Chapter 10 Mensuration (Additional Questions)

The perimeter of a closed figure is defined as the total length of its boundary.

Therefore, the correct option is (B) Boundary.

Given:

Length of the rectangle, $l = 10$ cm

Width of the rectangle, $w = 5$ cm

To Find:

Perimeter of the rectangle.

Solution:

The formula for the perimeter of a rectangle is:

$P = 2 \times (\text{length} + \text{width})$

$P = 2 \times (l + w)$

Substitute the given values of length and width:

$P = 2 \times (10 \text{ cm} + 5 \text{ cm})$

$P = 2 \times (15 \text{ cm})$

$P = 30 \text{ cm}$

The perimeter of the rectangle is $30$ cm.

Therefore, the correct option is (B) $30$ cm.

A square is a closed figure with four equal sides.

The perimeter of any closed figure is the length of its boundary.

For a square with side length 's', the perimeter is the sum of the lengths of its four sides.

Perimeter = side + side + side + side

Perimeter = $s + s + s + s$

Perimeter = $4 \times s$

The formula for the perimeter of a square with side 's' is $4s$.

Therefore, the correct option is (C) $4s$.

Given:

The lengths of the sides of the triangular park are $12$ m, $15$ m, and $18$ m.

To Find:

The perimeter of the triangular park.

Solution:

The perimeter of a triangle is the sum of the lengths of its three sides.

Perimeter = Sum of the lengths of the three sides

Perimeter = $12$ m $+ 15$ m $+ 18$ m

Perimeter = $45$ m

The perimeter of the triangular park is $45$ m.

Therefore, the correct option is (B) $45$ m.

Given:

Length of the rectangular field, $l = 50$ m

Width of the rectangular field, $w = 30$ m

Number of rounds of wire for fencing = $3$

To Find:

Total length of wire needed for fencing.

Solution:

First, we need to find the perimeter of the rectangular field, as the wire is used along the boundary.

The formula for the perimeter of a rectangle is:

$P = 2 \times (l + w)$

Substitute the given values of length and width:

$P = 2 \times (50 \text{ m} + 30 \text{ m})$

$P = 2 \times (80 \text{ m})$

$P = 160 \text{ m}$

The perimeter of the rectangular field is $160$ m.

The farmer wants to fence the field with 3 rounds of wire. So, the total length of wire needed is 3 times the perimeter.

Total length of wire = Number of rounds $\times$ Perimeter

Total length of wire = $3 \times 160 \text{ m}$

To calculate $3 \times 160$:

Total length of wire = $480$ m

The total length of wire needed is $480$ m.

Therefore, the correct option is (C) $480$ m.

Given:

Side length of the square field, $s = 20$ m

To Find:

Area of the square field.

Solution:

The formula for the area of a square with side length 's' is:

Area $= \text{side} \times \text{side}$

Area $= s \times s$

Area $= s^2$

Substitute the given side length, $s = 20$ m:

Area $= (20 \text{ m})^2$

Area $= 20 \text{ m} \times 20 \text{ m}$

Area $= 400$ $m^2$

The area of the square field is $400$ $m^2$.

Therefore, the correct option is (C) $400$ $m^2$.

The area of a closed figure is the measure of the region enclosed by its boundary.

Perimeter is the length of the boundary.

Length is a one-dimensional measure.

Volume is the space occupied by a three-dimensional object.

Therefore, the amount of surface enclosed by a closed figure is called its Area.

The correct option is (C) Area.

Given:

Length of the rectangle = $l$

Width of the rectangle = $w$

To Find:

The formula for the area of the rectangle.

Solution:

The area of a rectangle represents the space it occupies within its boundary. It is calculated by multiplying its length and its width.

Area of rectangle = Length $\times$ Width

Area $= l \times w$

Therefore, the formula for the area of a rectangle with length 'l' and width 'w' is $l \times w$.

The correct option is (C) $l \times w$.

Given:

Length of the classroom floor, $L = 8$ m

Width of the classroom floor, $W = 6$ m

Side length of the square tile, $s = 0.5$ m

To Find:

Number of tiles needed to cover the floor.

Solution:

First, we need to find the area of the classroom floor.

Area of floor = Length $\times$ Width

Area of floor $= L \times W$

Area of floor $= 8 \text{ m} \times 6 \text{ m}$

Area of floor $= 48 \text{ m}^2$

Next, we find the area of one square tile.

Area of one square tile = Side $\times$ Side

Area of one square tile $= s \times s = s^2$

Area of one square tile $= (0.5 \text{ m})^2$

Area of one square tile $= 0.5 \text{ m} \times 0.5 \text{ m}$

Area of one square tile $= 0.25 \text{ m}^2$

The number of tiles needed to cover the floor is the ratio of the area of the floor to the area of one tile.

Number of tiles = $\frac{\text{Area of floor}}{\text{Area of one tile}}$

Number of tiles $= \frac{48 \text{ m}^2}{0.25 \text{ m}^2}$

To simplify the division, multiply the numerator and denominator by 100:

Number of tiles $= \frac{48 \times 100}{0.25 \times 100}$

Number of tiles $= \frac{4800}{25}$

Performing the division:

$4800 \div 25 = 192$

Number of tiles $= 192$

The number of tiles needed to cover the classroom floor is 192.

Therefore, the correct option is (B) 192.

We are asked to identify the unit typically used for measuring the area of a large field.

Let's examine the given options:

• (A) cm (centimeter) is a unit of length.
• (B) $cm^2$ (square centimeter) is a unit of area, suitable for small areas like paper or small objects.
• (C) $m^2$ (square meter) is a unit of area, suitable for moderate to large areas like rooms, houses, or fields.
• (D) km (kilometer) is a unit of length.

Area is measured in square units. Therefore, options (A) and (D) are incorrect as they are units of length.

Comparing $cm^2$ and $m^2$, a square meter is significantly larger than a square centimeter ($1$ $m^2$ = $100 \times 100$ $cm^2$ = $10,000$ $cm^2$).

For a large field, a larger unit of area is required. While units like hectares ($10,000$ $m^2$) or square kilometers ($km^2$) are often used for very large land areas, among the given options, $m^2$ is the most appropriate unit for measuring the area of a large field compared to $cm^2$.

Therefore, the correct option is (C) $m^2$.

Given:

Perimeter of the square = $48$ cm

To Find:

The length of the side of the square.

Solution:

The formula for the perimeter of a square with side length 's' is:

Perimeter $= 4 \times \text{side}$

Perimeter $= 4s$

We are given that the perimeter is $48$ cm. So, we can set up the equation:

To find the side 's', we need to divide the perimeter by 4.

Divide both sides of the equation (i) by 4:

$\frac{48 \text{ cm}}{4} = \frac{4s}{4}$

$\frac{\cancel{48}^{12} \text{ cm}}{\cancel{4}_{1}} = s$

$12 \text{ cm} = s$

The length of the side of the square is $12$ cm.

The length of its side is $12$ cm.

Therefore, the correct option is (A) $12$ cm.

Given:

Area of the rectangle = $60$ $cm^2$

Length of the rectangle, $l = 12$ cm

To Find:

The width of the rectangle, $w$.

Solution:

The formula for the area of a rectangle is:

Area $= \text{length} \times \text{width}$

Area $= l \times w$

We are given the area and the length. We can use the formula to find the width:

To find 'w', we need to divide the area by the length.

Divide both sides of equation (i) by $12$ cm:

$\frac{60 \text{ cm}^2}{12 \text{ cm}} = \frac{12 \text{ cm} \times w}{12 \text{ cm}}$

$\frac{\cancel{60}^{5} \text{ cm}^2}{\cancel{12}_{1} \text{ cm}} = w$

$5 \text{ cm} = w$

The width of the rectangle is $5$ cm.

The width of the rectangle is $5$ cm.

Therefore, the correct option is (A) $5$ cm.

Given:

The figure is a regular pentagon.

Side length of the regular pentagon, $s = 7$ cm

To Find:

The perimeter of the regular pentagon.

Solution:

A regular pentagon is a polygon with 5 equal sides and 5 equal angles.

