Chapter 6 Mensuration (Class 6 - Maths NCERT Exemplar Solutions)

Solution:

The perimeter of a figure is the total length of its outer boundary. Let us calculate the length of all the outer edges starting from the bottom-left corner and moving clockwise:

1. Left vertical side of the small square = $1\text{ cm}$

2. Top horizontal side of the small square = $1\text{ cm}$

3. Vertical segment connecting the two squares: Since the larger square has a side of $3\text{ cm}$ and the smaller square has a side of $1\text{ cm}$, the remaining vertical part is $3\text{ cm} - 1\text{ cm} = 2\text{ cm}$.

4. Top horizontal side of the large square = $3\text{ cm}$

5. Right vertical side of the large square = $3\text{ cm}$

6. Bottom horizontal side (combined) = $3\text{ cm} + 1\text{ cm} = 4\text{ cm}$

Total Perimeter = Sum of all outer sides

Total Perimeter = $1\text{ cm} + 1\text{ cm} + 2\text{ cm} + 3\text{ cm} + 3\text{ cm} + 4\text{ cm}$

Total Perimeter = $14\text{ cm}$

Thus, the correct option is (B).

Solution:

(a) False.

To raise a boundary wall around a house, one needs to know the length of the boundary, which is the Perimeter of the land, not its Area.

(b) True.

A sports track is built along the boundary of the ground. Therefore, the person must find the Perimeter to determine the length of the track.

Solution:

(a) Perimeter of a triangle = Sum of all its sides.

Perimeter = $4.5\text{ cm} + 6.02\text{ cm} + 5.38\text{ cm}$

The perimeter is $15.9\text{ cm}$.

(b) Area of a square = $\text{Side} \times \text{Side}$

Area = $5\text{ cm} \times 5\text{ cm}$

Area = $25\text{ cm}^2$.

Given:

For Bhavna: Side of square field = $80\text{ m}$, Number of rounds = $10$.

For Sushmita: Length of rectangular field = $150\text{ m}$, Breadth = $60\text{ m}$, Number of rounds = $8$.

To Find: Who covers more distance and by how much.

Solution:

Distance covered by Bhavna:

Perimeter of the square field = $4 \times \text{Side}$

Perimeter = $4 \times 80\text{ m} = 320\text{ m}$

Total distance = $10 \times 320\text{ m} = 3200\text{ m}$

Distance covered by Sushmita:

Perimeter of the rectangular field = $2 \times (\text{Length} + \text{Breadth})$

Perimeter = $2 \times (150\text{ m} + 60\text{ m}) = 2 \times 210\text{ m} = 420\text{ m}$

Total distance = $8 \times 420\text{ m}$

Total distance = $3360\text{ m}$

Comparison:

Sushmita covers more distance than Bhavna.

Difference in distance = $3360\text{ m} - 3200\text{ m} = 160\text{ m}$

Sushmita covers $160\text{ m}$ more distance.

Given:

Perimeter of the rectangular field = $800\text{ m}$.

Length ($l$) is thrice its breadth ($b$).

To Find: The length of the field.

Solution:

Let the breadth of the field be $x$ metres.

Then, the length of the field = $3x$ metres.

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

$800 = 2 \times 4x$

$800 = 8x$

Length of the field = $3x = 3 \times 100\text{ m} = 300\text{ m}$

The length of the field is $300\text{ m}$.

Given:

Total cost of fencing = $\textsf{₹}\! 12000$.

Cost of fencing per metre = $\textsf{₹}\! 30$.

To Find: Area of the square field.

Solution:

Perimeter of the square field = $\frac{\text{Total Cost}}{\text{Cost per metre}}$

Perimeter = $\frac{12000}{30} = 400\text{ m}$

We know that Perimeter of a square = $4 \times \text{Side}$

$\text{Side} = \frac{400}{4} = 100\text{ m}$

Now, Area of the square field = $\text{Side} \times \text{Side}$

$\text{Area} = 100\text{ m} \times 100\text{ m} = 10000\text{ m}^2$

The area of the square field is $10000\text{ m}^2$.

Given:

Length of the room = $4\text{ m}$, Breadth of the room = $3\text{ m}$.

Side of the square tile = $20\text{ cm}$.

To Find: Number of tiles required.

Solution:

First, we convert all dimensions to the same unit (cm).

Length of the floor = $4 \times 100\text{ cm} = 400\text{ cm}$

Breadth of the floor = $3 \times 100\text{ cm} = 300\text{ cm}$

Area of the floor = $\text{Length} \times \text{Breadth} = 400\text{ cm} \times 300\text{ cm} = 120000\text{ cm}^2$

Area of one square tile = $\text{Side} \times \text{Side} = 20\text{ cm} \times 20\text{ cm} = 400\text{ cm}^2$

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

Number of tiles = $\frac{120000}{400}$

Number of tiles = $\frac{\cancel{120000}^{300}}{\cancel{400}_{1}}$

The number of tiles required is $300$.

Example 8

Given: A figure (Fig. 6.2) with the following side lengths:

$AB = 4$, $BC = 2$, $CD = 2$, $DE = 3$, $EF = 1$, $IH = 4$, $IJ = 3$ and $AJ = 3$.

To Find: The area of the figure by splitting it into four vertical rectangles.

Construction Required:

Split the figure into four vertical rectangles by drawing vertical lines. As per the dimensions provided:

1. Rectangle 1 has dimensions $4 \times 2$.

2. Rectangle 2 has dimensions $6 \times 1$.

3. Rectangle 3 has dimensions $3 \times 2$.

4. Rectangle 4 has dimensions $4 \times 2$.

Solution:

To find the total area, we calculate the area of each individual rectangle and then add them together.

Area of a rectangle = $\text{length} \times \text{breadth}$

Now, we find the Total Area by adding the results of equations (i), (ii), (iii), and (iv):

Total Area = $\text{Area of Rectangle 1} + \text{Area of Rectangle 2} $$ + \text{Area of Rectangle 3} + \text{Area of Rectangle 4}$

Total Area = $8 + 6 + 6 + 8$

Thus, the total area of the figure is $28 \text{ sq units}$.

Alternate Solution

We can also find the area by splitting the figure into horizontal rectangles. However, based on the specific vertical split provided, the calculation above is the most direct method to achieve the result as per the requirements.

Analysis of Perimeter

Given: Four figures (i), (ii), (iii), and (iv) are each composed of 6 unit squares. A unit square has a side length of 1 unit.

To Find: The figure with the smallest perimeter.

Solution:

The perimeter of any shape is the total length of its outer boundary. Since each square is a unit square, we calculate the perimeter by counting the number of unit segments along the outside of each figure.

For Figure (i):

This figure forms a rectangle with a length of 3 units and a breadth of 2 units.

Perimeter of rectangle = $2 \times (\text{length} + \text{breadth})$

For Figure (ii):

By counting the external unit segments of the boundary:

Top edges = $2 \text{ units}$, Bottom edges = $1 \text{ unit}$, Left edges = $4 \text{ units}$, Right edges = $4 \text{ units}$, and internal steps = $1 \text{ unit}$.

