Key Concept: Perimeter of a Square

b) 20 m
[Solution Description]
The perimeter of a square is calculated using:
$\text{Perimeter} = 4 \times \text{side length}$
Given: side length = 5 m
Step 1: Multiply the side length by 4
$4 \times 5 \text{ m} = 20 \text{ m}$
Thus, the perimeter of the square is 20 m.

Key Concept: Triangle, Equilateral Triangle

c) 12 cm
[Solution Description]
Let the third side be $x$. The perimeter of any triangle is the sum of its sides: $8 + 12 + x = 32$.
Solve for $x$:
$20 + x = 32$
Subtract 20 from both sides: $x = 32 - 20 = 12$ cm.
Since the triangle is not equilateral, all sides cannot be equal. The given sides are 8 cm and 12 cm, so the third side must be different from these. However, $x = 12$ cm here, meaning two sides are equal, but the triangle is not equilateral (since an equilateral triangle has all sides equal). Thus, this satisfies the condition.

Key Concept: Perimeter of Rectangle and Square

c) Assertion is true, but Reason is false.
[Solution Description]
First, calculate the perimeter of the rectangle using the formula $\text{Perimeter of a rectangle} = 2 \times (l + b)$. Here, length ($l$) = 12 cm and breadth ($b$) = 8 cm. So, perimeter = $2 \times (12 + 8) = 40 \, \text{cm}$.
For the perimeter of the square to be equal to that of the rectangle, we use the formula $\text{Perimeter of a square} = 4 \times \text{side}$. Setting this equal to 40 cm, we get side = $10 \, \text{cm}$.
The reason states that the perimeter of any quadrilateral can be made equal to that of a square by adjusting the side lengths. This is false because not all quadrilaterals can have their perimeters adjusted to match a square's perimeter just by changing side lengths—for example, a quadrilateral with fixed angles may not allow such adjustments.
Therefore, the assertion is true, but the reason is false.

Key Concept: Perimeter Calculation

c) 40 cm
[Solution Description]
The perimeter of a pentagon is the sum of all its side lengths.
Perimeter = 7 cm + 8 cm + 10 cm + 6 cm + 9 cm
Perimeter = 40 cm

Key Concept: Perimeter Application

c) 40 m
[Solution Description]
For a rectangle, opposite sides are equal. Therefore, the perimeter is calculated as:
Perimeter = 2 $\times$ (Length + Width)
Perimeter = 2 $\times$ (12 m + 8 m)
Perimeter = 2 $\times$ 20 m
Perimeter = 40 m

Key Concept: Perimeter of irregular polygon

b) 46.5 cm
[Solution Description]
First, identify the longest side and reduce it by 25\%. Then, recalculate the perimeter with the modified side length.
Step 1: Identify the longest side:
The given side lengths are 7 cm, 9 cm, 12 cm, 8 cm, and 14 cm. The longest side is 14 cm.
Step 2: Reduce the longest side by 25\%:
$25\% \text{ of } 14\,\text{cm} = 0.25 \times 14\,\text{cm} = 3.5\,\text{cm}$
$\text{New length} = 14\,\text{cm} - 3.5\,\text{cm} = 10.5\,\text{cm}$
Step 3: Calculate the new perimeter:
Replace the longest side (14 cm) with the new length (10.5 cm) and sum all sides:
$\text{New perimeter} = 7\,\text{cm} + 9\,\text{cm} + 12\,\text{cm} + 8\,\text{cm} + 10.5\,\text{cm} = 46.5\,\text{cm}$

Key Concept: Perimeter of Regular Polygon

c) 7 units
[Solution Description]
A regular pentagon has five equal sides.
Given the perimeter is 35 units, the length of one side $s$ is calculated as:
$5s = 35$
$s = 7$ units.
Therefore, the length of one side of the pentagon is 7 units.

Key Concept: Perimeter of Rectangle

b) 200 square meters
[Solution Description]
Let the breadth of the rectangle be $b$ meters. Then, the length is $2b$ meters.
The perimeter of a rectangle is given by:
$2(l + b) = 60$
Substituting the values:
$2(2b + b) = 60$
Simplify:
$2(3b) = 60$
$6b = 60$
$b = 10$ meters
Thus, the length is $2b = 20$ meters.
The area of the rectangle is $l \times b = 20 \times 10 = 200$ square meters.