The perimeter of any polygon is the sum of the lengths of its sides.

For a regular pentagon with side length 's', the perimeter is the sum of the lengths of its 5 equal sides.

Perimeter = side + side + side + side + side

Perimeter = $s + s + s + s + s$

Perimeter = $5 \times s$

Substitute the given side length, $s = 7$ cm:

Perimeter = $5 \times 7 \text{ cm}$

Perimeter = $35 \text{ cm}$

The perimeter of the regular pentagon is $35$ cm.

Therefore, the correct option is (B) $35$ cm.

Area is a measure of the two-dimensional space occupied by a shape. Units of area are derived from units of length by squaring them.

Let's examine each option:

• (A) cm (centimeter) is a unit of length.
• (B) $m^2$ (square meter) is a unit derived from the meter (a unit of length) by squaring. It is a unit of area.
• (C) km (kilometer) is a unit of length.
• (D) $cm^2$ (square centimeter) is a unit derived from the centimeter (a unit of length) by squaring. It is a unit of area.

Units used for measuring area are typically in the form of (unit of length)$^2$.

Based on the analysis, $m^2$ and $cm^2$ are units used for measuring area.

Therefore, the correct options are (B) $m^2$ and (D) $cm^2$.

The perimeter of any closed figure is the total length of its boundary.

For a polygon (whether regular or irregular), the boundary is formed by its sides.

Therefore, the perimeter of an irregular polygon is the sum of the lengths of its sides.

Let an irregular polygon have sides with lengths $s_1, s_2, s_3, \dots, s_n$.

The perimeter $P$ is given by:

$P = s_1 + s_2 + s_3 + \dots + s_n = \sum_{i=1}^{n} s_i$

Thus, the perimeter of an irregular polygon is the sum of the lengths of its Sides.

The correct option is (B) Sides.

Given:

Length of the rectangular piece of land, $l = 20$ m

Width of the rectangular piece of land, $w = 15$ m

Cost of fencing per metre = $\textsf{₹}$ 50

To Find:

Total cost of fencing the land.

Solution:

Fencing is done along the boundary of the land, so we need to calculate the perimeter of the rectangular land.

The formula for the perimeter of a rectangle is:

Perimeter $= 2 \times (\text{length} + \text{width})$

Perimeter $= 2 \times (l + w)$

Substitute the given values of length and width:

Perimeter $= 2 \times (20 \text{ m} + 15 \text{ m})$

Perimeter $= 2 \times (35 \text{ m})$

Perimeter $= 70 \text{ m}$

The perimeter of the rectangular land is $70$ m.

The cost of fencing is $\textsf{₹}$ 50 per metre.

Total cost of fencing = Perimeter $\times$ Cost per metre

Total cost of fencing = $70 \text{ m} \times \textsf{₹} \ 50/\text{m}$

Total cost of fencing = $70 \times 50$ $\textsf{₹}$

Total cost of fencing = $3500$ $\textsf{₹}$

The total cost of fencing the land is $\textsf{₹}$ 3500.

Therefore, the correct option is (B) $\textsf{₹}$ 3500.

Given:

Length of the rectangular piece of land, $l = 20$ m

Width of the rectangular piece of land, $w = 15$ m

To Find:

The area of the rectangular land.

Solution:

The formula for the area of a rectangle is:

Area $= \text{length} \times \text{width}$

Area $= l \times w$

Substitute the given values of length and width:

Area $= 20 \text{ m} \times 15 \text{ m}$

Area $= 300 \text{ m}^2$

The area of the rectangular land is $300$ $m^2$.

Therefore, the correct option is (C) $300$ $m^2$.

Assertion (A): Two squares with the same perimeter must have the same area.

Evaluation of Assertion (A):

Let the side length of the first square be $s_1$ and the side length of the second square be $s_2$.

The perimeter of the first square is $P_1 = 4s_1$.

The perimeter of the second square is $P_2 = 4s_2$.

If the perimeters are the same, $P_1 = P_2$, then:

Dividing both sides by 4, we get:

So, two squares with the same perimeter must have the same side length.

The area of the first square is $A_1 = s_1^2$.

The area of the second square is $A_2 = s_2^2$.

Since $s_1 = s_2$ from equation (i), it follows that $s_1^2 = s_2^2$.

Thus, $A_1 = A_2$.

Therefore, Assertion (A) is True.

Reason (R): If the perimeters are equal, their sides must be equal, and thus their areas will also be equal.

Evaluation of Reason (R):

As shown in the evaluation of Assertion (A), if the perimeters of two squares are equal ($4s_1 = 4s_2$), it mathematically implies that their sides must be equal ($s_1 = s_2$).

Also, if the sides of two squares are equal ($s_1 = s_2$), it implies that their areas must be equal ($s_1^2 = s_2^2$).

The reason correctly states the relationship between equal perimeters leading to equal sides and equal sides leading to equal areas for squares.

Therefore, Reason (R) is True.

Relationship between A and R:

Reason (R) provides the logical steps that explain why two squares with the same perimeter must have the same area. The equality of perimeters implies the equality of sides, and the equality of sides implies the equality of areas. Reason (R) directly supports and justifies Assertion (A).

Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).

Therefore, the correct option is (A) Both A and R are true and R is the correct explanation of A.

We need to match each shape with its correct perimeter formula:

(i) Square: A square has 4 equal sides. The perimeter is the sum of the lengths of its sides, which is $s+s+s+s = 4s$.

Match (i) with (d).

(ii) Rectangle: A rectangle has 4 sides with opposite sides equal in length. The perimeter is the sum of the lengths of its sides, which is $l+w+l+w = 2l+2w = 2(l+w)$.

Match (ii) with (a).

(iii) Equilateral Triangle: An equilateral triangle has 3 equal sides. The perimeter is the sum of the lengths of its sides, which is $s+s+s = 3s$.

Match (iii) with (c).

(iv) Regular Hexagon: A regular hexagon is a polygon with 6 equal sides. The perimeter is the sum of the lengths of its sides, which is $s+s+s+s+s+s = 6s$.

Match (iv) with (b).

The correct matching is:

(i) - (d)

(ii) - (a)

(iii) - (c)

(iv) - (b)

Checking the given options, this matching corresponds to option (A).

Therefore, the correct option is (A) (i)-(d), (ii)-(a), (iii)-(c), (iv)-(b).

We need to match each shape with its correct perimeter formula:

(i) Square: A square has 4 equal sides. The perimeter is the sum of the lengths of its sides, which is $s+s+s+s = 4s$.

Match (i) with (d).

(ii) Rectangle: A rectangle has 4 sides with opposite sides equal in length. The perimeter is the sum of the lengths of its sides, which is $l+w+l+w = 2l+2w = 2(l+w)$.

Match (ii) with (a).

(iii) Equilateral Triangle: An equilateral triangle has 3 equal sides. The perimeter is the sum of the lengths of its sides, which is $s+s+s = 3s$.

Match (iii) with (c).

(iv) Regular Hexagon: A regular hexagon is a polygon with 6 equal sides. The perimeter is the sum of the lengths of its sides, which is $s+s+s+s+s+s = 6s$.

Match (iv) with (b).

The correct matching is:

(i) - (d)

(ii) - (a)

(iii) - (c)

(iv) - (b)

Checking the given options, this matching corresponds to option (A).

Therefore, the correct option is (A) (i)-(d), (ii)-(a), (iii)-(c), (iv)-(b).

Given:

The side of a square is doubled.

To Find:

How the area of the square changes.

Solution:

Let the original side length of the square be $s$.

The original area of the square is $A_{original} = s \times s = s^2$.

Now, the side of the square is doubled. The new side length is $2 \times s = 2s$.

The new area of the square is $A_{new} = (\text{new side}) \times (\text{new side}) = (2s) \times (2s)$.

$A_{new} = 4s^2$

We can compare the new area to the original area:

$A_{new} = 4s^2 = 4 \times (s^2) = 4 \times A_{original}$

The new area is 4 times the original area.

This means the area is quadrupled.

When the side of a square is doubled, its area becomes 4 times the original area.