For Figure (iii):

By counting the unit segments along the outer boundary of this shape:

For Figure (iv):

By counting the unit segments along the outer boundary of this shape:

Comparison:

Comparing the perimeters obtained from (i), (ii), (iii), and (iv):

The smallest perimeter is $10 \text{ units}$, which corresponds to Figure (i).

Therefore, the correct option is (D).

Given:

Side of the square park $ABCD = 100\text{ m}$.

Dimensions of each rectangular flower bed = $10\text{ m} \times 5\text{ m}$.

The flower beds are located at the corners of the square park.

To Find:

The length of the boundary (perimeter) of the remaining park.

Solution:

The length of the boundary of a shape is its perimeter.

Initially, the perimeter of the square park $ABCD$ is calculated as:

When a rectangular shape is removed from the corner of a square, the boundary length replaced inside the park is equal to the boundary length removed from the outside.

Specifically, for each rectangular flower bed of $10\text{ m} \times 5\text{ m}$:

The segments of the original square boundary removed are one length ($10\text{ m}$) and one breadth ($5\text{ m}$).

The new segments added to the boundary inside the park are also one length ($10\text{ m}$) and one breadth ($5\text{ m}$).

Therefore, there is no change in the total length of the boundary.

Length of the boundary of the remaining park = $400\text{ m}$.

The correct option is (B).

Given:

Original side of the square ($s$) = $10\text{ cm}$.

The side is doubled, so the new side ($s'$) = $2 \times 10\text{ cm} = 20\text{ cm}$.

To Find:

How many times the new perimeter is compared to the original perimeter.

Solution:

First, let us calculate the original perimeter ($P_1$):

Now, let us calculate the new perimeter ($P_2$) with the doubled side:

To find the ratio:

The new perimeter becomes 2 times the original perimeter.

Alternate Solution:

Let the side of the square be $s$.

Perimeter $P = 4s$.

If the side is doubled, the new side is $2s$.

New Perimeter $P' = 4(2s) = 8s$.

Comparing $P'$ and $P$:

$P' = 2P$.

The correct option is (A).

Given:

Length of the rectangular sheet ($l$) = $20\text{ cm}$.

Breadth of the rectangular sheet ($b$) = $10\text{ cm}$.

A rectangular piece of size $5\text{ cm} \times 2\text{ cm}$ is cut from one corner.

To Find:

The correct statement regarding the change in Area and Perimeter of the remaining sheet.

Solution:

1. Regarding Area:

The area of the original rectangular sheet is calculated as:

Since a rectangular piece is removed from the sheet, the area of the remaining sheet will definitely decrease. Specifically, the new area will be $200\text{ cm}^2 - (5 \times 2)\text{ cm}^2 = 190\text{ cm}^2$.

2. Regarding Perimeter:

The perimeter is the total length of the boundary. The original perimeter is:

When a rectangle is cut from a corner, the segments that were removed from the outer boundary ($5\text{ cm}$ horizontal and $2\text{ cm}$ vertical) are replaced by two new segments of the same lengths inside the sheet to form the new boundary.

Thus, the total length of the boundary (perimeter) remains the same ($60\text{ cm}$).

Therefore, Perimeter remains same but area changes.

The correct option is (A).

Given:

Two regular hexagons are provided.

Perimeter of each regular hexagon = $30\text{ cm}$.

To Find:

The perimeter of the new figure formed by joining them.

Solution:

A regular hexagon has $6$ equal sides. Let the side length be $s$.

When two hexagons are joined along one side, that side becomes internal and is not part of the boundary (perimeter).

The new figure has $5$ sides from the first hexagon and $5$ sides from the second hexagon on its boundary.

The correct option is (D).

Given:

Four figures (i), (ii), (iii), and (iv) are shown.

Condition: Figures (ii), (iii), and (iv) have all sides equal. Figure (i) does not have all sides equal.

To Find:

Which of the following is a regular polygon.

Solution:

A polygon is said to be a regular polygon if it satisfies two conditions:

1. All its sides are of equal length (Equilateral).

2. All its interior angles are of equal measure (Equiangular).

Let us evaluate each figure:

(i) Rectangle: In this figure, sides are not equal (as per the question). Thus, it is not a regular polygon.

(ii) Square: All sides are equal and each interior angle is $90^\circ$. It satisfies both conditions. Thus, it is a regular polygon.

(iii) Rhombus: All sides are equal, but the interior angles are not equal (opposite angles are equal, but adjacent angles are different). Thus, it is not a regular polygon.

(iv) Plus sign shape (Decagon): Although all sides are equal, the interior angles are not all the same. It has some internal angles of $90^\circ$ and some reflex angles of $270^\circ$. Thus, it is not a regular polygon.

Therefore, figure (ii) is the only regular polygon.

The correct option is (B).

Given:

Length of each side segment = $2\text{ cm}$.

Number of sides for shape (A) = $14$.

Number of sides for shape (B) = $8$.

Number of sides for shape (C) = $10$.

Number of sides for shape (D) = $12$.

To Find:

Perimeter of each shape and matching with Column II.

Solution:

The perimeter of a polygon is the total length of its boundary. Since all sides are of equal length, we use the formula:

$\text{Perimeter} = \text{Number of sides} \times \text{Length of each side}$

For Shape (A):

For Shape (B):

For Shape (C):

For Shape (D):

Matching Result Table:

Given:

Figures in Column I with specified side lengths.

To Find:

The perimeter of each figure.

Solution:

(A) Rectangle:

Length ($l$) = $6$, Breadth ($b$) = $4$.

This matches with (iii).

(B) Square:

Side ($s$) = $5$.

This matches with (iii).

(C) Equilateral Triangle:

Side ($s$) = $6$.

This matches with (ii).

(D) Isosceles Triangle:

Sides are $4$, $4$, and $2$.

This matches with (i).

Final Result:

The matching pairs are: A - (iii), B - (iii), C - (ii), D - (i).

Given:

A shaded figure inside a grid with vertices labeled A, B, C, D, E, F, G, H, M, and N.

To Find:

The sequence of line segments that make up the perimeter of the shaded portion.

Solution:

The perimeter of a figure is the total length of its boundary. To find the perimeter of the shaded portion, we follow the outer boundary line starting from point A and moving clockwise until we return to point A.

The path is as follows:

1. From A to B (AB)

2. From B to M (BM)

3. From M to D (MD)

4. From D to E (DE)

5. From E to N (EN)

6. From N to G (NG)

7. From G to H (GH)

8. From H to A (HA)

Therefore, the complete expression for the perimeter is:

AB + BM + MD + DE + EN + NG + GH + HA

Solution:

The amount of surface or region enclosed within the boundary of a closed plane figure is defined as its Area.

Answer: Area

Given:

Length of the rectangle ($l$) = $5\text{ cm}$

Breadth of the rectangle ($b$) = $3\text{ cm}$

To Find:

Area of the rectangle.

Solution:

Answer: $15\text{ sq cm}$

Given:

Length of rectangle ($l$) = $6$ units

Breadth of rectangle ($b$) = $2$ units

Perimeter of Square = Perimeter of Rectangle

Solution:

(a) Finding Area of Rectangle:

(b) Finding Area of Square:

First, find the perimeter of the rectangle:

Since the perimeters are the same, Perimeter of Square = $16$ units.