Key Concept: Regular Polygon, Generalization

a) 6 cm
[Solution Description]
The perimeter of the square is $4 \times 9 = 36$ cm. Since the hexagon has the same perimeter, let the side of the hexagon be $s$. A regular hexagon has 6 sides, so:
$6 \times s = 36$. Solving for $s$, we get $s = \frac{36}{6} = 6$ cm.

Key Concept: Triangle Perimeter, Mixed Units

b) $11s + 11d$
[Solution Description]
The first triangle's perimeter is $7s + 5d$. The second triangle's perimeter is $4s + 6d$. The total perimeter when combined is $(7s + 5d) + (4s + 6d) = 11s + 11d$.

Key Concept: Triangle Perimeter, Equilateral Triangle

c) 26 cm
[Solution Description]
Let the third side be $x$ cm. The perimeter of the triangle is given by $5 + 8 + x = 13 + x$. According to the problem, the perimeter is twice the third side, so $13 + x = 2x$. Solving for $x$, we get $x = 13$ cm. The perimeter is $2x = 26$ cm.

Key Concept: Equilateral triangle perimeter, Triangle inequality

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
First, let's analyze the Assertion. The perimeter of an equilateral triangle with side $a$ is $3a$. For the scalene triangle, the perimeter is $a + b + c$. Given $a = b + c$, the perimeter of the scalene triangle becomes $a + a = 2a$.
Now, compare the perimeters:
$3a > 2a \text{ for } a > 0$
So, the Assertion is true.
Next, examine the Reason. The triangle inequality states that for any triangle with sides $x$, $y$, and $z$, the sum of any two sides must be greater than the third side, i.e., $x + y > z$, $y + z > x$, and $z + x > y$. This is a fundamental property of triangles, so the Reason is true.
However, does the Reason correctly explain the Assertion? The Reason provides a general property of triangles but does not directly justify why the perimeter of the equilateral triangle must be greater than that of the scalene triangle in this specific case. The Assertion follows from algebraic comparison ($3a > 2a$) rather than the triangle inequality itself.
Therefore, both the Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Key Concept: Perimeter of an equilateral triangle

c) 12 cm
[Solution Description]
The perimeter of an equilateral triangle is given by $3 \times \text{length of a side}$. Let the length of one side be $x$ cm. Given the perimeter is 36 cm, we can write:
$3x = 36$
Solving for $x$:
$x = \frac{36}{3} = 12 \text{ cm}$
Thus, the length of one side is 12 cm.

Key Concept: Distance covered

c) 750 m
[Solution Description]
The perimeter of the equilateral triangle is $3 \times \text{side length} = 3 \times 50 \text{ m} = 150 \text{ m}$. For 5 complete rounds, the total distance covered is:
$5 \times 150 \text{ m} = 750 \text{ m}$
Thus, the total distance covered is 750 meters.

Key Concept: Perimeter comparison, Equilateral triangle properties

c) $36\sqrt{3}$ cm$^2$
[Solution Description]
First find perimeter of square: $4 \times 9 = 36$ cm. Since perimeters are equal, equilateral triangle has perimeter 36 cm. Each side is $36/3 = 12$ cm. Area of equilateral triangle is $\frac{\sqrt{3}}{4} \times \text{side}^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}$ cm$^2$.

Key Concept: Perimeter of Regular Polygon, Understanding Generalization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a regular hexagon, all six sides are equal in length. According to the general formula for the perimeter of a regular polygon, Perimeter = Number of sides $\times$ length of one side. Here, the number of sides is 6 and the length of one side is 5 cm. Thus, the perimeter is correctly calculated as $6 \times 5$ cm = 30 cm. Both the assertion (A) and reason (R) are true, and the reason (R) correctly explains why the assertion (A) holds true.