Therefore, the correct option is (C) It is quadrupled (multiplied by 4).

Given:

Two squares of side length $5$ cm are joined along one side.

To Find:

The perimeter of the resulting shape.

Solution:

When two squares of side $5$ cm are joined along one side, the resulting shape is a rectangle.

Let the side of each square be $s = 5$ cm.

When they are joined along one side, the length of the resulting rectangle will be the sum of the sides of the two squares that are now adjacent and form the length.

Length of the rectangle, $l = s + s = 5 \text{ cm} + 5 \text{ cm} = 10 \text{ cm}$.

The width of the resulting rectangle will be the side of the square that is not used for joining.

Width of the rectangle, $w = s = 5 \text{ cm}$.

The resulting shape is a rectangle with length $10$ cm and width $5$ cm.

The perimeter of a rectangle is given by the formula:

Perimeter $= 2 \times (l + w)$

Substitute the values of length and width:

Perimeter $= 2 \times (10 \text{ cm} + 5 \text{ cm})$

Perimeter $= 2 \times (15 \text{ cm})$

Perimeter $= 30 \text{ cm}$

Alternatively, consider the sides that form the perimeter of the combined shape. Each square has 4 sides. When two squares are joined along one side, one side from each square becomes an internal boundary and is not part of the perimeter of the combined shape. So, the sides contributing to the perimeter are the remaining sides of the two squares.

Number of sides on the perimeter = (Number of sides of first square) + (Number of sides of second square) - 2 $\times$ (Number of joined sides)

Number of sides on the perimeter = $4 + 4 - 2 \times 1 = 8 - 2 = 6$

The lengths of these 6 sides are equal to the side length of the original squares, which is $5$ cm.

Perimeter = Sum of the lengths of the sides on the boundary

Perimeter = $5 \text{ cm} + 5 \text{ cm} + 5 \text{ cm} + 5 \text{ cm} + 5 \text{ cm} + 5 \text{ cm}$

Perimeter = $6 \times 5 \text{ cm}$

Perimeter = $30 \text{ cm}$

The perimeter of the shape formed by joining two squares of side 5 cm along one side is $30$ cm.

Therefore, the correct option is (A) $30$ cm.

Let's evaluate each statement:

(A) Area is measured in square units.

Area represents the amount of surface enclosed by a 2D figure. It is calculated by multiplying two lengths, so its units are squared (e.g., $cm^2$, $m^2$, $km^2$). This statement is True.

(B) Perimeter is measured in linear units.

Perimeter is the total length of the boundary of a 2D figure. It is a linear measure (e.g., cm, m, km). This statement is True.

(C) Two rectangles with the same area must have the same perimeter.

Let's consider a counterexample.

Rectangle 1: Length $l_1 = 12$ cm, Width $w_1 = 5$ cm.

Area $A_1 = l_1 \times w_1 = 12 \text{ cm} \times 5 \text{ cm} = 60$ $cm^2$.

Perimeter $P_1 = 2(l_1 + w_1) = 2(12 \text{ cm} + 5 \text{ cm}) = 2(17 \text{ cm}) = 34$ cm.

Rectangle 2: Length $l_2 = 10$ cm, Width $w_2 = 6$ cm.

Area $A_2 = l_2 \times w_2 = 10 \text{ cm} \times 6 \text{ cm} = 60$ $cm^2$.

Perimeter $P_2 = 2(l_2 + w_2) = 2(10 \text{ cm} + 6 \text{ cm}) = 2(16 \text{ cm}) = 32$ cm.

Here, we have two rectangles with the same area ($60$ $cm^2$) but different perimeters ($34$ cm and $32$ cm).

Therefore, this statement is False.

(D) The perimeter of a regular polygon is the number of sides multiplied by the length of one side.

A regular polygon has equal sides and equal angles. If 'n' is the number of sides and 's' is the length of each side, the perimeter is the sum of the lengths of all n sides, which is $n \times s$. This statement is True.

The false statement is (C).

Therefore, the correct option is (C) Two rectangles with the same area must have the same perimeter.

We are given that the length ($l$) and width ($w$) of a rectangle are measured in metres (m).

The formula for the area of a rectangle is:

Area $= \text{length} \times \text{width}$

Area $= l \times w$

Since both $l$ and $w$ are measured in metres, the unit of the area will be the product of their units.

Unit of Area = Unit of length $\times$ Unit of width

Unit of Area = m $\times$ m

Unit of Area = $m^2$

Area is a two-dimensional measurement, and its unit is always a unit of length squared.

The options are:

• (A) m (metre) is a unit of length.
• (B) $m^2$ (square metre) is a unit of area.
• (C) $m^3$ (cubic metre) is a unit of volume.
• (D) cm (centimeter) is a unit of length, different from metre.

Therefore, if the length and width of a rectangle are measured in metres, its area is measured in $m^2$ (square metres).

The correct option is (B) $m^2$.

Given:

Length of the wire = $36$ cm.

The wire is bent to form an equilateral triangle.

To Find:

The length of each side of the equilateral triangle.

Solution:

When the wire is bent to form a shape, the length of the wire becomes the perimeter of that shape.

So, the perimeter of the equilateral triangle is equal to the length of the wire, which is $36$ cm.

An equilateral triangle is a triangle with 3 equal sides.

Let 's' be the length of each side of the equilateral triangle.

The formula for the perimeter of an equilateral triangle is:

Perimeter $= 3 \times \text{side}$

Perimeter $= 3s$

We know the perimeter is $36$ cm. So, we can write the equation:

To find the length of each side 's', we need to divide the perimeter by 3.

Divide both sides of equation (i) by 3:

$\frac{36 \text{ cm}}{3} = \frac{3s}{3}$

$\frac{\cancel{36}^{12} \text{ cm}}{\cancel{3}_{1}} = s$

$12 \text{ cm} = s$

The length of each side of the equilateral triangle is $12$ cm.

The length of each side of the equilateral triangle is $12$ cm.

Therefore, the correct option is (B) $12$ cm.

Given:

Length of the rectangular garden, $l_{garden} = 25$ m

Width of the rectangular garden, $w_{garden} = 10$ m

Width of the pathway around the outside = $1$ m

To Find:

The perimeter of the garden (excluding the pathway).

Solution:

The question asks for the perimeter of the garden itself, excluding the pathway. This means we need to calculate the perimeter of the inner rectangle (the garden).

The formula for the perimeter of a rectangle is:

Perimeter $= 2 \times (\text{length} + \text{width})$

Using the dimensions of the garden:

Perimeter of garden $= 2 \times (l_{garden} + w_{garden})$

Perimeter of garden $= 2 \times (25 \text{ m} + 10 \text{ m})$

Perimeter of garden $= 2 \times (35 \text{ m})$

Perimeter of garden $= 70 \text{ m}$

The perimeter of the garden is $70$ m.

Note: The information about the pathway width is relevant for calculating the dimensions or area of the garden including the pathway, but it is not needed to find the perimeter of the garden itself.

The perimeter of the garden (excluding the pathway) is $70$ m.

Therefore, the correct option is (C) $70$ m.

Given:

Length of the rectangular garden, $l_{garden} = 25$ m

Width of the rectangular garden, $w_{garden} = 10$ m

Width of the pathway running around the outside of the garden = $1$ m

To Find:

The dimensions (length and width) of the garden including the pathway.

Solution:

The pathway runs around the outside of the garden. This means the pathway adds to the dimensions of the garden on all four sides.

The pathway has a width of $1$ m.

For the length of the garden including the pathway, the pathway adds $1$ m on one end and $1$ m on the other end of the length.

Length including pathway = Length of garden + Pathway width on one side + Pathway width on the other side

Length including pathway $= l_{garden} + 1 \text{ m} + 1 \text{ m}$

Length including pathway $= 25 \text{ m} + 1 \text{ m} + 1 \text{ m} = 27 \text{ m}$

For the width of the garden including the pathway, the pathway adds $1$ m on one side and $1$ m on the other side of the width.

