Perimeter And AreaClass 6 Mathematics NCERT Solutions

Given: Total race distance = $350 \text{ m}$ Inner track: Square with side $100 \text{ m}$. Perimeter = $4 \times 100 = 400 \text{ m}$. Outer track: Square with side $150 \text{ m}$. Perimeter = $4 \times 150 = 600 \text{ m}$. Finish line: Center of one side (let's assume the top side) for both tracks.

To Find: Starting position 'A' on the inner track. Starting position 'B' on the outer track.

Logic: The starting point must be located $350 \text{ m}$ before the finish line along the track. We will trace back $350 \text{ m}$ from the finish line in a counter-clockwise direction to find the start.

Runner A (Inner Track): Finish line is at the center of the top side, which is $50 \text{ m}$ from the top-right corner.

Starting Position A: The starting point for the runner on the inner track is the top-left corner. (To verify: Running clockwise from top-left corner, the distance is $50 \text{ m}$ along the top side to the finish. This is not 350m. Let's assume the runner runs the full track. The race is 350m. The perimeter is 400m. The start must be 400-350=50m before the finish line. 50m before the center of the top side is the top-left corner. The path is: from top-left, run 50m to finish, then 100m right side, 100m bottom side, 100m left side. Total = 350m. This is correct.)

Runner B (Outer Track): Finish line is at the center of the top side, which is $75 \text{ m}$ from the top-right corner.

Starting Position B: The starting point for the runner on the outer track is on the bottom side, $125 \text{ m}$ from the bottom-right corner (or $25 \text{ m}$ from the bottom-left corner).

Analysis: The figure shows a shape on a square grid. The perimeter is composed of straight lines (along the grid lines) and diagonal lines (across the grid squares).

The given shape has 6 straight sides and 3 diagonal sides. If we incorrectly assume that diagonal lines also have a length of 1 unit, the perimeter would be $6 \times 1 + 3 \times 1 = 9$ units. This is what Akshi calculated.

However, the diagonal lines are longer than the straight lines. Correct perimeter = $(6 \times 1) + (3 \times \sqrt{2})$ units $= 6 + 3\sqrt{2} \approx 6 + 3(1.414) = 6 + 4.242 = 10.242$ units.

Since $10.242 > 9$, Toshi is correct.

Final Answer: I think Toshi is correct. The perimeter will be more than 9 units because the length of a diagonal of a square is always greater than the length of its side.

Let: 's' represent the length of one straight unit (the side of a small square). 'd' represent the length of one diagonal unit (the diagonal of a small square).

By counting the sides of each figure from the textbook images:

Generalization for the Perimeter of a Regular Polygon: A regular polygon is a polygon with all sides of equal length and all angles of equal measure.

Let a regular polygon have '$n$' sides. Let the length of each side be '$L$'.

The perimeter of the polygon is the sum of the lengths of all its sides. Since all '$n$' sides have the same length '$L$', the perimeter can be calculated by multiplication.

General Formula: Perimeter of a regular polygon = (Number of sides) $\times$ (Length of one side) $P = n \times L$

Final Answer: In general, the perimeter of a regular polygon is the product of the number of its sides and the length of one side.

Given: Perimeter $P = 14 \text{ cm}$ Breadth $b = 2 \text{ cm}$

Formula: Perimeter of a rectangle, $P = 2 \times (\text{length} + \text{breadth}) = 2(l + b)$

Solution: $14 = 2 \times (l + 2)$ $\frac{14}{2} = l + 2$ $7 = l + 2$ $l = 7 - 2 = 5 \text{ cm}$

Final Answer: The length is $5 \text{ cm}$.

Given: Perimeter $P = 20 \text{ cm}$

Formula: Perimeter of a square, $P = 4 \times \text{side} = 4s$

Solution: $20 = 4 \times s$ $s = \frac{20}{4} = 5 \text{ cm}$

Final Answer: The length of a side is $5 \text{ cm}$.