Now, calculate the area of the square:

Answers: (a) $12$, (b) $16$

Solution:

(a) We know that $1$ metre is equal to $100$ centimetres.

Answer: $100$

(b) $1$ square centimetre is the area of a square with side $1\text{ cm}$.

Answer: $1$

(c) $1$ square metre is the area of a square with side $1\text{ m}$.

Since $1\text{ m} = 100\text{ cm}$:

Answer: $1$, $100$

(d) From the previous part:

Answer: $10,000$

Given:

Original length of rectangle = $l$

Original breadth of rectangle = $b$

New length = $\frac{l}{2}$

New breadth = $2b$

Solution:

Area of the original rectangle is given by:

Area of the new rectangle is given by:

Since $\text{Area}_1 = \text{Area}_2$, the area remains the same.

Answer: True

Given:

Side of the original square = $s$

New side of the square = $2s$

Solution:

Area of the original square is:

Area of the new square is:

The new area is $4$ times the original area, not double ($2$ times).

Answer: False

Given:

Shape is a regular octagon (Number of sides, $n = 8$).

Side of the octagon ($s$) = $6\text{ cm}$.

Solution:

The calculated perimeter is $48\text{ cm}$, but the statement says $36\text{ cm}$.

Answer: False

Solution:

Fencing is done along the boundary of a field. The total length of the boundary of a closed figure is its perimeter.

Therefore, to calculate the amount of fencing material required, the farmer must find the perimeter.

Answer: True

Solution:

A compound wall is built along the boundary of the property. To determine the length of the wall, one needs to calculate the perimeter, not the area.

Area represents the surface region inside the boundary, which is irrelevant for the length of the wall.

Answer: False

Solution:

Painting is an activity that covers the entire surface of the wall. The measure of this surface is the area of the wall.

The cost of painting is usually calculated per square unit of area. Perimeter only gives the length of the boundary.

Answer: False

Solution:

A frame is placed around the edges (boundary) of a picture. To find the length of the framing material needed, we must calculate the total length of the boundary.

This measure is the perimeter of the picture.

Answer: True

Given:

Total perimeter of the design = $28\text{ cm}$.

The design consists of four regular hexagons joined together.

To Find:

The length of each side of the regular hexagon.

Solution:

First, we need to count the number of side segments that form the outer boundary (perimeter) of the given design in Fig. 6.11.

By observing the figure:

1. The top hexagon contributes $3$ side segments to the boundary.

2. The bottom hexagon contributes $3$ side segments to the boundary.

3. The left hexagon contributes $4$ side segments to the boundary.

4. The right hexagon contributes $4$ side segments to the boundary.

Total number of sides on the boundary = $3 + 3 + 4 + 4 = 14$ sides.

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

Thus, the length of each side of the hexagon is $2\text{ cm}$.

Given:

Perimeter of the isosceles triangle = $50\text{ cm}$.

Length of one of the equal sides = $18\text{ cm}$.

To Find:

Length of the third side.

Solution:

In an isosceles triangle, two sides are equal in length.

Let the equal sides be $a = 18\text{ cm}$ and the third side be $b$.

The length of the third side is $14\text{ cm}$.

Given:

Perimeter of the rectangle = $40\text{ cm}$.

Condition: Length ($l$) is three times the breadth ($b$).

To Find:

Length ($l$) and Width (Breadth, $b$) of the rectangle.

Solution:

Let the breadth of the rectangle be $x\text{ cm}$.

According to the question, the length is:

We know the formula for the perimeter of a rectangle:

Substituting the values:

Now, find the length using equation (i):

The dimensions of the rectangle are: Length = $15\text{ cm}$ and Width = $5\text{ cm}$.

Given:

Length of the rectangular lawn ($l$) = $10\text{ m}$

Width (breadth) of the rectangular lawn ($b$) = $4\text{ m}$

Sides to be fenced: Two smaller sides and one longer side.

Entrance gap = $1\text{ m}$

To Find:

The total length of the fencing required.

Solution:

According to the problem, the fencing is done on three sides of the rectangle: two smaller sides (widths) and one longer side (length).

However, there is a gap of $1\text{ m}$ left for the entrance. This gap should be subtracted from the total boundary length to find the actual length of the fence.

The length of the fencing is $17\text{ m}$.

Given:

A unit of area represented by one small rectangle $\square$.

A figure (Fig. 6.13) composed of these unit rectangles.

To Find:

The total area of the region in terms of the given unit.

Solution:

The total area of the region is the sum of all the unit rectangles contained within the boundary of Fig. 6.13.

Looking at the figure, it is divided into two vertical columns:

1. The left column consists of $8$ unit rectangles stacked vertically.

2. The right column consists of $5$ unit rectangles stacked vertically.

Since each small rectangle represents one unit of area, the area of the region is $13$ units.

Given:

Distance around the square field (Perimeter) = $200$ rods.

Length of one rod = $140\text{ cm}$.

To Find:

The length of the side of the field in metres.

Solution:

First, calculate the total distance around the field (Perimeter) in centimetres.

Now, convert the perimeter into metres ($1\text{ m} = 100\text{ cm}$):

Since the field is square, all four sides are equal. Let the side be $s$.

The side of the field is $70\text{ m}$.

Given:

Total distance covered in $4$ rounds = $6\text{ km}$.

Condition: Length ($l$) = $2 \times$ Breadth ($b$).

To Find:

The length of the field ($l$).

Solution:

First, convert the total distance into metres ($1\text{ km} = 1000\text{ m}$):

Calculate the distance covered in one round (Perimeter):

Let the breadth of the field be $x\text{ m}$. Then the length is $2x\text{ m}$.

Now, calculate the length:

The length of the field is $500\text{ m}$.

Alternate Solution:

We know Perimeter $P = 1500\text{ m}$ and $l = 2b$.

Substitute $b = \frac{l}{2}$ into the perimeter formula:

Given:

Three squares are joined together with sides $4\text{ cm}$, $10\text{ cm}$ and $3\text{ cm}$ as shown in Fig. 6.14.

To Find:

The perimeter of the resulting figure.

Solution:

The perimeter of the figure is the sum of all its outer boundaries. Let us calculate the lengths of the horizontal and vertical boundaries separately.

1. Horizontal Boundaries:

The bottom boundary is the sum of the bases of the three squares:

$\text{Bottom length} = 4\text{ cm} + 10\text{ cm} + 3\text{ cm} = 17\text{ cm}$

The top boundary consists of the top sides of each square:

$\text{Top length} = 4\text{ cm} + 10\text{ cm} + 3\text{ cm} = 17\text{ cm}$

2. Vertical Boundaries:

The leftmost outer vertical side is $4\text{ cm}$.

The rightmost outer vertical side is $3\text{ cm}$.