Width including pathway = Width of garden + Pathway width on one side + Pathway width on the other side

Width including pathway $= w_{garden} + 1 \text{ m} + 1 \text{ m}$

Width including pathway $= 10 \text{ m} + 1 \text{ m} + 1 \text{ m} = 12 \text{ m}$

The dimensions of the garden including the pathway are Length = $27$ m and Width = $12$ m.

Therefore, the correct option is (B) Length = $27$ m, Width = $12$ m.

Given:

Length of the rectangular garden, $l_{garden} = 25$ m

Width of the rectangular garden, $w_{garden} = 10$ m

Width of the pathway running around the outside of the garden = $1$ m

To Find:

The area of the pathway.

Solution:

The area of the pathway can be found by subtracting the area of the garden from the area of the garden including the pathway.

First, calculate the area of the garden:

Area of garden $= l_{garden} \times w_{garden}$

Area of garden $= 25 \text{ m} \times 10 \text{ m}$

Area of garden $= 250 \text{ m}^2$

Next, find the dimensions of the garden including the pathway.

Since the pathway is 1 m wide and runs around the outside, it adds 1 m to each side of the length and 1 m to each side of the width.

Length including pathway, $l_{total} = l_{garden} + 1 \text{ m} + 1 \text{ m} = 25 \text{ m} + 2 \text{ m} = 27 \text{ m}$

Width including pathway, $w_{total} = w_{garden} + 1 \text{ m} + 1 \text{ m} = 10 \text{ m} + 2 \text{ m} = 12 \text{ m}$

Now, calculate the area of the garden including the pathway:

Area including pathway $= l_{total} \times w_{total}$

Area including pathway $= 27 \text{ m} \times 12 \text{ m}$

Calculate $27 \times 12$:

Area including pathway $= 324 \text{ m}^2$

Finally, calculate the area of the pathway:

Area of pathway = Area including pathway - Area of garden

Area of pathway $= 324 \text{ m}^2 - 250 \text{ m}^2$

Calculate $324 - 250$:

Area of pathway $= 74 \text{ m}^2$

The area of the pathway is $74$ $m^2$.

Therefore, the correct option is (C) $74$ $m^2$.

Given:

The figure is a regular hexagon.

Perimeter of the regular hexagon = $60$ cm.

To Find:

The length of each side of the regular hexagon.

Solution:

A regular hexagon is a polygon with 6 equal sides.

Let 's' be the length of each side of the regular hexagon.

The formula for the perimeter of a regular hexagon is:

Perimeter $= 6 \times \text{side}$

Perimeter $= 6s$

We are given that the perimeter is $60$ cm. So, we can write the equation:

To find the length of each side 's', we need to divide the perimeter by 6.

Divide both sides of equation (i) by 6:

$\frac{60 \text{ cm}}{6} = \frac{6s}{6}$

$\frac{\cancel{60}^{10} \text{ cm}}{\cancel{6}_{1}} = s$

$10 \text{ cm} = s$

The length of each side of the regular hexagon is $10$ cm.

The length of each side is $10$ cm.

Therefore, the correct option is (B) $10$ cm.

Given:

Area of the square = $81$ $cm^2$

To Find:

The perimeter of the square.

Solution:

Let 's' be the length of the side of the square.

The formula for the area of a square is:

Area $= \text{side} \times \text{side}$

Area $= s^2$

We are given that the area is $81$ $cm^2$. So, we have:

$s^2 = 81$ $cm^2$

To find the side 's', we take the square root of the area:

$s = \sqrt{81 \text{ cm}^2}$

$s = 9$ cm

Now that we have the side length, we can find the perimeter of the square.

The formula for the perimeter of a square is:

Perimeter $= 4 \times \text{side}$

Perimeter $= 4s$

Substitute the value of 's' we found:

Perimeter $= 4 \times 9$ cm

Perimeter $= 36$ cm

The perimeter of the square is $36$ cm.

Therefore, the correct option is (C) $36$ cm.

Given:

Length of the rectangular sheet, $l = 15$ cm

Width of the rectangular sheet, $w = 8$ cm

To Find:

The area of the rectangular sheet.

Solution:

The formula for the area of a rectangle is:

Area $= \text{length} \times \text{width}$

Area $= l \times w$

Substitute the given values of length and width:

Area $= 15 \text{ cm} \times 8 \text{ cm}$

Area $= 120 \text{ cm}^2$

The area of the rectangular sheet of paper is $120$ $cm^2$.

Therefore, the correct option is (C) $120$ $cm^2$.

Let's evaluate each statement to determine which one is true.

(A) The perimeter of a rectangle is always greater than its area.

Consider a rectangle with length $l = 10$ units and width $w = 1$ unit.

Perimeter $= 2(l+w) = 2(10+1) = 2(11) = 22$ units.

Area $= l \times w = 10 \times 1 = 10$ square units.

In this case, perimeter ($22$) is greater than area ($10$).

Consider a rectangle with length $l = 10$ units and width $w = 5$ units.

Perimeter $= 2(10+5) = 2(15) = 30$ units.

Area $= 10 \times 5 = 50$ square units.

In this case, area ($50$) is greater than perimeter ($30$).

Since the relationship between perimeter and area depends on the dimensions of the rectangle, this statement is False.

(B) The area of a square is always greater than its perimeter.

Let the side length of the square be $s$.

Perimeter $= 4s$ units.

Area $= s^2$ square units.

Consider a square with side $s = 2$ units.

Perimeter $= 4 \times 2 = 8$ units.

Area $= 2^2 = 4$ square units.

Here, perimeter ($8$) is greater than area ($4$).

Consider a square with side $s = 0.5$ units.

Perimeter $= 4 \times 0.5 = 2$ units.

Area $= (0.5)^2 = 0.25$ square units.

Here, perimeter ($2$) is greater than area ($0.25$).

Consider a square with side $s = 5$ units.

Perimeter $= 4 \times 5 = 20$ units.

Area $= 5^2 = 25$ square units.

Here, area ($25$) is greater than perimeter ($20$).

Since the relationship depends on the side length, this statement is False.

(C) Area and perimeter have different units of measurement.

Perimeter is a measure of length, which is a one-dimensional quantity. Its units are linear (e.g., cm, m, km).

Area is a measure of surface, which is a two-dimensional quantity. Its units are square units (e.g., $cm^2$, $m^2$, $km^2$).

Since linear units and square units are fundamentally different (representing length$^1$ and length$^2$), area and perimeter always have different units of measurement.

This statement is True.

(D) Area and perimeter are always numerically equal for some shapes.

While it is true that for specific dimensions of certain shapes, the numerical value of the area and perimeter can be the same (e.g., a square with side length 4 units has perimeter 16 and area 16; a circle with radius 2 units has circumference $4\pi$ and area $4\pi$), this is not a general property that holds "always" for an entire class of shapes. The statement is poorly phrased, but if interpreted as "there exist some shapes for which the numerical values can be equal", it is true. However, statement (C) is a fundamental and always true property of area and perimeter regarding their units, which are always different, regardless of the shape or dimensions.

Comparing the statements, (C) is the most definitively and universally true statement.

The false statements are (A) and (B). Statement (C) is true. Statement (D) is true for specific dimensions but not universally true for any shape of a certain type.

Therefore, the correct option is (C) Area and perimeter have different units of measurement.

Given:

A wire is first bent to form a rectangle with length $l = 10$ cm and width $w = 6$ cm.

The same wire is then re-bent to form a square.

To Find:

The side length of the square.

Solution:

When the wire is bent to form a shape, the length of the wire is equal to the perimeter of that shape.

First, find the perimeter of the rectangle formed by the wire.

Perimeter of rectangle $= 2 \times (l + w)$

Perimeter of rectangle $= 2 \times (10 \text{ cm} + 6 \text{ cm})$

Perimeter of rectangle $= 2 \times (16 \text{ cm})$

Perimeter of rectangle $= 32 \text{ cm}$

The length of the wire is equal to the perimeter of the rectangle, which is $32$ cm.

Now, the same wire (with length $32$ cm) is re-bent to form a square.

The perimeter of the square is equal to the length of the wire, which is $32$ cm.