Given: Perimeter $P = 12 \text{ m}$ Length $l = 3 \text{ m}$

Formula: Perimeter of a rectangle, $P = 2(l + b)$

Solution: $12 = 2 \times (3 + b)$ $\frac{12}{2} = 3 + b$ $6 = 3 + b$ $b = 6 - 3 = 3 \text{ m}$

Final Answer: The breadth is $3 \text{ m}$.

Given: Length of rectangle, $l = 5 \text{ cm}$ Breadth of rectangle, $b = 3 \text{ cm}$

To Find: The length of a side of the square formed from the same wire.

Solution: First, we find the length of the wire, which is equal to the perimeter of the rectangle.

Formula (Perimeter of rectangle): $P = 2(l + b)$ $P = 2(5 + 3) = 2(8) = 16 \text{ cm}$

The length of the wire is $16 \text{ cm}$.

Now, this wire is bent to form a square. The perimeter of the square will be equal to the length of the wire.

Formula (Perimeter of square): $P_{\text{square}} = 4 \times \text{side} = 4s$ $16 = 4s$ $s = \frac{16}{4} = 4 \text{ cm}$

Final Answer: The length of a side of the square will be $4 \text{ cm}$.

Given: Perimeter of the triangle, $P = 55 \text{ cm}$ Length of the first side, $s_1 = 20 \text{ cm}$ Length of the second side, $s_2 = 14 \text{ cm}$

To Find: The length of the third side, $s_3$.

Formula: Perimeter of a triangle, $P = s_1 + s_2 + s_3$

Solution: $55 = 20 + 14 + s_3$ $55 = 34 + s_3$ $s_3 = 55 - 34$ $s_3 = 21 \text{ cm}$

Final Answer: The length of the third side of the triangle is $21 \text{ cm}$.

Given: Length of the rectangular park, $l = 150 \text{ m}$ Breadth of the rectangular park, $b = 120 \text{ m}$ Cost of fencing = ₹ 40 per metre

To Find: The total cost of fencing the park.

Solution: First, we need to find the perimeter of the park, which is the total length to be fenced.

Formula (Perimeter of rectangle): $P = 2(l + b)$ $P = 2(150 + 120) = 2(270) = 540 \text{ m}$

The total length of the fence required is $540 \text{ m}$.

Now, we calculate the total cost. Total Cost = Length of fence $\times$ Cost per metre $= 540 \text{ m} \times ₹ 40/\text{m}$ $= ₹ 21600$

Final Answer: The cost of fencing the rectangular park would be ₹ 21,600.

Given: Total length of the string = $36 \text{ cm}$. This length will be the perimeter of the shapes formed.

a. A square A square has 4 equal sides. Perimeter of a square, $P = 4 \times \text{side} = 4s$ $36 = 4s$ $s = \frac{36}{4} = 9 \text{ cm}$ Answer a: The length of each side of the square will be $9 \text{ cm}$.

b. A triangle with all sides of equal length (an equilateral triangle) An equilateral triangle has 3 equal sides. Perimeter of an equilateral triangle, $P = 3 \times \text{side} = 3s$ $36 = 3s$ $s = \frac{36}{3} = 12 \text{ cm}$ Answer b: The length of each side of the triangle will be $12 \text{ cm}$.

c. A hexagon with sides of equal length (a regular hexagon) A regular hexagon has 6 equal sides. Perimeter of a regular hexagon, $P = 6 \times \text{side} = 6s$ $36 = 6s$ $s = \frac{36}{6} = 6 \text{ cm}$ Answer c: The length of each side of the hexagon will be $6 \text{ cm}$.

Given: Length of the rectangular field, $l = 230 \text{ m}$ Breadth of the rectangular field, $b = 160 \text{ m}$ Number of rounds of rope = 3

To Find: The total length of rope needed.

Solution: First, we find the perimeter of the field, which is the length of rope needed for one round.

Formula (Perimeter of rectangle): $P = 2(l + b)$ $P = 2(230 + 160) = 2(390) = 780 \text{ m}$

The length of rope for one round is $780 \text{ m}$.