There are also vertical segments where the squares of different heights meet:

The difference in height between the middle square and the left square:

The difference in height between the middle square and the right square:

3. Total Perimeter:

$\text{Perimeter} = \text{Sum of all outer segments}$

$\text{Perimeter} = 17 \text{ (bottom)} + 4 \text{ (top left)} + 10 \text{ (top middle)} $$ + 3 \text{ (top right)} + 4 \text{ (left)} + 3 \text{ (right)} + 6 \text{ (gap 1)} + 7 \text{ (gap 2)}$

$\text{Perimeter} = 17 + 17 + 4 + 3 + 6 + 7$

$\text{Perimeter} = 54\text{ cm}$

Final Answer: The perimeter of the figure is $54\text{ cm}$.

Given:

In Fig. 6.15, all triangles are equilateral. The largest triangle $ABC$ has a side $AB = 8$ units. Subsequent triangles are formed by taking the midpoints of the sides.

To Find:

The perimeter of the figure.

Solution:

The figure consists of a central equilateral triangle of side $8$ units, with branches of smaller triangles attached to each side. Based on the midpoint property, the side lengths of the triangles at each level are:

Side of Level 1 triangle = $8$ units

Side of Level 2 triangle = $4$ units

Side of Level 3 triangle = $2$ units

Side of Level 4 triangle = $1$ unit

The perimeter is the sum of the outer boundary segments. We calculate the contribution of each triangle level to the total perimeter for all three sides of the central triangle:

1. Contribution of the $8$-unit triangle:

On each side of $8$ units, $4$ units are covered by the base of the next triangle. The outer part is $8 - 4 = 4$ units.

$\text{Total contribution} = 3 \times (8 - 4) = 12\text{ units}$

2. Contribution of the $4$-unit triangles:

There are 3 such triangles. For each, one side ($4$ units) is completely outer, and another side has $2$ units covered by the next triangle ($4 - 2 = 2$ units outer).

$\text{Total contribution} = 3 \times (4 + 2) = 18\text{ units}$

3. Contribution of the $2$-unit triangles:

There are 3 such triangles. For each, one side ($2$ units) is outer, and another side has $1$ unit covered by the next triangle ($2 - 1 = 1$ unit outer).

$\text{Total contribution} = 3 \times (2 + 1) = 9\text{ units}$

4. Contribution of the $1$-unit triangles:

There are 3 such triangles. For each, two sides are outer boundaries.

$\text{Total contribution} = 3 \times (1 + 1) = 6\text{ units}$

Total Perimeter:

$\text{Total Perimeter} = 12 + 18 + 9 + 6$

Final Answer: The perimeter of the figure is $45\text{ units}$.

Given:

Length of rectangular field ($l$) = $250\text{ m}$

Width of rectangular field ($w$) = $150\text{ m}$

Total distance to cover in second case = $4\text{ km} = 4000\text{ m}$

To Find:

1. Distance covered in 3 rounds.

2. Number of rounds required to cover $4\text{ km}$.

Solution:

First, we find the distance covered in one round, which is equal to the perimeter of the rectangular field.

$\text{Perimeter} = 2 \times (250\text{ m} + 150\text{ m})$

$\text{Perimeter} = 2 \times 400\text{ m}$

$\text{Perimeter} = 800\text{ m}$

Case 1: Distance covered in 3 rounds

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

$\text{Distance} = 3 \times 800\text{ m}$

$\text{Distance} = 2400\text{ m}$

Converting to kilometers ($1000\text{ m} = 1\text{ km}$):

$\text{Distance} = 2.4\text{ km}$

Case 2: Number of rounds for $4\text{ km}$

We know that $4\text{ km} = 4000\text{ m}$.

$\text{Number of rounds} = \frac{4000}{800}$

$\text{Number of rounds} = \frac{\cancel{4000}^{5}}{\cancel{800}_{1}}$

$\text{Number of rounds} = 5$

Final Answer: Anuradha ran $2400\text{ m}$ (or $2.4\text{ km}$) in 3 rounds. She should run $5$ times around the field to cover $4\text{ km}$.

Given:

Total distance covered = $4\text{ km}$

Number of rounds = $10$

To Find:

The length of the track (side of the square track).

Solution:

First, we convert the total distance into metres:

$4\text{ km} = 4 \times 1000\text{ m} = 4000\text{ m}$

The distance covered in $10$ rounds is the perimeter of the square track multiplied by $10$. Therefore, the perimeter of the square track is:

$\text{Perimeter} = \frac{\text{Total Distance}}{\text{Number of rounds}}$

$\text{Perimeter} = \frac{\cancel{4000}^{400}}{\cancel{10}_{1}}\text{ m}$

$\text{Perimeter} = 400\text{ m}$

Since the track is a square, the perimeter is given by $4 \times \text{side}$.

$\text{side} = \frac{\cancel{400}^{100}}{\cancel{4}_{1}}$

$\text{side} = 100\text{ m}$

Final Answer: The length of the track (each side) is $100\text{ m}$.

Given:

Dimensions of Molly’s lawn = $12\text{ m} \times 8\text{ m}$

Dimensions of Dolly’s lawn = $15\text{ m} \times 5\text{ m}$

To Find:

Total fencing required for both lawns.

Solution:

Fencing is built around the boundary, so we need to find the perimeter of both rectangular lawns.

For Molly’s Lawn:

$\text{Length} (l_1) = 12\text{ m}, \text{ Breadth} (b_1) = 8\text{ m}$

$\text{Perimeter} = 2 \times (l_1 + b_1)$

$\text{Perimeter} = 2 \times (12 + 8) = 2 \times 20 = 40\text{ m}$

For Dolly’s Lawn:

$\text{Length} (l_2) = 15\text{ m}, \text{ Breadth} (b_2) = 5\text{ m}$

$\text{Perimeter} = 2 \times (l_2 + b_2)$

$\text{Perimeter} = 2 \times (15 + 5) = 2 \times 20 = 40\text{ m}$

Total Fencing Required:

$\text{Total fencing} = \text{Perimeter of Molly's lawn} $$ + \text{Perimeter of Dolly's lawn}$

$\text{Total fencing} = 80\text{ m}$

Final Answer: Total fencing required for both lawns is $80\text{ m}$.

Given:

Perimeter of a regular pentagon = $1540\text{ cm}$

To Find:

Length of each side.

Solution:

A regular pentagon has five equal sides.

$1540 = 5 \times \text{side}$

$\text{side} = \frac{1540}{5}$

$\text{side} = 308\text{ cm}$

Final Answer: The length of each side of the regular pentagon is $308\text{ cm}$.

Given:

Perimeter of the triangle = $28\text{ cm}$

Length of one side = $8\text{ cm}$

To Find:

All possible sides of an isosceles triangle with these measurements.

Solution:

An isosceles triangle has two equal sides. Let the sides of the triangle be $a, b,$ and $c$. We are given one side is $8\text{ cm}$. There are two possible cases:

Case 1: The given side ($8\text{ cm}$) is the unequal side (base).

Let the two equal sides be $x$.

$x + x + 8 = 28$

$2x = 28 - 8$

$2x = 20$

$x = 10\text{ cm}$

In this case, the sides are $(10\text{ cm}, 10\text{ cm}, 8\text{ cm})$. Since $10 + 8 > 10$, this triangle is possible.