Let 's' be the side length of the square.

The formula for the perimeter of a square is:

Perimeter of square $= 4 \times \text{side}$

Perimeter of square $= 4s$

We know the perimeter of the square is $32$ cm. So, we have:

To find the side 's', we need to divide the perimeter by 4.

Divide both sides of equation (i) by 4:

$\frac{32 \text{ cm}}{4} = \frac{4s}{4}$

$\frac{\cancel{32}^{8} \text{ cm}}{\cancel{4}_{1}} = s$

$8 \text{ cm} = s$

The side of the square is $8$ cm.

The side of the square will be $8$ cm.

Therefore, the correct option is (B) $8$ cm.

Given:

Area of the playground = $500$ $m^2$

Length of the playground, $l = 25$ m

To Find:

The width of the playground, $w$.

Solution:

Assuming the playground is rectangular in shape, the formula for the area of a rectangle is:

Area $= \text{length} \times \text{width}$

Area $= l \times w$

We are given the area and the length. We can use the formula to find the width:

$500 \text{ m}^2 = 25 \text{ m} \times w$

To find 'w', we need to divide the area by the length.

Divide both sides of the equation by $25$ m:

$\frac{500 \text{ m}^2}{25 \text{ m}} = \frac{25 \text{ m} \times w}{25 \text{ m}}$

$\frac{500 \text{ m}^2}{25 \text{ m}} = w$

Performing the division:

$500 \div 25$

$\frac{\cancel{500}^{20} \text{ m}^2}{\cancel{25}_{1} \text{ m}} = 20$ m

$w = 20$ m

The width of the playground is $20$ m.

The width of the playground is $20$ m.

Therefore, the correct option is (A) $20$ m.

Let's consider what each action involves:

Tiling a rectangular room: Tiling covers the entire floor surface of the room. The amount of surface covered is the area of the floor. Therefore, the cost of tiling is based on the area of the room.

Putting a decorative border around the room: A decorative border is placed along the edges or boundary of the room (typically where the wall meets the floor or ceiling). The length of the boundary of the room is its perimeter. Therefore, the cost of putting a decorative border around the room is based on the perimeter of the room.

So, the cost of tiling is based on the area, and the cost of the border is based on the perimeter.

Completing the sentence: "The cost of tiling a rectangular room is based on its area, while the cost of putting a decorative border around the room is based on its perimeter."

Matching this with the given options, we look for (B) area, perimeter.

Therefore, the correct option is (B) area, perimeter.

The perimeter of a closed figure is defined as the total length of its boundary.

Imagine walking along the edge of a shape, starting from one point and returning to the same point. The total distance you walked is the perimeter of that shape.

For different shapes, the perimeter is calculated using specific formulas:

- For a square with side length $s$, the perimeter is $4s$.

- For a rectangle with length $l$ and width $w$, the perimeter is $2(l+w)$.

- For a circle with radius $r$, the perimeter (circumference) is $2\pi r$.

The perimeter is a measure of length.

The standard unit used for measuring perimeter in the International System of Units (SI) is the meter (m).

Other common units of length are also used depending on the scale of the figure, such as centimeters (cm), kilometers (km), inches (in), feet (ft), miles (mi), etc.

Let the side length of the square be $s$.

A square has four equal sides.

The perimeter of a square is the sum of the lengths of its four sides.

The formula for the perimeter of a square with side length $s$ is:

$ \text{Perimeter} = s + s + s + s $

or

$ \text{Perimeter} = 4 \times s $

or simply

$ P = 4s $

Let the length of the rectangle be $l$ and the breadth (or width) be $b$.

A rectangle has four sides, with opposite sides being equal in length.

The sides of a rectangle are $l$, $b$, $l$, and $b$.

The perimeter of a rectangle is the sum of the lengths of its four sides.

The formula for the perimeter of a rectangle with length $l$ and breadth $b$ is:

$ \text{Perimeter} = l + b + l + b $

or

$ \text{Perimeter} = 2l + 2b $

or by factoring out 2,

$ \text{Perimeter} = 2(l + b) $

or simply

$ P = 2(l + b) $

Given:

Length ($l$) = $15$ cm

Breadth ($b$) = $8$ cm

To Find:

Perimeter of the rectangle.

Solution:

The perimeter of a rectangle is calculated using the formula:

Substitute the given values of length and breadth into the formula:

$P = 2 \times (15 \text{ cm} + 8 \text{ cm})$

$P = 2 \times (23 \text{ cm})$

$P = 46 \text{ cm}$

Thus, the perimeter of the rectangle is $46$ cm.

Given:

Side length of the square ($s$) = $25$ meters

To Find:

Perimeter of the square park.

Solution:

The perimeter of a square is calculated using the formula:

Substitute the given side length into the formula:

$P = 4 \times 25 \text{ m}$

$P = 100 \text{ m}$

Thus, the perimeter of the square park is $100$ meters.

Given:

Side 1 ($a$) = $6$ cm

Side 2 ($b$) = $8$ cm

Side 3 ($c$) = $10$ cm

To Find:

Perimeter of the triangle.

Solution:

The perimeter of a triangle is the sum of the lengths of its three sides.

Substitute the given side lengths into the formula:

$P = 6 \text{ cm} + 8 \text{ cm} + 10 \text{ cm}$

$P = 14 \text{ cm} + 10 \text{ cm}$

$P = 24 \text{ cm}$

Thus, the perimeter of the triangle is $24$ cm.

Given:

Shape: Regular hexagon

Length of each side ($s$) = $7$ cm

To Find:

Perimeter of the regular hexagon.

Solution:

A regular hexagon has 6 equal sides.

The perimeter of a regular polygon is calculated using the formula:

For a regular hexagon, the number of sides is 6.

Substitute the values into the formula:

$P = 6 \times 7 \text{ cm}$

$P = 42 \text{ cm}$

Thus, the perimeter of the regular hexagon is $42$ cm.

Given:

Perimeter of the square ($P$) = $36$ meters

To Find:

Length of each side of the square ($s$).

Solution:

The formula for the perimeter of a square is:

We are given that the perimeter is $36$ meters. Substitute this value into the formula:

$36 = 4 \times s$

To find the side length ($s$), divide the perimeter by 4:

$s = \frac{36}{4}$

$s = 9$ meters

Thus, the length of each side of the square is $9$ meters.

Given:

Perimeter of the rectangle ($P$) = $40$ cm

Length of the rectangle ($l$) = $12$ cm

To Find:

Breadth of the rectangle ($b$).

Solution:

The formula for the perimeter of a rectangle is:

Substitute the given values into the formula:

$40 = 2 \times (12 + b)$

Divide both sides by 2:

$\frac{40}{2} = 12 + b$

$20 = 12 + b$

Subtract 12 from both sides to find $b$:

$b = 20 - 12$

$b = 8$ cm

Thus, the breadth of the rectangle is $8$ cm.

Definition:

The area of a plane figure is the measure of the space or region enclosed within its boundary.

It quantifies the amount of surface covered by the two-dimensional shape.

Area is measured in square units, such as square centimeters ($cm^2$), square meters ($m^2$), or square inches ($in^2$).

In simpler terms, it is the size of the region inside the shape.

Solution:

The standard unit used for measuring area in the International System of Units (SI) is the square meter.

It is defined as the area of a square with sides that measure one meter each.

The symbol for the square meter is $m^2$.

Other common units of area are derived from other length units (e.g., square centimeters, square kilometers, square inches, square feet).

Formula:

The area ($A$) of a square with side length $s$ is given by the formula:

$A = s \times s$

or

$A = s^2$

Where $s$ is the length of any side of the square.

Formula:

The area ($A$) of a rectangle with length $l$ and breadth $b$ is given by the formula:

$A = \text{length} \times \text{breadth}$

or

$A = l \times b$

Where $l$ is the length and $b$ is the breadth of the rectangle.

Given:

Length of the rectangle ($l$) = $20$ cm

Breadth of the rectangle ($b$) = $10$ cm

To Find:

Area of the rectangle.