For 3 rounds, the total length of rope is: Total Length = Perimeter $\times$ Number of rounds $= 780 \text{ m} \times 3$ $= 2340 \text{ m}$

Final Answer: The total length of rope needed is $2340 \text{ m}$.

Given: Akshi's track is a rectangle with length $l = 70 \text{ m}$ and breadth $b = 40 \text{ m}$. Number of rounds = 5

To Find: The total distance Akshi covered.

Solution: First, find the perimeter of Akshi's track, which is the distance covered in one round.

Formula (Perimeter of rectangle): $P = 2(l + b)$ $P = 2(70 + 40) = 2(110) = 220 \text{ m}$

The distance covered in one round is $220 \text{ m}$.

Total distance covered in 5 rounds: Total Distance = Distance per round $\times$ Number of rounds $= 220 \text{ m} \times 5$ $= 1100 \text{ m}$

Final Answer: The total distance Akshi has covered in 5 rounds is $1100 \text{ m}$.

Given: Toshi's track is a rectangle with length $l = 50 \text{ m}$ and breadth $b = 40 \text{ m}$ (inferred from the diagram in the 'Matha Pachchi!' section). Number of rounds = 7 From the previous question, Akshi ran $1100 \text{ m}$.

To Find: The total distance Toshi covered and who ran longer.

Solution: First, find the perimeter of Toshi's track.

Formula (Perimeter of rectangle): $P = 2(l + b)$ $P = 2(50 + 40) = 2(90) = 180 \text{ m}$

The distance Toshi covers in one round is $180 \text{ m}$.

Total distance covered by Toshi in 7 rounds: Total Distance = Distance per round $\times$ Number of rounds $= 180 \text{ m} \times 7$ $= 1260 \text{ m}$

Now, we compare the distances: Akshi's distance = $1100 \text{ m}$ Toshi's distance = $1260 \text{ m}$

Since $1260 > 1100$, Toshi ran a longer distance.

Final Answer: The total distance Toshi has covered is $1260 \text{ m}$. Toshi ran a longer distance.

Assumptions: Let the starting point be the bottom-left corner of each rectangular track. Let the runners move in a clockwise direction.

Akshi's Track: Perimeter = $220 \text{ m}$ (Sides: Bottom=70m, Right=40m, Top=70m, Left=40m)

a. Akshi after 250 m: $250 \text{ m} = 1 \times 220 \text{ m} + 30 \text{ m}$. This means Akshi completes one full round and runs an additional $30 \text{ m}$. Position 'A' is $30 \text{ m}$ from the starting point along the bottom side.

b. Akshi after 500 m: $500 \text{ m} = 2 \times 220 \text{ m} + 60 \text{ m}$. This means Akshi completes two full rounds and runs an additional $60 \text{ m}$. Position 'B' is $60 \text{ m}$ from the starting point along the bottom side.

c. Akshi after 1000 m: $1000 \text{ m} = 4 \times 220 \text{ m} + 120 \text{ m}$. Akshi has finished 4 full rounds. For her position, she runs an additional $120 \text{ m}$: $70 \text{ m}$ (bottom side) + $40 \text{ m}$ (right side) + $10 \text{ m}$ (top side). Position 'C' is on the top side, $10 \text{ m}$ from the top-right corner.

Toshi's Track: Perimeter = $180 \text{ m}$ (Sides: Bottom=50m, Right=40m, Top=50m, Left=40m)

d. Toshi after 250 m: $250 \text{ m} = 1 \times 180 \text{ m} + 70 \text{ m}$. This means Toshi completes one full round and runs an additional $70 \text{ m}$: $50 \text{ m}$ (bottom side) + $20 \text{ m}$ (right side). Position 'X' is on the right side, $20 \text{ m}$ up from the bottom-right corner.

e. Toshi after 500 m: $500 \text{ m} = 2 \times 180 \text{ m} + 140 \text{ m}$. This means Toshi completes two full rounds and runs an additional $140 \text{ m}$: $50 \text{ m}$ (bottom) + $40 \text{ m}$ (right) + $50 \text{ m}$ (top). Position 'Y' is at the top-left corner of her track.

f. Toshi after 1000 m: $1000 \text{ m} = 5 \times 180 \text{ m} + 100 \text{ m}$. Toshi has finished 5 full rounds. For her position, she runs an additional $100 \text{ m}$: $50 \text{ m}$ (bottom) + $40 \text{ m}$ (right) + $10 \text{ m}$ (top). Position 'Z' is on the top side, $10 \text{ m}$ from the top-right corner.