Case 2: The given side ($8\text{ cm}$) is one of the two equal sides.

Let the equal sides be $8\text{ cm}$ and $8\text{ cm}$, and the third side be $y$.

$8 + 8 + y = 28$

$16 + y = 28$

$y = 28 - 16$

$y = 12\text{ cm}$

In this case, the sides are $(8\text{ cm}, 8\text{ cm}, 12\text{ cm})$. Since $8 + 8 > 12$ (sum of two sides is greater than the third), this triangle is also possible.

Final Answer: The possible sets of sides for the isosceles triangle are $(10\text{ cm}, 10\text{ cm}, 8\text{ cm})$ and $(8\text{ cm}, 8\text{ cm}, 12\text{ cm})$.

Given:

Total length of aluminium strip (Perimeter) = $40\text{ cm}$

Dimensions must be natural numbers.

To Find:

All possible pairs of length ($l$) and breadth ($b$) for the rectangular frames.

Solution:

The length of the strip is equal to the perimeter of the rectangle.

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

$l + b = 20$

We need to find pairs of natural numbers $(l, b)$ such that their sum is $20$. By convention, we usually take $l \ge b$.

The possible pairs are:

Final Answer: There are 10 possible rectangular frames with dimensions (in cm): (19, 1), (18, 2), (17, 3), (16, 4), (15, 5), (14, 6), (13, 7), (12, 8), (11, 9), and (10, 10).

Given:

Shape of the base = Regular hexagon

Perimeter of the base = $60\text{ cm}$

To Find:

Length of each side of the base.

Solution:

A regular hexagon is a polygon with six equal sides.

Substituting the given perimeter:

$60\text{ cm} = 6 \times \text{side}$

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

$\text{side} = 10\text{ cm}$

Final Answer: The length of each side of the base is $10\text{ cm}$.

Given:

Total number of display boards = $24$

Length of one board ($l$) = $1\text{ m } 50\text{ cm} = 1.5\text{ m}$

Breadth of one board ($b$) = $1\text{ m}$

Total length of aluminium strip available = $100\text{ m}$

To Find:

1. Number of boards that can be framed with $100\text{ m}$ strip.

2. Length of strip required for the remaining boards.

Solution:

The framing is done along the boundary, so we first find the perimeter of one display board.

$\text{Perimeter} = 2 \times (1.5\text{ m} + 1\text{ m})$

$\text{Perimeter} = 2 \times 2.5\text{ m}$

$\text{Perimeter of one board} = 5\text{ m}$

1. Number of boards framed:

$\text{Number of boards} = \frac{\text{Total strip available}}{\text{Perimeter of one board}}$

$\text{Number of boards} = \frac{\cancel{100}^{20}}{\cancel{5}_{1}}$

$\text{Number of boards} = 20$

2. Length required for remaining boards:

$\text{Remaining boards} = 24 - 20 = 4$

$\text{Strip required for 4 boards} = 4 \times \text{Perimeter of one board}$

$\text{Strip required} = 4 \times 5\text{ m} = 20\text{ m}$

Final Answer: $20$ boards will be framed using the available strip, and $20\text{ m}$ more strip is required for the remaining boards.

Given:

Total number of boards = $24$

Length of one board ($l$) = $1.5\text{ m}$

Breadth of one board ($b$) = $1\text{ m}$

Breadth of the cloth = $120\text{ cm} = 1.2\text{ m}$

To Find:

1. Total area of cloth required (in square metres).

2. Length of the cloth if its breadth is $1.2\text{ m}$.

Solution:

To cover the boards, we need to calculate the area.

$\text{Area of one board} = 1.5\text{ m} \times 1\text{ m} = 1.5\text{ m}^2$

1. Total area for 24 boards:

$\text{Total Area} = 24 \times 1.5\text{ m}^2$

$\text{Total Area} = 36\text{ m}^2$

2. Length of the cloth:

The total area of the cloth must equal the total area of the boards.

$\text{Area of cloth} = \text{Length of cloth} \times \text{Breadth of cloth}$

$36\text{ m}^2 = \text{Length} \times 1.2\text{ m}$

$\text{Length} = \frac{36}{1.2}$

To simplify, multiply numerator and denominator by $10$:

$\text{Length} = \frac{\cancel{360}^{30}}{\cancel{12}_{1}}$

$\text{Length} = 30\text{ m}$

Final Answer: Total $36\text{ sq m}$ of cloth is required, and the length of the cloth will be $30\text{ m}$.

Given:

The outer sides of the park from Fig. 6.16 are $200\text{ m}$, $300\text{ m}$, $80\text{ m}$, $300\text{ m}$, $200\text{ m}$ and $260\text{ m}$.

Rate of fencing = $\textsf{₹} 20$ per metre.

Dimensions of rectangular flower bed: Length ($l$) = $100\text{ m}$, Breadth ($b$) = $80\text{ m}$.

Rate of manuring = $\textsf{₹} 50$ per square metre.

To Find:

1. Length of the outer boundary (Perimeter).

2. Total cost of fencing the park.

3. Cost of manuring the flower bed.

Solution:

1. Length of the outer boundary:

The length of the outer boundary is the sum of all external sides of the park.

$\text{Perimeter} = 200\text{ m} + 300\text{ m} + 80\text{ m} + 300\text{ m} + 200\text{ m} + 260\text{ m}$

$\text{Length of outer boundary} = 1340\text{ m}$

2. Total cost of fencing:

$\text{Cost of fencing} = 1340 \times 20$

$\text{Cost of fencing} = \textsf{₹} 26,800$

3. Cost of manuring the flower bed:

First, calculate the area of the rectangular flower bed.

$\text{Area of flower bed} = l \times b$

$\text{Area} = 100\text{ m} \times 80\text{ m} = 8000\text{ m}^2$

$\text{Cost of manuring} = 8000 \times 50$

$\text{Cost of manuring} = \textsf{₹} 4,00,000$

Final Answer: The length of the outer boundary is $1340\text{ m}$. The total cost of fencing is $\textsf{₹} 26,800$. The cost of manuring the flower bed is $\textsf{₹} 4,00,000$.

Given:

The sides of the park from Fig. 6.17 are $150\text{ m}$, $100\text{ m}$, $120\text{ m}$, $180\text{ m}$, $270\text{ m}$ and $280\text{ m}$.

Total cost of fencing = $\textsf{₹} 55,000$.

To Find:

Cost of fencing per metre.

Solution:

First, we find the total perimeter (length of fencing) of the park by adding all its outer sides.

$\text{Perimeter} = 150 + 100 + 120 + 180 + 270 + 280$

$\text{Total Perimeter} = 1100\text{ m}$

Now, we find the cost per metre.

$\text{Cost per metre} = \frac{55000}{1100}$

$\text{Cost per metre} = \frac{\cancel{550}^{50}}{\cancel{11}_{1}}$

$\text{Cost per metre} = \textsf{₹} 50$

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

Given:

Rectangle $ABCD$ is composed of unit squares (side = $1$ unit).

By counting the small squares in Fig. 6.18:

Length of the rectangle ($l$) = $10$ units

Breadth of the rectangle ($b$) = $6$ units

To Find:

(a) Perimeter of the rectangle $ABCD$.