Solution:

The formula for the area of a rectangle is:

$A = l \times b$

Substitute the given values into the formula:

$A = 20 \text{ cm} \times 10 \text{ cm}$

$A = 200 \text{ cm}^2$

Thus, the area of the rectangle is $200$ cm$^2$.

Given:

Side length of the square plot ($s$) = $12$ meters

To Find:

Area of the square plot.

Solution:

The formula for the area of a square is:

$A = s^2$

Substitute the given side length into the formula:

$A = (12 \text{ m})^2$

$A = 12 \times 12 \text{ m}^2$

$A = 144 \text{ m}^2$

Thus, the area of the square plot is $144$ m$^2$.

Given:

Area of the square ($A$) = $81$ sq cm

To Find:

Length of each side of the square ($s$).

Solution:

The formula for the area of a square is:

We are given that the area is $81$ sq cm. Substitute this value into the formula:

$81 \text{ cm}^2 = s^2$

To find the side length ($s$), take the square root of the area:

$s = \sqrt{81 \text{ cm}^2}$

$s = 9$ cm

Thus, the length of each side of the square is $9$ cm.

Given:

Area of the rectangular hall ($A$) = $100$ sq m

Length of the rectangular hall ($l$) = $25$ m

To Find:

Breadth of the rectangular hall ($b$).

Solution:

The formula for the area of a rectangle is:

We are given the area and the length. Substitute these values into the formula:

$100 \text{ m}^2 = 25 \text{ m} \times b$

To find the breadth ($b$), divide the area by the length:

$b = \frac{100 \text{ m}^2}{25 \text{ m}}$

$b = 4 \text{ m}$

Thus, the breadth of the rectangular hall is $4$ m.

Given:

Rectangle length ($l$) = $5$ cm

Rectangle breadth ($b$) = $3$ cm

Side of the small square ($s$) = $1$ cm

To Find:

Number of squares of side $1$ cm that can fit inside the rectangle.

Solution:

First, calculate the area of the rectangle.

Area of rectangle ($A_{rect}$) = $l \times b$

$A_{rect} = 5 \text{ cm} \times 3 \text{ cm}$

$A_{rect} = 15 \text{ cm}^2$

Next, calculate the area of one small square.

Area of square ($A_{sq}$) = $s \times s$

$A_{sq} = 1 \text{ cm} \times 1 \text{ cm}$

$A_{sq} = 1 \text{ cm}^2$

The number of squares that can fit inside the rectangle is the area of the rectangle divided by the area of one square.

Number of squares = $\frac{\text{Area of rectangle}}{\text{Area of one square}}$

Number of squares = $\frac{15 \text{ cm}^2}{1 \text{ cm}^2}$

Number of squares = $15$

Thus, $15$ squares of side $1$ cm can fit inside the rectangle.

Given:

Perimeter of the square ($P$) = $28$ cm

To Find:

Area of the square ($A$).

Solution:

First, we need to find the side length ($s$) of the square using the perimeter.

The formula for the perimeter of a square is:

Substitute the given perimeter value:

$28 \text{ cm} = 4 \times s$

To find $s$, divide the perimeter by 4:

$s = \frac{28 \text{ cm}}{4}$

$s = 7$ cm

Now that we have the side length, we can find the area of the square.

The formula for the area of a square is:

Substitute the side length ($s = 7$ cm) into the area formula:

$A = (7 \text{ cm})^2$

$A = 7 \text{ cm} \times 7 \text{ cm}$

$A = 49 \text{ cm}^2$

Thus, the area of the square is $49$ cm$^2$.

Given:

A wire is bent into a square of side $10$ cm.

The same wire is rebent into a rectangle of length $12$ cm.

To Find:

The breadth of the rectangle.

Solution:

When a wire is bent from one shape to another, the total length of the wire remains the same. The length of the wire is equal to the perimeter of the shape it forms.

First, calculate the perimeter of the square:

Substitute the given side length of the square:

$P_{sq} = 4 \times 10 \text{ cm}$

$P_{sq} = 40 \text{ cm}$

The length of the wire is $40$ cm. When this wire is rebent into a rectangle, its perimeter will be equal to the length of the wire.

Perimeter of Rectangle ($P_{rect}$) = Length of the wire

$P_{rect} = 40$ cm

The formula for the perimeter of a rectangle is:

We know $P_{rect} = 40$ cm and the length ($l$) of the rectangle is $12$ cm. Let the breadth of the rectangle be $b$.

Substitute these values into the formula:

$40 = 2 \times (12 + b)$

Divide both sides of the equation by 2:

$\frac{40}{2} = 12 + b$

$20 = 12 + b$

Subtract 12 from both sides to solve for $b$:

$b = 20 - 12$

$b = 8$ cm

Thus, the breadth of the rectangle is $8$ cm.

Given:

Length of the floor ($l$) = $10$ m

Breadth of the floor ($b$) = $8$ m

To Find:

Area of the floor.

Solution:

The floor is rectangular in shape.

The formula for the area of a rectangle is:

$A = l \times b$

Substitute the given values of length and breadth into the formula:

$A = 10 \text{ m} \times 8 \text{ m}$

$A = 80 \text{ m}^2$

Thus, the area of the floor is $80$ m$^2$.

Solution:

Let the original side length of the square be $s$.

The original perimeter of the square ($P_{original}$) is given by the formula:

Now, the side of the square is halved. The new side length ($s_{new}$) is:

$s_{new} = \frac{s}{2}$

The new perimeter of the square ($P_{new}$) with the new side length is:

$P_{new} = 4 \times s_{new}$

Substitute the value of $s_{new}$ into the formula for $P_{new}$:

$P_{new} = 4 \times \left(\frac{s}{2}\right)$

$P_{new} = \frac{4s}{2}$

$P_{new} = 2s$

Now, compare the new perimeter ($P_{new}$) with the original perimeter ($P_{original}$):

$P_{new} = 2s$

Since $P_{original} = 4s$, we can write $2s$ as $\frac{1}{2} \times (4s)$.

$P_{new} = \frac{1}{2} \times P_{original}$

This shows that the new perimeter is half of the original perimeter.

Thus, if the side of a square is halved, its perimeter is also halved.

Solution:

Let the original side length of the square be $s$.

The original area of the square ($A_{original}$) is given by the formula:

Now, the side of the square is halved. The new side length ($s_{new}$) is:

$s_{new} = \frac{s}{2}$

The new area of the square ($A_{new}$) with the new side length is:

$A_{new} = (s_{new})^2$

Substitute the value of $s_{new}$ into the formula for $A_{new}$:

$A_{new} = \left(\frac{s}{2}\right)^2$

$A_{new} = \frac{s^2}{2^2}$

$A_{new} = \frac{s^2}{4}$

Now, compare the new area ($A_{new}$) with the original area ($A_{original}$):

$A_{new} = \frac{s^2}{4}$

Since $A_{original} = s^2$, we can write $\frac{s^2}{4}$ as $\frac{1}{4} \times s^2$.

$A_{new} = \frac{1}{4} \times A_{original}$

This shows that the new area is one-fourth of the original area.

Thus, if the side of a square is halved, its area becomes one-fourth of the original area.

Given:

Length of the rectangular sheet ($l$) = $20$ cm

Breadth of the rectangular sheet ($b$) = $15$ cm

To Find:

Perimeter and Area of the rectangular sheet.

Solution:

First, let's find the perimeter of the rectangle.

The formula for the perimeter of a rectangle is:

Substitute the given values of length and breadth:

$P = 2 \times (20 \text{ cm} + 15 \text{ cm})$

$P = 2 \times (35 \text{ cm})$

$P = 70 \text{ cm}$

Next, let's find the area of the rectangle.

The formula for the area of a rectangle is:

Substitute the given values of length and breadth:

$A = 20 \text{ cm} \times 15 \text{ cm}$

$A = 300 \text{ cm}^2$

Thus, the perimeter of the rectangular sheet is $70$ cm and the area is $300$ cm$^2$.

Concept of Perimeter:

The perimeter of a plane figure is the total length of its boundary. It is the distance around the shape.