Given: Area of the rectangular garden, $A = 300 \text{ sq m}$ Length of the garden, $l = 25 \text{ m}$

To Find: The width of the garden, $w$.

Formula: Area of a rectangle, $A = \text{length} \times \text{width} = l \times w$

Solution: $300 = 25 \times w$ $w = \frac{300}{25}$ $w = 12 \text{ m}$

Final Answer: The width of the garden is $12 \text{ m}$.

Given: Length of the plot, $l = 500 \text{ m}$ Width of the plot, $w = 200 \text{ m}$ Rate of tiling = ₹ 8 per 100 sq m

To Find: The total cost of tiling the plot.

Solution: First, calculate the total area of the plot.

Formula (Area of rectangle): $A = l \times w$ $A = 500 \text{ m} \times 200 \text{ m} = 100,000 \text{ sq m}$

The rate is given per hundred sq m. So, we find how many 'hundred sq m' units are in the total area. Number of units = $\frac{\text{Total Area}}{100} = \frac{100,000}{100} = 1000$ units.

Now, calculate the total cost. Total Cost = Number of units $\times$ Rate per unit $= 1000 \times ₹ 8$ $= ₹ 8000$

Final Answer: The cost of tiling the plot is ₹ 8,000.

Given: Length of the grove, $l = 100 \text{ m}$ Width of the grove, $w = 50 \text{ m}$ Area required per tree = $25 \text{ sq m}$

To Find: The maximum number of trees that can be planted.

Solution: First, calculate the total area of the grove.

Formula (Area of rectangle): $A = l \times w$ $A = 100 \text{ m} \times 50 \text{ m} = 5000 \text{ sq m}$

Now, divide the total area by the area required for each tree to find the number of trees. Number of trees = $\frac{\text{Total Area}}{\text{Area per tree}}$ $= \frac{5000}{25} = 200$

Final Answer: The maximum number of trees that can be planted is 200.

a. First Figure (L-shape) We can split this L-shaped figure into two rectangles.

b. Second Figure (Plus-shape) We can split this figure into a central rectangle and two side rectangles.

Analysis of Tangram Pieces: A standard tangram set has 7 pieces. Let the area of the smallest triangle (pieces C and E) be 1 unit.

Final Answer: There are three groups of pieces with the same area: {A, B}, {C, E}, and {D, F, G}.

Comparison of Areas: Let the area of Shape C (small triangle) be 1 unit.

How many times bigger is D than C? Ratio of areas = $\frac{\text{Area(D)}}{\text{Area(C)}} = \frac{2}{1} = 2$. Shape D is twice as big as Shape C.

Final Answer: Shape D is twice as big as Shape C. Shapes C and E are equal in area, and Shape D has an area equal to the sum of the areas of Shape C and Shape E.

Reasoning: If we take the two small triangles (C and E), they can be arranged to perfectly cover the square (F). They can also be arranged to perfectly cover the medium triangle (D). Therefore, their areas must be equal.

Final Answer: Neither shape has more area. Shape D and Shape F have the same area.

Reasoning: Similar to the previous question, the two small triangles (C and E) can be arranged to form the square (F) and can also be rearranged to form the parallelogram (G). Since they are composed of the same smaller units, their areas are equal.

Final Answer: Neither shape has more area. Shape F and Shape G have the same area.

Ratio of Areas: Ratio = $\frac{\text{Area(A)}}{\text{Area(G)}} = \frac{4 \text{ small triangles}}{2 \text{ small triangles}} = 2$.

This means the area of Shape A is twice the area of Shape G.

Final Answer: The area of Shape A is twice as big as the area of Shape G.