(b) Area of the rectangle $ABCD$.

(c) Ten parts of equal area.

(d) Perimeter of each part.

Solution:

(a) Perimeter of the rectangle $ABCD$:

$\text{Perimeter} = 2 \times (10 + 6) = 2 \times 16$

$\text{Perimeter} = 32\text{ units}$

(b) Area of the rectangle $ABCD$:

$\text{Area} = 10 \times 6 = 60\text{ square units}$

(c) Division into ten parts of equal area:

Total Area = $60$ sq units. To divide it into $10$ equal parts:

$\text{Area of each part} = \frac{60}{10} = 6\text{ square units}$

Each shaded part must consist of $6$ unit squares. The figure already shows two such parts.

(d) Perimeter of each part:

Let's calculate the perimeter of the two already shaded parts in Fig. 6.18.

For the lower shaded portion (staircase shape):

Number of unit squares = $6$

Tracing the outer boundary edges: $1 \text{ (top)} + 1 + 1 + 1 + 1 + 2 + 1 + 2 + 1 + 1 + 1 + 1 = 14\text{ units}$.

For the upper shaded portion (cross-like shape):

Number of unit squares = $6$

Tracing the outer boundary edges: $1 \text{ (top)} + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12\text{ units}$.

Since $14 \neq 12$, we conclude that the perimeters of the parts are not all equal, even though their areas are the same.

Final Answer: (a) $32\text{ units}$, (b) $60\text{ sq units}$, (d) Perimeter of top part is $14\text{ units}$ and bottom part is $12\text{ units}$. They are not equal.

Given:

Side of each square tile = $15\text{ cm}$

From Fig 6.19, we count the number of tiles:

Number of tiles along the length ($PO$ or $MN$) = $7$

Number of tiles along the breadth ($PM$ or $ON$) = $4$

To Find:

The area of the wall $MNOP$.

Solution:

First, let us find the actual dimensions of the wall in cm.

$\text{Length of the wall} = 7 \times 15\text{ cm} = 105\text{ cm}$

$\text{Breadth of the wall} = 4 \times 15\text{ cm} = 60\text{ cm}$

Now, we calculate the area of the rectangular wall.

$\text{Area} = 105\text{ cm} \times 60\text{ cm}$

$\text{Area} = 6300\text{ cm}^2$

Final Answer: The area of the wall is $6300\text{ cm}^2$.

Given:

Length of the rectangular field ($l$) = $120\text{ cm}$

Relation between length and breadth: $l = 6 \times \text{breadth } (b)$

To Find:

1. Breadth of the field ($b$)

2. Perimeter of the field

Solution:

1. Finding the breadth ($b$):

According to the given condition:

$120 = 6 \times b$

$b = \frac{120}{6}$

$b = 20\text{ cm}$

2. Finding the perimeter:

The formula for the perimeter of a rectangle is:

Substituting the values of $l$ and $b$:

$\text{Perimeter} = 2 \times (120 + 20)$

$\text{Perimeter} = 2 \times 140$

$\text{Perimeter} = 280\text{ cm}$

Final Answer: The breadth of the field is $20\text{ cm}$ and the perimeter is $280\text{ cm}$.

Given:

Dimensions of Anmol's chart paper = $90\text{ cm} \times 40\text{ cm}$

Dimensions of Abhishek's chart paper = $50\text{ cm} \times 70\text{ cm}$

To Find:

Which chart paper has more area and the difference between them.

Solution:

Area of Anmol's chart paper:

$\text{Area}_1 = \text{length} \times \text{breadth}$

$\text{Area}_1 = 90 \times 40 = 3600\text{ cm}^2$

Area of Abhishek's chart paper:

$\text{Area}_2 = \text{length} \times \text{breadth}$

$\text{Area}_2 = 50 \times 70 = 3500\text{ cm}^2$

Comparison:

Comparing both areas, we see that $3600\text{ cm}^2 > 3500\text{ cm}^2$.

$\text{Difference} = 100\text{ cm}^2$

Final Answer: Anmol's chart paper will cover more area by $100\text{ cm}^2$.

Given:

Length of path ($L$) = $60\text{ m} = 6000\text{ cm}$

Width of path ($W$) = $3\text{ m} = 300\text{ cm}$

Side of square tile ($s$) = $25\text{ cm}$

To Find:

1. Number of tiles in one row along the width.

2. Number of such rows along the length.

3. Total number of tiles used.

Solution:

1. Number of tiles in one row along width:

$\text{Tiles along width} = \frac{\text{Width of path}}{\text{Side of tile}}$

$\text{Tiles along width} = \frac{300}{25} = 12$

2. Number of such rows (along the length):

$\text{Number of rows} = \frac{\text{Length of path}}{\text{Side of tile}}$

$\text{Number of rows} = \frac{6000}{25} = 240$

3. Total number of tiles used:

$\text{Total tiles} = \text{Tiles along width} \times \text{Number of rows}$

$\text{Total tiles} = 12 \times 240 = 2880$

Alternate Solution:

Total number of tiles = $\frac{\text{Area of path}}{\text{Area of one tile}}$

$\text{Total tiles} = \frac{6000 \times 300}{25 \times 25}$

$\text{Total tiles} = \frac{\cancel{6000}^{240} \times \cancel{300}^{12}}{\cancel{25}_{1} \times \cancel{25}_{1}} = 240 \times 12 = 2880$

Final Answer: There are $12$ tiles in a row along the width, $240$ rows, and a total of $2880$ tiles were used.

Given:

Side of square slab = $90\text{ cm} = 0.9\text{ m}$

Area of floor = $81\text{ m}^2$

To Find:

Number of square slabs.

Solution:

First, we find the area of one square slab in square metres.

$\text{Area of one slab} = \text{side} \times \text{side}$

$\text{Area of one slab} = 0.9\text{ m} \times 0.9\text{ m} = 0.81\text{ m}^2$

Now, we find the total number of slabs required:

$\text{Number of slabs} = \frac{\text{Total area of floor}}{\text{Area of one slab}}$

$\text{Number of slabs} = \frac{81}{0.81}$

To simplify, multiply numerator and denominator by $100$:

$\text{Number of slabs} = \frac{8100}{81}$

$\text{Number of slabs} = 100$

Final Answer: A total of $100$ square slabs are needed.

Given:

Length of rectangular field ($l$) = $8\text{ m}$

Breadth of rectangular field ($b$) = $2\text{ m}$

Perimeter of square field = Perimeter of rectangular field

To Find:

Compare the areas of both fields and find which is greater.

Solution:

1. Calculations for the Rectangular Field:

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

$\text{Perimeter} = 2 \times (8 + 2) = 2 \times 10 = 20\text{ m}$

$\text{Area of rectangle} = l \times b$

$\text{Area of rectangle} = 8 \times 2 = 16\text{ m}^2$

2. Calculations for the Square Field:

The perimeter of the square field is also $20\text{ m}$.