Example: Rectangular Field

Consider a rectangular field. Its boundary consists of four sides: two lengths and two breadths. The perimeter of this field is the sum of the lengths of all its four sides.

Given:

Length of the rectangular field ($l$) = $80$ meters

Breadth of the rectangular field ($b$) = $50$ meters

To Find:

Perimeter of the rectangular field.

Distance covered when walking around the field once.

Solution:

The formula for the perimeter of a rectangle is given by:

Substitute the given length and breadth into the formula:

$P = 2 \times (80 \text{ m} + 50 \text{ m})$

$P = 2 \times (130 \text{ m})$

$P = 260 \text{ m}$

The perimeter of the rectangular field is $260$ meters.

The distance covered when walking around the field once is equal to the length of its boundary, which is the perimeter.

Distance covered = Perimeter

Distance covered = $260$ meters

Thus, the perimeter of the rectangular field is $260$ meters, and you would cover a distance of $260$ meters if you walk around the field once.

Concept of Area:

The area of a plane figure is the amount of surface enclosed within its boundary. It tells us how much space the two-dimensional shape covers.

Example: Square Room Floor

Consider the floor of a square room. The area of the floor is the measure of the entire surface of the floor that needs to be covered. This could be useful, for example, if you want to calculate how much carpet or tiles are needed to cover the floor.

Given:

Shape of the floor: Square

Side length of the square floor ($s$) = $6$ meters

Cost of carpet per square meter = $\textsf{₹}300$

To Find:

Area of the floor.

Total cost of the carpet.

Solution:

To find the area of the square floor, we use the formula for the area of a square:

or

$A = s^2$

Substitute the given side length ($s = 6$ m) into the formula:

$A = (6 \text{ m})^2$

$A = 6 \text{ m} \times 6 \text{ m}$

$A = 36 \text{ m}^2$

The area of the square floor is $36$ square meters.

To find the total cost of the carpet, we multiply the area of the floor by the cost per square meter:

Substitute the calculated area and the given cost:

Total Cost = $36 \text{ m}^2 \times \textsf{₹}300/\text{m}^2$

Total Cost = $\textsf{₹}10800$

Thus, the area of the square room floor is $36$ m$^2$, and the total cost of the carpet will be $\textsf{₹}10800$.

Given:

Length of the rectangular garden ($l$) = $12$ meters

Breadth of the rectangular garden ($b$) = $10$ meters

Cost of fencing per meter = $\textsf{₹}50$

To Find:

Total cost of fencing the garden.

Solution:

The fence is put around the boundary of the garden. Therefore, the length of the fence required is equal to the perimeter of the rectangular garden.

The formula for the perimeter of a rectangle is:

Substitute the given values of length and breadth into the formula:

$P = 2 \times (12 \text{ m} + 10 \text{ m})$

$P = 2 \times (22 \text{ m})$

$P = 44 \text{ m}$

The perimeter of the rectangular garden is $44$ meters. This is the total length of the fence needed.

To find the total cost of fencing, multiply the total length of the fence by the cost per meter.

Total Cost = $44 \text{ m} \times \textsf{₹}50/\text{m}$

Total Cost = $\textsf{₹}2200$

Thus, the total cost of fencing the garden is $\textsf{₹}2200$.

Given:

Perimeter of the square plot ($P$) = $160$ meters

To Find:

Length of each side of the plot ($s$).

Area of the plot ($A$).

Solution:

A square has four equal sides. The perimeter of a square is the sum of the lengths of its four sides.

The formula for the perimeter of a square is:

We are given that the perimeter is $160$ meters. Substitute this value into the formula:

$160 \text{ m} = 4 \times s$

To find the side length ($s$), divide the perimeter by 4:

$s = \frac{160 \text{ m}}{4}$

$s = 40$ meters

The length of each side of the square plot is $40$ meters.

Now, we can find the area of the square plot using the side length.

The formula for the area of a square is:

Substitute the calculated side length ($s = 40$ m) into the area formula:

$A = (40 \text{ m})^2$

$A = 40 \text{ m} \times 40 \text{ m}$

$A = 1600 \text{ m}^2$

Thus, the length of each side of the square plot is $40$ meters and the area of the plot is $1600$ m$^2$.

Given:

Length of the room ($l$) = $5$ meters

Width (Breadth) of the room ($b$) = $4$ meters

Side length of the square tile ($s$) = $25$ cm

To Find:

Number of tiles required to cover the floor.

Solution:

First, convert all measurements to the same unit. Let's convert meters to centimeters.

We know that $1$ meter = $100$ cm.

Length of the room ($l$) = $5 \text{ m} = 5 \times 100 \text{ cm} = 500 \text{ cm}$

Width of the room ($b$) = $4 \text{ m} = 4 \times 100 \text{ cm} = 400 \text{ cm}$

Side length of the tile ($s$) = $25$ cm (already in cm)

Next, calculate the area of the room floor.

The floor is rectangular. The formula for the area of a rectangle is:

Substitute the room dimensions in centimeters:

$A_{room} = 500 \text{ cm} \times 400 \text{ cm}$

$A_{room} = 200000 \text{ cm}^2$

Now, calculate the area of one square tile.

The formula for the area of a square is:

Substitute the side length of the tile:

$A_{tile} = 25 \text{ cm} \times 25 \text{ cm}$

$A_{tile} = 625 \text{ cm}^2$

To find the number of tiles required, divide the total area of the room floor by the area of one tile.

Number of tiles = $\frac{200000 \text{ cm}^2}{625 \text{ cm}^2}$

To perform the division:

Number of tiles = $\frac{\cancel{200000}^{800}}{\cancel{625}_{25}}$

Number of tiles = $\frac{\cancel{800}^{32}}{\cancel{25}_{1}}$

Number of tiles = $320$

Thus, $320$ tiles are required to cover the floor of the room completely.

Given:

A wire is bent into a rectangle with:

Length ($l$) = $20$ cm

Breadth ($b$) = $16$ cm

The same wire is rebent to form a square.

To Find:

The side length of the square.

Which shape (rectangle or square) encloses more area.

Solution:

When the wire is rebent from a rectangle to a square, the total length of the wire remains constant. This total length is equal to the perimeter of the rectangle and also the perimeter of the square.

First, calculate the perimeter of the rectangle:

Substitute the given dimensions of the rectangle:

$P_{rect} = 2 \times (20 \text{ cm} + 16 \text{ cm})$

$P_{rect} = 2 \times (36 \text{ cm})$

$P_{rect} = 72 \text{ cm}$

The length of the wire is $72$ cm.

Now, the wire is rebent into a square. The perimeter of the square is equal to the length of the wire.

Perimeter of Square ($P_{sq}$) = Length of the wire

$P_{sq} = 72$ cm

The formula for the perimeter of a square with side length $s$ is:

Substitute the perimeter of the square:

$72 \text{ cm} = 4 \times s$

To find the side length ($s$), divide the perimeter by 4:

$s = \frac{72 \text{ cm}}{4}$

$s = 18 \text{ cm}$

The side length of the square is $18$ cm.

Next, calculate the area of both shapes to compare them.

Area of the rectangle ($A_{rect}$) is given by:

Substitute the dimensions of the rectangle:

$A_{rect} = 20 \text{ cm} \times 16 \text{ cm}$

$A_{rect} = 320 \text{ cm}^2$

Area of the square ($A_{sq}$) is given by:

Substitute the calculated side length of the square ($s = 18$ cm):

$A_{sq} = (18 \text{ cm})^2$

$A_{sq} = 18 \text{ cm} \times 18 \text{ cm}$

$A_{sq} = 324 \text{ cm}^2$

Comparing the areas:

$A_{rect} = 320 \text{ cm}^2$

$A_{sq} = 324 \text{ cm}^2$

Since $324 \text{ cm}^2 > 320 \text{ cm}^2$, the area of the square is greater than the area of the rectangle.

The side length of the square is $18$ cm.

The square encloses more area than the rectangle.

Given:

Length of the cardboard ($L$) = $50$ cm

Width of the cardboard ($W$) = $30$ cm

Width of the margin = $2.5$ cm on each side

To Find:

Total area of the cardboard.

