times\; 16\; \backslash text\{\; cm\}^2\; =\; 96\; \backslash text\{\; cm\}^2$

The cube is cut into 1 cm cubes. The total number of 1 cm cubes is $4 \times 4 \times 4 = 64$ cubes.

Each small cube has a side length of 1 cm.

The surface area of one small cube is $6 \times (1 \text{ cm})^2 = 6 \text{ cm}^2$.

The total surface area of all 64 cut-out cubes is $64 \times 6 \text{ cm}^2 = 384 \text{ cm}^2$.

The ratio of the surface area of the original cube to the total surface area of the cut-out cubes is:

$$\backslash frac\{\backslash text\{Surface\; Area\; (Original)\}\}\{\backslash text\{Total\; Surface\; Area\; (Cut-out\; Cubes)\}\}\}\; =\; \backslash frac\{96\; \backslash text\{\; cm\}^2\}\{384\; \backslash text\{\; cm\}^2\}$$

Simplifying the ratio:

$$\backslash frac\{96\}\{384\}\; =\; \backslash frac\{1\}\{4\}$$

Thus, the required ratio is 1:4.

Let the side length of the original square sheet be 'a' units.

The area of the original square is $a^2$ square units.

When a circle of the maximum possible size is cut from this square, the diameter of the circle will be equal to the side of the square, which is 'a' units. The radius of this circle is $\backslash frac\{a\}\{2\}$ units.

Now, a square of the maximum possible size is cut from this circle. The diagonal of this inscribed square will be equal to the diameter of the circle, which is 'a' units.

Let the side length of this inner square be 's'. Using the Pythagorean theorem for the diagonal of a square ($d^2 = s^2 + s^2 = 2s^2$), we have:

$$a^2\; =\; 2s^2$$

Solving for $s^2$, which represents the area of the inner square:

$$s^2\; =\; \backslash frac\{a^2\}\{2\}$$

The area of the final square is $\backslash frac\{a^2\}\{2\}$ square units.

Comparing this to the area of the original square ($a^2$), we find that the area of the final square is $\backslash frac\{1\}\{2\}$ of the original square's area.

The area of a triangle is given by the formula: $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash text\{base\}\; \backslash times\; \backslash text\{height\}$.

Consider a rectangle with length 'l' and width 'w'.

To form the largest possible triangle within this rectangle, we can choose one side of the rectangle as the base of the triangle and the corresponding height of the rectangle as the height of the triangle.

For example, if we take the length 'l' as the base of the triangle, the maximum height the triangle can have within the rectangle is the width 'w'.

Area of the triangle = $\backslash frac\{1\}\{2\}\; \backslash times\; \backslash text\{base\}\; \backslash times\; \backslash text\{height\}\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; l\; \backslash times\; w\; =\; \backslash frac\{lw\}\{2\}$.

Alternatively, if we take the width 'w' as the base, the maximum height would be the length 'l', giving the same area $\backslash frac\{wl\}\{2\}$.

Therefore, the area of the largest triangle that can be fitted into a rectangle of length l and width w is $\backslash frac\{lw\}\{2\}$.