Key Concept: Perimeter of Regular Polygons, Number of Sides and Side Length

b) 5
[Solution Description]
Let the original number of sides be $n$ and the side length be $s$. Given:
$n \times s = 60$
New number of sides: $n + 2$, new perimeter:
$(n + 2) \times s = 84$
Subtract first equation from second:
$2s = 24 \Rightarrow s = 12 \text{ cm}$
Substitute $s = 12$ into first equation:
$n \times 12 = 60 \Rightarrow n = 5$

Key Concept: Perimeter of Equilateral Triangle vs Square, Comparison

c) 3 cm
[Solution Description]
Perimeter of square = $4 \times 9 = 36$ cm. Since perimeters are equal, perimeter of equilateral triangle = 36 cm.
$\text{Side length of triangle} = \frac{36}{3} = 12 \text{ cm}$
Difference in side lengths = $12 - 9 = 3$ cm.

Key Concept: Perimeter of irregular polygon, grid-based

b) $8s + 4d$
[Solution Description]
The figure has an outer boundary consisting of 8 straight lines (each of length $s$) and 4 diagonal lines (each of length $d$). The perimeter is calculated by summing all the boundary lengths: $8s + 4d$.

Key Concept: Perimeter of complex polygon, combining straight and diagonal lines

b) $4s + 4d$
[Solution Description]
Original perimeter: $5s + 3d$. Replacing one straight side with a diagonal changes one $s$ to $d$, so the new perimeter becomes $(5s - s) + (3d + d) = 4s + 4d$.

Key Concept: Perimeter using straight and diagonal lines

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the perimeter of a 3x3 grid made of 9 unit squares is 12 units. To verify this, we calculate the perimeter step by step:
1. A single unit square has a perimeter of 4 units (since each side is 1 unit).
2. When squares are arranged in a 3x3 grid, they share sides with adjacent squares.
3. In a 3x3 grid, there are 12 shared sides (each internal square shares its sides with adjacent squares).
4. For every shared side, 2 units are removed from the total perimeter (1 unit for each adjacent square's side).
5. Total reduction in perimeter due to shared sides = 12 shared sides * 2 units = 24 units.
6. Total perimeter of all 9 squares if not joined = 9 squares * 4 units = 36 units.
7. Final perimeter = Total unjoined perimeter - Reduction due to shared sides = 36 - 24 = 12 units.
Therefore, the assertion is true.
The reason states that each shared side reduces the perimeter by 2 units. This is also true because when two squares share a side, both squares lose their individual 1-unit side, effectively reducing the perimeter by 2 units (1 unit per square).
Moreover, the reason correctly explains why the perimeter of the 3x3 grid is 12 units, as demonstrated in the calculation above.

Key Concept: Smallest perimeter with unit squares

b) 12 units
[Solution Description]
The smallest perimeter occurs when the squares are arranged in a $3 \times 3$ square grid. The resulting figure will have dimensions $3 \text{ units} \times 3 \text{ units}$. The perimeter of this square is $4 \times 3 = 12 \text{ units}$.

Key Concept: Perimeter Exploration, Real-World Application

c) 26 units
[Solution Description]
To maximize perimeter, arrange squares in a single row or column to expose the most sides. With 12 squares:
\begin{itemize}

Key Concept: Change in perimeter

c) The perimeter increases.
[Solution Description]
First, find the side length of the original square. Since the perimeter is $24 \text{ units}$, each side is $24 \text{ units} \div 4 = 6 \text{ units}$. Attaching a unit square (side length $1 \text{ unit}$) to one of the sides changes the figure. The attachment can either increase the perimeter or leave it unchanged, but never decrease it. Specifically, if the unit square is attached externally, the perimeter increases by $2 \text{ units}$ (the new perimeter becomes $24 \text{ units} + 2 \text{ units} = 26 \text{ units}$). If attached internally (flush with the side), the perimeter remains $24 \text{ units}$. However, among the options, only increasing is plausible for typical attachments.

Key Concept: Perimeter of Rectangular Running Tracks

d) Assertion is false, but Reason is true.
[Solution Description]
The perimeter of the outer track (Akshi's track) is calculated as $2 \times (70 + 40) = 220$ m. In 5 rounds, Akshi covers $5 \times 220 = 1100$ m.
Assuming the inner track has dimensions smaller than the outer track (but not provided explicitly in the syllabus), it is implied that the inner track has a smaller perimeter. If the inner track had length $60$ m and breadth $30$ m (for example), its perimeter would be $2 \times (60 + 30) = 180$ m. Toshi would then cover $7 \times 180 = 1260$ m, which is more than Akshi's distance.
Since the actual dimensions of the inner track are not specified, the assertion could be false if the inner track does not have a smaller perimeter. However, based on common understanding, the reason is true (outer tracks generally have larger perimeters). Thus, the assertion may not always be true, but the reason is correct in most cases.