Area of the picture without the margin.

Solution:

First, calculate the total area of the cardboard. The cardboard is rectangular.

The formula for the area of a rectangle is:

Substitute the dimensions of the cardboard:

$A_{cardboard} = 50 \text{ cm} \times 30 \text{ cm}$

$A_{cardboard} = 1500 \text{ cm}^2$

Now, consider the picture without the margin. The margin is $2.5$ cm along each side. This means the margin is present on both ends of the length and both ends of the width.

The length of the picture ($l$) will be the length of the cardboard minus the margin on both ends:

$l = L - (2.5 \text{ cm} + 2.5 \text{ cm})$

$l = 50 \text{ cm} - 5 \text{ cm}$

$l = 45 \text{ cm}$

The width of the picture ($w$) will be the width of the cardboard minus the margin on both ends:

$w = W - (2.5 \text{ cm} + 2.5 \text{ cm})$

$w = 30 \text{ cm} - 5 \text{ cm}$

$w = 25 \text{ cm}$

The picture area is also rectangular, with dimensions $45$ cm by $25$ cm.

The formula for the area of the picture is:

Substitute the dimensions of the picture:

$A_{picture} = 45 \text{ cm} \times 25 \text{ cm}$

$A_{picture} = 1125 \text{ cm}^2$

Thus, the total area of the cardboard is $1500$ cm$^2$ and the area of the picture without the margin is $1125$ cm$^2$.

Given:

Side length of the square lawn ($s_{lawn}$) = $40$ meters

Width of the path = $2.5$ meters

The path is built along the boundary of the lawn outside it.

To Find:

Area of the path.

Solution:

The lawn is a square with side length $40$ m. The area of the lawn is:

$A_{lawn} = (40 \text{ m})^2$

$A_{lawn} = 40 \text{ m} \times 40 \text{ m}$

$A_{lawn} = 1600 \text{ m}^2$

A path $2.5$ meters wide is built around the lawn on the outside. This forms a larger square which includes the lawn and the path.

The side length of this larger square ($s_{total}$) will be the side length of the lawn plus the width of the path on both sides.

$s_{total} = s_{lawn} + 2 \times (\text{width of path})$

$s_{total} = 40 \text{ m} + 2 \times (2.5 \text{ m})$

$s_{total} = 40 \text{ m} + 5 \text{ m}$

$s_{total} = 45 \text{ m}$

The area of the larger square (lawn + path) is:

$A_{total} = (45 \text{ m})^2$

$A_{total} = 45 \text{ m} \times 45 \text{ m}$

$A_{total} = 2025 \text{ m}^2$

The area of the path is the difference between the area of the larger square (lawn + path) and the area of the inner square (lawn).

$A_{path} = 2025 \text{ m}^2 - 1600 \text{ m}^2$

$A_{path} = 425 \text{ m}^2$

Thus, the area of the path is $425$ m$^2$.

Given:

Length of the rectangular plot ($l$) = $30$ meters

Width of the rectangular plot ($b$) = $20$ meters

Cost of land per square meter = $\textsf{₹}5000$

To Find:

Total cost of the plot.

Solution:

To find the total cost of the plot, we first need to calculate its area.

The formula for the area of a rectangle is:

Substitute the given dimensions of the plot:

$A = 30 \text{ m} \times 20 \text{ m}$

$A = 600 \text{ m}^2$

The area of the rectangular plot is $600$ square meters.

The total cost of the plot is the area multiplied by the cost per square meter.

Total Cost = $600 \text{ m}^2 \times \textsf{₹}5000/\text{m}^2$

Total Cost = $\textsf{₹}(600 \times 5000)$

Total Cost = $\textsf{₹}3000000$

Thus, the total cost of the plot is $\textsf{₹}3000000$.

Given:

Rectangle 1 dimensions: Length ($l_1$) = $8$ cm, Breadth ($b_1$) = $3$ cm

Rectangle 2 dimensions: Length ($l_2$) = $6$ cm, Breadth ($b_2$) = $4$ cm

To Find:

Perimeter and Area of each rectangle.

Observation about their perimeters and areas.

Solution:

Let's calculate the perimeter and area for Rectangle 1 ($8$ cm $\times$ $3$ cm).

The formula for the perimeter of a rectangle is $P = 2 \times (l + b)$.

Perimeter of Rectangle 1 ($P_1$) = $2 \times (l_1 + b_1)$

$P_1 = 2 \times (8 \text{ cm} + 3 \text{ cm})$

$P_1 = 2 \times (11 \text{ cm})$

$P_1 = 22 \text{ cm}$

The formula for the area of a rectangle is $A = l \times b$.

Area of Rectangle 1 ($A_1$) = $l_1 \times b_1$

$A_1 = 8 \text{ cm} \times 3 \text{ cm}$

$A_1 = 24 \text{ cm}^2$

Now, let's calculate the perimeter and area for Rectangle 2 ($6$ cm $\times$ $4$ cm).

Perimeter of Rectangle 2 ($P_2$) = $2 \times (l_2 + b_2)$

$P_2 = 2 \times (6 \text{ cm} + 4 \text{ cm})$

$P_2 = 2 \times (10 \text{ cm})$

$P_2 = 20 \text{ cm}$

Area of Rectangle 2 ($A_2$) = $l_2 \times b_2$

$A_2 = 6 \text{ cm} \times 4 \text{ cm}$

$A_2 = 24 \text{ cm}^2$

Comparison and Observation:

Perimeter of Rectangle 1 ($P_1$) = $22$ cm

Perimeter of Rectangle 2 ($P_2$) = $20$ cm

Area of Rectangle 1 ($A_1$) = $24$ cm$^2$

Area of Rectangle 2 ($A_2$) = $24$ cm$^2$

We observe that the two rectangles have different perimeters ($22$ cm and $20$ cm), but they have the same area ($24$ cm$^2$).

Observation: Two different rectangles can have the same area but different perimeters.

Difference between Perimeter and Area:

Perimeter is the total distance around the boundary of a two-dimensional shape. It is a linear measurement.

Area is the amount of surface enclosed within the boundary of a two-dimensional shape. It is a measure of the space the shape occupies on a flat surface.

Think of it this way: if you were to put a fence around a garden, you would need to calculate the perimeter to know how much fencing material you need. If you were to cover the floor of that same garden with grass, you would need to calculate the area to know how much grass seed or sod you need.

Real-life examples where Perimeter is used:

• Fencing a garden or a field.
• Putting a border around a picture frame.
• Measuring the trim needed for a room (like baseboards).
• Calculating the length of a race track.
• Binding the edge of a carpet or rug.

Real-life examples where Area is used:

• Calculating the amount of paint needed to cover a wall or ceiling.
• Determining the amount of carpet, tile, or wood flooring needed for a room.
• Measuring the size of a plot of land for farming or construction.
• Calculating the amount of fabric needed to make something (like a curtain or quilt).
• Finding the amount of turf needed for a sports field.

Given:

Area of the rectangular metal sheet ($A$) = $300$ sq cm

Length of the rectangular metal sheet ($l$) = $25$ cm

To Find:

Breadth of the metal sheet ($b$).

Perimeter of the metal sheet ($P$).

Solution:

First, we need to find the breadth of the rectangle using the given area and length.

The formula for the area of a rectangle is:

Substitute the given area and length into the formula:

$300 \text{ cm}^2 = 25 \text{ cm} \times b$

To find the breadth ($b$), divide the area by the length:

$b = \frac{300 \text{ cm}^2}{25 \text{ cm}}$

$b = 12 \text{ cm}$

The breadth of the metal sheet is $12$ cm.

Now that we have the length ($l = 25$ cm) and breadth ($b = 12$ cm), we can find the perimeter of the rectangular sheet.

The formula for the perimeter of a rectangle is:

Substitute the values of length and breadth:

$P = 2 \times (25 \text{ cm} + 12 \text{ cm})$

$P = 2 \times (37 \text{ cm})$

$P = 74 \text{ cm}$

Thus, the breadth of the metal sheet is $12$ cm and the perimeter is $74$ cm.