Given: Let the area of Shape C = 1 unit.

Total Area of the big square: Total Area = Sum of the areas of all 7 pieces = $4 + 4 + 1 + 2 + 1 + 2 + 2 = 16$ units.

Since 1 unit is the area of Shape C, the total area is 16 times the area of Shape C.

Final Answer: The area of the big square is 16 times the area of Shape C.

Reasoning: When we rearrange shapes without adding or removing any material, the total area remains constant. This is the principle of conservation of area. The rectangle is formed using the exact same 7 tangram pieces that formed the original square.

Area Calculation: The area of the original square was found to be 16 times the area of Shape C. Since the rectangle is made from the same pieces, its area must also be the sum of the areas of those pieces. Therefore, the area of the rectangle will be the same as the area of the square.

Final Answer: The area of the rectangle will be 16 times the area of Shape C. The reason is that the rectangle is composed of the same set of pieces as the original square, so their total areas must be identical.

Explanation: For a fixed area, different shapes can have different perimeters. Generally, the most 'compact' shape, like a square or circle, has the smallest perimeter for a given area.

Let the area be 16 square units. The square would be $4 \times 4$, with a perimeter of $4+4+4+4 = 16$ units. A possible rectangle with area 16 is $8 \times 2$. Its perimeter would be $2(8+2) = 20$ units.

Since $20 \neq 16$, the perimeters are different.

Final Answer: The perimeters of the square and the rectangle are different. For a given area, a square is the rectangle with the minimum possible perimeter. Any non-square rectangle with the same area will have a larger perimeter because it is 'stretched out' more.

a. First Figure (House shape) This shape can be divided into a rectangle at the bottom and a triangle at the top.

b. Second Figure (Arrow shape) This shape can be divided into a rectangle on the left and a triangle on the right.

c. Third Figure (Rocket shape) This shape can be divided into a central rectangle, a triangle on top, and two smaller rectangles at the bottom sides.

d. Fourth Figure This shape can be divided into a square at the bottom and a triangle at the top.

e. Fifth Figure (Plus sign) This shape can be divided into a central square and four surrounding rectangles.

Given: Area of first rectangle, $A_1 = 5 \text{ m} \times 10 \text{ m} = 50 \text{ sq m}$. Area of second rectangle, $A_2 = 2 \text{ m} \times 7 \text{ m} = 14 \text{ sq m}$.

To Find: Dimensions of a new rectangle whose area is $A_1 + A_2$.

Solution: Total Area, $A_{\text{total}} = A_1 + A_2 = 50 + 14 = 64 \text{ sq m}$.

We need to find two numbers (length and width) that multiply to 64. There are several possible answers.

Final Answer: One possible set of dimensions is 8 m and 8 m.

Given: Area of the garden, $A = 1000 \text{ sq m}$. Length of the garden, $l = 50 \text{ m}$.

To Find: The width of the garden, $w$.

Formula: Area = length $\times$ width

Solution: $1000 = 50 \times w$ $w = \frac{1000}{50}$ $w = 20 \text{ m}$

Final Answer: The width of the garden is $20 \text{ m}$.

Given: Floor dimensions: $l = 5 \text{ m}$, $w = 4 \text{ m}$. Carpet dimensions: square with side $s = 3 \text{ m}$.

To Find: The area of the floor that is not carpeted.

Final Answer: The area of the floor that is not carpeted is $11 \text{ sq m}$.

Given: Garden dimensions: $l = 15 \text{ m}$, $w = 12 \text{ m}$. Flower bed dimensions: $l_f = 2 \text{ m}$, $w_f = 1 \text{ m}$. Number of flower beds = 4.

To Find: The area available for the lawn.

Final Answer: The area available for laying down a lawn is $172 \text{ sq m}$.

Given: Area(A) = 18 sq units. Area(B) = 20 sq units. Perimeter(A) > Perimeter(B).

To Find: Examples of two such shapes (rectangles).

Solution: Let's find possible dimensions and perimeters for Shape A and Shape B.