Key Concept: Perimeter calculation, Multi-step problem

a) Inner runner starts 50 m before finish line, outer runner starts 250 m before
[Solution Description]
First calculate perimeter of inner track: $4 \times 100 = 400$ m. For 350 m run, the runner completes $\frac{350}{400} = 0.875$ of the track. Starting position should be 0.125 (1-0.875) of perimeter before finish line: $0.125 \times 400 = 50$ m back. For outer track (perimeter $4 \times 150 = 600$ m): $\frac{350}{600} \approx 0.583$. Starting position is $1-0.583 = 0.417$ of perimeter before finish line: $0.417 \times 600 \approx 250$ m back.

Key Concept: Perimeter, Running tracks, Multi-round distance calculation

d) Assertion is false, but Reason is true.
[Solution Description]
To solve this, we need to determine whether the assertion is correct based on the reason provided.
1. **Calculate Akshi's total distance:**
Given the outer track has length 70 m and breadth 40 m, its perimeter is $2 \times (70 + 40) = 220$ m.
For 5 rounds, total distance = $220 \times 5 = 1100$ m.
2. **Analyze Toshi's scenario:**
Since the inner track must be entirely within the outer track, its length and breadth must both be less than 70 m and 40 m respectively (assuming same shape). Let’s assume the inner track has dimensions 60 m (length) and 30 m (breadth). Then, its perimeter would be $2 \times (60 + 30) = 180$ m.
For 7 rounds, total distance = $180 \times 7 = 1260$ m.
This shows that even though the perimeter of the outer track is larger, Toshi could cover more distance if his track's perimeter is not significantly smaller. Hence, the assertion is situation-dependent and not necessarily always true. However, the reason correctly states that the outer track's perimeter is larger but fails to consider the number of rounds factor. Thus, the assertion is false in this context, but the reason about the perimeter comparison is logically sound.

Key Concept: Total rope length for fencing

c) 1920 meters
[Solution Description]
To find the total length of rope needed, first calculate the perimeter of the square field, then multiply by the number of rounds of rope.
Step 1: Calculate the perimeter of the square.
The perimeter of a square is $4 \times \text{side length}$.
$4 \times 120\, \text{m} = 480\, \text{m}$
Step 2: Multiply the perimeter by the number of rounds.
$4 \times 480\, \text{m} = 1920\, \text{m}$
Therefore, the total length of rope needed is 1920 meters.

Key Concept: Perimeter, Rope fencing

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason:
1. Calculate the perimeter of the square field:
$\text{Perimeter} = 4 \times 50\,m = 200\,m$
2. Total rope needed for 5 rounds:
$5 \times 200\,m = 1000\,m$
The total rope matches the assertion value.
The reason correctly explains how the perimeter is calculated, which is fundamental in determining the total rope length.
Therefore, both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Perimeter of a rectangle

b) 700 meters
[Solution Description]
To find the perimeter of the rectangle, we use the formula: $2 \times (\text{length} + \text{breadth})$. Here, the length is 200 meters and the breadth is 150 meters.
Step 1: Add the length and breadth.
$200\, \text{m} + 150\, \text{m} = 350\, \text{m}$
Step 2: Multiply by 2 to get the perimeter.
$2 \times 350\, \text{m} = 700\, \text{m}$
Therefore, the perimeter of the park is 700 meters.

Key Concept: Perimeter, Cost of fencing

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Calculate the perimeter of the rectangular garden.
$Perimeter = 2 \times (length + breadth)$
$Perimeter = 2 \times (25\,m + 40\,m) = 2 \times 65\,m = 130\,m$
Step 2: Calculate the total cost of fencing.
$Total\,Cost = Perimeter \times Cost\,per\,metre$
$Total\,Cost = 130\,m \times `100 = `13,000$
The Assertion states that the total cost of fencing is `13,000. From the calculation, this is true. The Reason explains the correct method to find the total cost, which aligns with the solution steps. Therefore, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Perimeter, Cost of fencing

c) `35,200`
[Solution Description]
First, find the circumference of the circular garden using the formula $C = 2 \pi r$, where $r = 28$ m.
$C = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176$ m.
Next, calculate the total length of fencing for 4 rounds: $176 \times 4 = 704$ m.
Finally, compute the total cost: $704 \times 50 = `35,200$.