$\text{side} = \frac{20}{4} = 5\text{ m}$

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

$\text{Area of square} = 5 \times 5 = 25\text{ m}^2$

Comparison:

Area of square field = $25\text{ m}^2$

Area of rectangular field = $16\text{ m}^2$

Since $25\text{ m}^2 > 16\text{ m}^2$:

Final Answer: The square field has the greater area.

Given:

Distance covered in one round (Perimeter) = $800\text{ m}$

To Find:

Area of the square park.

Solution:

The distance covered in one round of a square park is equal to its perimeter.

$800 = 4 \times \text{side}$

$\text{side} = \frac{800}{4}$

$\text{side} = 200\text{ m}$

Now, we find the area of the square park:

$\text{Area} = 200 \times 200$

$\text{Area} = 40,000\text{ m}^2$

Final Answer: The area of the park is $40,000\text{ m}^2$.

Given:

Initial side of the square ($s_1$) = $5\text{ cm}$

Solution:

1. Initial Area:

$\text{Initial Area} = s_1 \times s_1$

$\text{Initial Area} = 5 \times 5 = 25\text{ cm}^2$

2. Area when side is doubled:

New side ($s_2$) = $2 \times 5 = 10\text{ cm}$

$\text{New Area} = s_2 \times s_2$

$\text{New Area} = 10 \times 10 = 100\text{ cm}^2$

3. Comparison:

$\text{Increase in area} = \frac{\text{New Area}}{\text{Initial Area}}$

$\text{Increase in area} = \frac{\cancel{100}^4}{\cancel{25}_1}$

$\text{Increase in area} = 4\text{ times}$

Final Answer: If the side of the square is doubled, its area increases by $4$ times.

Given:

Dimensions of rectangular card = $8\text{ cm} \times 5\text{ cm}$

Side of square chart paper = $60\text{ cm}$

To Find:

1. Number of complete cards that can be made.

2. Area of the chart paper left.

Solution:

1. Number of cards:

To find how many cards can be placed, we check how many fit along the length and width of the square paper.

Number of cards along one side of $60\text{ cm}$ ($8\text{ cm}$ side) = $\frac{60}{8} = 7.5 \approx 7$ cards

Number of cards along the other side of $60\text{ cm}$ ($5\text{ cm}$ side) = $\frac{60}{5} = 12$ cards

$\text{Total complete cards} = 7 \times 12 = 84$

2. Area left:

$\text{Area of square chart paper} = 60 \times 60 = 3600\text{ cm}^2$

$\text{Area of one rectangular card} = 8 \times 5 = 40\text{ cm}^2$

$\text{Area of 84 cards} = 84 \times 40$

$\text{Area of cards} = 3360\text{ cm}^2$

$\text{Remaining area} = \text{Total Area} - \text{Area of cards}$

$\text{Remaining area} = 240\text{ cm}^2$

Final Answer: Amita can make $84$ cards and the area left is $240\text{ cm}^2$.

Given:

Dimensions of one page = $15\text{ cm} \times 24\text{ cm}$

Advertisement size = Half page

Rate of advertising = $\textsf{₹} 300$ per $10\text{ cm}^2$

To Find:

Total amount to be paid.

Solution:

First, we calculate the area of the full page:

$\text{Area of full page} = 15 \times 24$

$\text{Area} = 360\text{ cm}^2$

$\text{Area of half-page advertisement} = \frac{360}{2} = 180\text{ cm}^2$

Now, calculate the cost:

$\text{Cost for } 10\text{ cm}^2 = \textsf{₹} 300$

$\text{Cost for } 1\text{ cm}^2 = \textsf{₹} \frac{300}{10} = \textsf{₹} 30$

$\text{Total Cost} = 180 \times 30$

$\text{Total Cost} = \textsf{₹} 5400$

Final Answer: The company has to pay $\textsf{₹} 5400$.

Given:

Perimeter of square garden = $48\text{ m}$

Area of flower bed = $18\text{ m}^2$

To Find:

1. Area not covered by flower bed.

2. Fractional part covered by flower bed.

3. Ratio of covered area to remaining area.

Solution:

1. Area of the garden:

$\text{Perimeter} = 4 \times \text{side} = 48\text{ m}$

$\text{side} = \frac{48}{4} = 12\text{ m}$

$\text{Area of garden} = 12 \times 12 = 144\text{ m}^2$

2. Area not covered (remaining area):

$\text{Remaining area} = 144 - 18 = 126\text{ m}^2$

3. Fractional part covered:

$\text{Fraction} = \frac{\text{Area of flower bed}}{\text{Total area of garden}}$

$\text{Fraction} = \frac{\cancel{18}^1}{\cancel{144}_8} = \frac{1}{8}$

4. Ratio of covered area to remaining area:

$\text{Ratio} = \text{Area of flower bed} : \text{Remaining area}$

$\text{Ratio} = 18 : 126$

Dividing both by $18$:

$\text{Ratio} = \frac{\cancel{18}^1}{\cancel{126}_7} = 1 : 7$

Final Answer: The area not covered is $126\text{ m}^2$, the fractional part covered is $\frac{1}{8}$, and the ratio is $1 : 7$.

Given:

Side of the square ($s$) = $15\text{ cm}$

One side of the rectangle ($l$) = $18\text{ cm}$

Perimeter of Square = Perimeter of Rectangle

To Find:

Area of the rectangle.

Solution:

First, we find the perimeter of the square.

$\text{Perimeter of Square} = 4 \times 15 = 60\text{ cm}$

Now, let the other side of the rectangle be $b$. According to the question:

$\text{Perimeter of Rectangle} = 60\text{ cm}$

$18 + b = \frac{\cancel{60}^{30}}{\cancel{2}_{1}}$

$18 + b = 30$

$b = 30 - 18 = 12\text{ cm}$

Now, we calculate the area of the rectangle:

$\text{Area} = 18 \times 12$

$\text{Area} = 216\text{ cm}^2$

Final Answer: The area of the rectangle is $216\text{ cm}^2$.

Given:

Side of each small square = $2\text{ cm}$

Total area of all squares = $28\text{ cm}^2$

To Find:

Original length of the wire.

Solution:

First, let's find the area of one small square.

$\text{Area of one square} = \text{side} \times \text{side} = 2 \times 2 = 4\text{ cm}^2$

Now, let's find the total number of small squares formed.

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

$\text{Number of squares} = \frac{\cancel{28}^{7}}{\cancel{4}_{1}} = 7$

Each square was made by bending one piece of wire. The length of wire used for one square is its perimeter.

$\text{Perimeter of one square} = 4 \times \text{side} = 4 \times 2 = 8\text{ cm}$

Since there are $7$ such squares, the original length of the wire was:

$\text{Total length} = \text{Number of squares} \times \text{Perimeter of one square}$

$\text{Total length} = 7 \times 8 = 56\text{ cm}$

Final Answer: The original length of the wire was $56\text{ cm}$.

Given:

The dimensions of the park are given in Fig. 6.17.

Rate of fertilizer = $1$ packet per $300\text{ m}^2$.

To Find:

1. Total area of the park by dividing it into two rectangles.

2. Number of packets of fertilizer required.

Solution:

Let's divide the park into two rectangles by extending the dotted line $AB$ horizontally.