We need to find a pair where Perimeter(A) > Perimeter(B). Let's compare.

Drawing: (A textual description of the drawing)

Assumption: The dimensions of the book page are not given. Let us assume a standard page size, for example, a page that is $24 \text{ cm}$ high and $18 \text{ cm}$ wide.

Given: Page height = $24 \text{ cm}$ Page width = $18 \text{ cm}$ Top and bottom margin = $1 \text{ cm}$ each Left and right margin = $1.5 \text{ cm}$ each

To Find: The perimeter of the border. This usually refers to the inner perimeter of the border.

Solution: First, we find the dimensions of the rectangular area inside the border.

Now, we calculate the perimeter of this inner rectangle. Formula: Perimeter = $2 (\text{length} + \text{width})$ Perimeter = $2 (15 \text{ cm} + 22 \text{ cm})$ = $2 (37 \text{ cm})$ = $74 \text{ cm}$.

Final Answer: Assuming a page size of $24 \text{ cm} \times 18 \text{ cm}$, the perimeter of the border is $74 \text{ cm}$.

Given: Outer rectangle dimensions: $12 \text{ units} \times 8 \text{ units}$.

To Find: Dimensions for an inner rectangle with half the area.

Let's analyze the situation: Let the original square have a side length of '$s$'.

When the square is folded in half and cut, we get two identical rectangles.

Now let's evaluate each statement:

Final Answer: Statement c is the one that is always true.

Method: We estimate the area by counting the number of squares covered by each figure on the grid paper, following these conventions:

Analysis of Charan's House Plan: The overall plot is a rectangle with width = $15 \text{ ft} + 20 \text{ ft} = 35 \text{ ft}$. The depth is given as $30 \text{ ft}$.

a. Find the missing measurements and area of each room: Assuming the provided dimensions are for the rooms.

b. Find out the area of his house. The area of the house is the area of the entire rectangular plot. Area of house (plot) = Total Width $\times$ Total Depth = $35 \text{ ft} \times 30 \text{ ft}$ = $1050 \text{ sq ft}$.

(Verification: Sum of all component areas = $240 + 180 + 45 + 45 + 60 = 570 \text{ sq ft}$. The remaining area $1050-570 = 480 \text{ sq ft}$ is unaccounted for, suggesting the diagram is a schematic and not a complete floor plan. However, based on the outer dimensions, the total area is 1050 sq ft.)

Analysis of Sharan's House Plan: The overall plot is a rectangle. Total width = $7+23+12 = 42 \text{ ft}$. Total depth = $10+15 = 25 \text{ ft}$.

b. Find out the area of his house. Area of house (plot) = Total Width $\times$ Total Depth = $42 \text{ ft} \times 25 \text{ ft}$ = $1050 \text{ sq ft}$.

Rectangles with Area 24 square units: The pairs of whole numbers that multiply to 24 are:

Now, let's calculate the perimeter for each rectangle using the formula $P = 2(l+w)$:

a. Which rectangle has the greatest perimeter? The rectangle with dimensions $1 \times 24$ has the greatest perimeter (50 units).

b. Which rectangle has the least perimeter? The rectangle with dimensions $4 \times 6$ has the least perimeter (20 units).

c. Rectangles with Area 32 sq cm: The pairs of whole numbers that multiply to 32 are:

Perimeters:

For an area of 32 sq cm, the greatest perimeter is 66 cm (for the $1 \times 32$ rectangle) and the least perimeter is 24 cm (for the $4 \times 8$ rectangle).

Prediction and Reasoning: Yes, it is possible to predict the shape.

1. What is the smallest perimeter possible? To get the smallest perimeter for a fixed area, the shape must be as compact as possible. With 9 squares, the most compact arrangement is a $3 \times 3$ square. Perimeter of a $3 \times 3$ square = $3+3+3+3 = 12$ units. Answer: The smallest possible perimeter is 12 units.