Key Concept: Cost of fencing

c) `7,800`
[Solution Description]
Length of the garden $(l) = 80$ m.
Width of the garden $(w) = 50$ m.
Perimeter of the rectangle $= 2 \times (l + w) = 2 \times (80 + 50) = 260$ m.
Cost per meter $= `30$.
Total cost of fencing $= 260 \text{ m} \times `30/\text{m} = `7,800$.

Key Concept: Perimeter after folding and cutting

b) 48 cm
[Solution Description]
The original square has a side length of $8 \text{ cm}$. Folding it in half creates a rectangle of size $8 \text{ cm} \times 4 \text{ cm}$. Cutting along the fold gives two rectangles, each of size $4 \text{ cm} \times 8 \text{ cm}$. The perimeter of one rectangle is $2(4 + 8) = 24 \text{ cm}$, so the total perimeter for both rectangles is $48 \text{ cm}$.

Key Concept: Perimeter After Joining Shapes

c) 24 cm
[Solution Description]
First, find the side length of the square. Since the perimeter of the square is $16 \text{ cm}$, each side is $\frac{16}{4} = 4 \text{ cm}$. Folding the square in half gives a rectangle of size $4 \text{ cm} \times 2 \text{ cm}$. Cutting along the fold results in two rectangles, each of size $2 \text{ cm} \times 4 \text{ cm}$. The perimeter of one rectangle is $2(2 + 4) = 12 \text{ cm}$. Thus, the combined perimeter for both rectangles is $12 \times 2 = 24 \text{ cm}$.

Key Concept: Perimeter after cutting and rearranging

a) 20 cm
[Solution Description]
The original rectangle has dimensions $6 \text{ cm} \times 4 \text{ cm}$. Cutting it into two equal pieces along the length gives two smaller rectangles, each of size $3 \text{ cm} \times 4 \text{ cm}$. When these are joined side by side along their widths, the new figure becomes a rectangle of size $6 \text{ cm} \times 4 \text{ cm}$ again. However, one side is now internal, so we exclude it from the perimeter calculation. The perimeter is calculated as: $6 + 4 + 6 + 4 = 20 \text{ cm}$.

Key Concept: Perimeter of a Rectangle

c) 6 cm
[Solution Description]
Given the perimeter of the rectangle is 36 cm and the length is 12 cm, we need to find the breadth.
Step 1: Use the perimeter formula for a rectangle: $Perimeter = 2 \times (length + breadth)$. Plugging in the known values: $36\,cm = 2 \times (12\,cm + breadth)$.
Step 2: Divide both sides by 2 to isolate the sum of length and breadth: $18\,cm = 12\,cm + breadth$.
Step 3: Subtract the length from both sides to find the breadth: $breadth = 18\,cm - 12\,cm = 6\,cm$.
Therefore, the breadth of the rectangle is 6 cm.

Key Concept: Perimeter of a rectangle, Conceptual reasoning

d) Assertion is false, but Reason is true.
[Solution Description]
Let us verify both statements step by step:
1. **Assertion Check**:
- Perimeter of the rectangle with sides $2x$ and $3x$:
$2 \times (2x + 3x) = 2 \times 5x = 10x.$
- Perimeter of the square with side length $5x$:
$4 \times 5x = 20x.$
- The assertion claims these perimeters are equal ($10x = 20x$), which is false because $10x \neq 20x$.
2. **Reason Check**:
- The reason states the correct formula for the perimeter of a rectangle:
$\text{Perimeter} = 2 \times (\text{length} + \text{breadth}).$
- Thus, the reason is true.
Conclusion: Assertion is false, but Reason is true.