1. Dimensions of the two rectangles:

Rectangle 1 (Top part): Length = $150\text{ m}$. Height = $280\text{ m} - 180\text{ m} = 100\text{ m}$.

Rectangle 2 (Bottom part): Length = $270\text{ m}$. Breadth = $180\text{ m}$.

2. Calculating Areas:

$\text{Area of Rectangle 1} = 150 \times 100 = 15,000\text{ m}^2$

$\text{Area of Rectangle 2} = 270 \times 180$

$\text{Area of Rectangle 2} = 48,600\text{ m}^2$

$\text{Total Area} = 15,000 + 48,600 = 63,600\text{ m}^2$

3. Calculating Fertilizer Packets:

$\text{Number of packets} = \frac{\text{Total Area}}{300}$

$\text{Number of packets} = \frac{63600}{300} = \frac{\cancel{636}^{212}}{\cancel{3}_{1}}$

$\text{Number of packets} = 212$

Final Answer: The total area of the park is $63,600\text{ m}^2$ and $212$ packets of fertilizer are required.

Given:

Area of the rectangular field = $1600\text{ m}^2$

Length ($l$) = $80\text{ m}$

To Find:

Perimeter of the field.

Solution:

First, find the breadth ($b$) using the area formula.

$1600 = 80 \times b$

$b = \frac{\cancel{1600}^{20}}{\cancel{80}_{1}} = 20\text{ m}$

Now, find the perimeter:

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

$\text{Perimeter} = 2 \times (80 + 20) = 2 \times 100$

$\text{Perimeter} = 200\text{ m}$

Final Answer: The perimeter of the field is $200\text{ m}$.

Given:

Area of one small square on chess board = $4\text{ cm}^2$

A standard chess board has $8 \times 8 = 64$ small squares.

Solution:

Total area of the board:

$\text{Total Area} = 64 \times 4 = 256\text{ cm}^2$

At the beginning of a game, there are $32$ chess men in total ($16$ white and $16$ black), each occupying one square.

(b) Area of squares occupied by chess men:

$\text{Number of occupied squares} = 32$

$\text{Occupied Area} = 32 \times 4 = 128\text{ cm}^2$

(a) Area of squares left unoccupied:

$\text{Unoccupied Area} = \text{Total Area} - \text{Occupied Area}$

$\text{Unoccupied Area} = 256 - 128 = 128\text{ cm}^2$

Final Answer: Total area is $256\text{ cm}^2$, unoccupied area is $128\text{ cm}^2$, and occupied area is $128\text{ cm}^2$.

(a) Solution:

Perimeter = $36\text{ cm}$. Since $P = 2(l + b)$, we have $l + b = 18\text{ cm}$. We find natural number pairs for $l$ and $b$ (taking $l \ge b$):

(b) Solution:

Area = $36\text{ cm}^2$. Since $A = l \times b$, we find factors of $36$:

Given:

Each small square has an area of $1\text{ cm}^2$.

Since $\text{Area} = \text{side} \times \text{side}$, the length of each side of a small square is $1\text{ cm}$.

To Find:

The area and perimeter of figures (i), (ii), and (iii).

Solution for Figure (i):

1. Area:

The area is the total number of unit squares contained in the figure.

By counting, the right column has $6$ squares and the left column has $5$ squares (with a gap of one square in between).

$\text{Area} = 11\text{ cm}^2$

2. Perimeter:

The perimeter is the sum of the lengths of all outer boundary segments.

Outer segments = $2\text{ (top)} + 6\text{ (right)} + 2\text{ (bottom)} $$ + 2\text{ (bottom-left)} $$ + 1\text{ (in)} + 1\text{ (up)} + 1\text{ (out)} + 3\text{ (top-left)}$

$\text{Perimeter} = 18\text{ cm}$

Solution for Figure (ii):

1. Area:

Counting the total number of unit squares in the shape:

Central square = $1$

Four arms, each consisting of $3$ squares = $4 \times 3 = 12$

$\text{Total Area} = 1 + 12 = 13\text{ cm}^2$

2. Perimeter:

The perimeter consists of the outer edges. Each of the $4$ outer ends has $3$ segments exposed ($1+1+1=3$). Additionally, there are $8$ inward corners where $2$ segments meet for each corner.

$\text{Perimeter} = (4 \times 3) + (8 \times 2)$

$\text{Perimeter} = 28\text{ cm}$

Solution for Figure (iii):

1. Area:

Counting the unit squares:

Left pillar = $6$ squares

Right pillar = $5$ squares

Connecting top bar = $2$ squares

$\text{Total Area} = 6 + 5 + 2 = 13\text{ cm}^2$

2. Perimeter:

Summing the outer and inner boundary segments:

$\text{Outer edges} = 4\text{ (top)} + 6\text{ (left)} + 5\text{ (right)} + 1\text{ (bottom-left)} $$ + 1\text{ (bottom-right)} = 17\text{ units}$

$\text{Inner edges} = 5\text{ (left-inner)} + 2\text{ (top-inner)} + 4\text{ (right-inner)} $$ = 11\text{ units}$

$\text{Perimeter} = 28\text{ cm}$

Summary:

Given:

Total area of the figure = $96\text{ cm}^2$

The figure is composed of several small identical squares as shown in Fig. 6.21.

To Find:

1. Area of each small square.

2. Perimeter of the figure.

Solution:

1. Finding the area of each small square:

First, we count the total number of small squares in the given figure row by row from top to bottom:

Top row = $2\text{ squares}$

Second row = $4\text{ squares}$

Third row = $5\text{ squares}$

Fourth row (the longest bar) = $9\text{ squares}$

Fifth row (bottom hanging squares) = $4\text{ squares}$

$\text{Total number of small squares} = 2 + 4 + 5 + 9 + 4 = 24$

$\text{Area of each small square} = \frac{96}{24}$

$\text{Area of each small square} = 4\text{ cm}^2$

2. Finding the perimeter of the figure:

To find the perimeter, we first need the length of the side ($s$) of each small square.

$\text{Area of square} = s^2$

$4\text{ cm}^2 = s^2$

$s = \sqrt{4} = 2\text{ cm}$

Now, we count the number of unit segments (sides of the small squares) that form the outer boundary of the figure:

Horizontal segments:

Topmost edges = $2$

Edges on steps and bar top = $1 (\text{right}) + 2 (\text{right wing}) + 2 (\text{left wing}) $$ + 1 (\text{left}) + 1 (\text{left}) = 7$

Bottom edges (including bar base and hanging square bases) = $9$

$\text{Total horizontal units} = 2 + 7 + 9 = 18$

Vertical segments:

Right side steps and bar end = $4$

Left side steps and bar end = $4$

Vertical sides of the 4 hanging squares = $4 \times 2 = 8$

$\text{Total vertical units} = 4 + 4 + 8 = 16$

Total Perimeter:

$\text{Total boundary segments} = 18 + 16 = 34\text{ units}$

$\text{Perimeter} = 34 \times 2\text{ cm}$

$\text{Perimeter} = 68\text{ cm}$

Final Answer: The area of each small square is $4\text{ cm}^2$ and the perimeter of the figure is $68\text{ cm}$.