2. What is the largest perimeter possible? To get the largest perimeter, the shape must be as 'stringy' or non-compact as possible. With 9 squares, this is achieved by arranging them in a single line, forming a $1 \times 9$ rectangle. Perimeter of a $1 \times 9$ rectangle = $2(1+9) = 2(10) = 20$ units. Answer: The largest possible perimeter is 20 units.

3. Make a figure with a perimeter of 18 units. We start with the largest perimeter (20) and make the shape more compact to reduce it. The $1 \times 9$ rectangle has a perimeter of 20. If we move one square from the end and attach it to the side of another square, we reduce the perimeter. For example, a $1 \times 8$ rectangle with one square attached to the side of one of the end squares. The perimeter of this shape is 18 units. Another example: A $2 \times 5$ rectangle uses 10 squares. We need 9. An 'L' shape made from a $2 \times 4$ rectangle with one square attached to the end of the short side. Let's count: 4+2+3+1+1+1 = 12. No. A cross shape (5 squares) has P=12. Adding 4 more squares in the corners gives a 3x3 square with P=12. Let's construct a shape with P=18. Take a $1 \times 8$ rectangle (P=18). Add the 9th square to the side of an inner square. This gives a shape with perimeter 18.

Reasoning: The perimeter changes based on how many sides of the squares are internal (touching other squares) versus external (forming the boundary). To maximize the perimeter, you minimize the internal sides (a straight line). To minimize the perimeter, you maximize the internal sides (a compact square). Perimeters between the minimum and maximum can often be achieved in multiple ways by arranging the squares differently.

Analysis of Adding a Square: When a new square is attached to a figure made of squares, its perimeter changes based on how many sides of the new square touch the existing figure. Let the side length of a square be 1 unit.

Observing the specific case: The new square is attached to a flat edge of the existing figure. One side of the new square covers one side of the original figure. The new square adds its other three sides to the perimeter.

Experimenting with Placement: a) Increases: Yes. As shown, attaching the square to any flat outer edge increases the perimeter by 2. The new perimeter would be $24 + 2 = 26$ units.

c) Stays the same: Yes. This happens if we place the new square into an 'inner corner' where it touches two sides of the original figure. By placing it there, we cover two sides of the original figure (perimeter decreases by 2) and add two new sides from the new square (perimeter increases by 2). The net change is $2 - 2 = 0$. The perimeter stays the same.

b) Decreases: Yes. This happens if we place the new square into a 'nook' or 'fjord' where it touches three sides of the original figure. By placing it there, we cover three sides of the original figure (perimeter decreases by 3) and add only one new side from the new square (perimeter increases by 1). The net change is $1 - 3 = -2$. The perimeter decreases by 2.

Analysis of the Pieces: The original rectangle is $6 \text{ cm} \times 4 \text{ cm}$. Arrangement 'a' has a perimeter of $28 \text{ cm}$. The simplest way to achieve this is if the original rectangle was cut into two smaller rectangles of size $6 \text{ cm} \times 2 \text{ cm}$. When these two pieces are joined along their short ($2 \text{ cm}$) sides, they form a new rectangle of size $12 \text{ cm} \times 2 \text{ cm}$. Perimeter of this new rectangle = $2(12 + 2) = 2(14) = 28 \text{ cm}$. This matches arrangement 'a'. So, we will assume the two pieces are identical rectangles of size $6 \text{ cm} \times 2 \text{ cm}$.

b. Perimeter of arrangement b: Arrangement 'b' shows the two $6 \text{ cm} \times 2 \text{ cm}$ pieces stacked on top of each other along their long ($6 \text{ cm}$) sides. This recreates the original $6 \text{ cm} \times 4 \text{ cm}$ rectangle. Perimeter = $2 (6 + 4) = 2(10) = 20 \text{ cm}$.

c. Perimeter of arrangement c: Arrangement 'c' shows the two $6 \text{ cm} \times 2 \text{ cm}$ pieces joined to form an 'L' shape. One piece is placed horizontally, and the other is placed vertically at one end. Let's trace the boundary lengths:

Arrange to form a figure with a perimeter of 22 cm: As calculated above, arrangement 'c' (the L-shape) already has a perimeter of $22 \text{ cm}$.