Key Concept: Perimeter of a rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the perimeter of a rectangle, we use the formula: $2 \times (length + breadth)$. Here, length = 5 cm and breadth = 3 cm. Substituting these values into the formula: $2 \times (5\ cm + 3\ cm) = 2 \times 8\ cm = 16\ cm$. Therefore, the Assertion is true. The Reason correctly explains how to calculate the perimeter of a rectangle, so it is also true and correctly explains the Assertion.

Key Concept: Perimeter of a square, Multi-step problem

b) 576 \$
[Solution Description]
First, find the side length of the square garden using the perimeter formula: $\text{Perimeter} = 4 \times \text{side length}$. Given Perimeter = 48 m.
So, $48 \text{ m} = 4 \times \text{side length}$. Thus, $\text{side length} = 48 \text{ m}/4 = 12 \text{ m}$.
The total distance to be tiled is equal to the perimeter, which is 48 m.
Number of tiles required = $\text{Total distance}/\text{Tile coverage} = 48 \text{ m}/0.5 \text{ m per tile} = 96 \text{ tiles}$.
Total cost = $96 \text{ tiles} \times 6 \text{ \$/tile} = 576 \text{ \$}$.

Key Concept: Perimeter of a square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
First, check if the assertion is true: Given a square with side length 5 cm, the perimeter is $4 \times 5 = 20$ cm. This matches the assertion. Now, check the reason: It states that the perimeter of any square is calculated by multiplying the length of one side by 4, which is correct as per the formula for the perimeter of a square. Furthermore, the reason correctly explains why the assertion is true, since the perimeter calculation in the assertion follows directly from the general rule stated in the reason.

Key Concept: Perimeter of a square

b) 20 meters
[Solution Description]
The formula for the perimeter of a square is $4 \times \text{side length}$. Given the side length is 5 meters, we calculate the perimeter as follows:
$4 \times 5 = 20 \text{ meters}$

Key Concept: Perimeter of a triangle, Mixed units

a) $d = 1, s = 2$
[Solution Description]
Step 1: Understand the relationship $s = 2d$.
Step 2: Express all sides in terms of $d$ or $s$. Here, we choose $d$:
$5d + 3s + \text{third side} = 15 \\ \\ 5d + 3(2d) + \text{third side} = 15 \\ \\ 5d + 6d + \text{third side} = 15 \\ \\ 11d + \text{third side} = 15$
Step 3: For a valid triangle, the third side must satisfy triangle inequalities. However, since $s = 2d$, substituting back gives:
$11d + \text{third side} = 15$
From the question’s context, the third side can be written as $4d$ (derived from $2s$):
$11d + 4d = 15 \\ \\ 15d = 15 \\ \\ d = 1$
Thus, $s = 2d = 2$.

Key Concept: Equilateral triangle perimeter

b) 18 cm
[Solution Description]
Since it is an equilateral triangle, all sides are equal. Perimeter is calculated as $3 \times \text{side length} = 3 \times 6\,\text{cm} = 18\,\text{cm}$.

Key Concept: Perimeter with mixed units

b) 6s + 3d
[Solution Description]
Add all the side lengths together: $2s + 3d + 4s = (2s + 4s) + 3d = 6s + 3d$.

Key Concept: Perimeter of a regular polygon, General formula application

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A regular heptagon has $7$ sides. Given the side length $(s)$ is $5$ cm, the perimeter $(P)$ can be calculated using the formula:
$P = n \times s = 7 \times 5\,\text{cm} = 35\,\text{cm}.$
The assertion states the perimeter is $35$ cm, which matches our calculation. The reason provides the correct general formula for calculating the perimeter of a regular polygon, and it correctly explains the assertion. Thus, both are true, and the reason justifies the assertion.

Key Concept: Generalization of perimeter for regular polygons

c) 9 cm
[Solution Description]
A regular pentagon has five equal sides. The perimeter of a regular pentagon is five times the length of one side. Let the length of one side be $s$. Then,
$\text{Perimeter} = 5 \times s$.
Given that the perimeter is 45 cm,
$5s = 45 \text{ cm}$.
Solving for $s$,
$s = \frac{45 \text{ cm}}{5} = 9 \text{ cm}$.

Key Concept: Perimeter of a regular polygon

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the perimeter of an equilateral triangle with each side measuring 5 cm is 15 cm. Since all sides are equal in an equilateral triangle, the perimeter is calculated as $3 \times \text{side length} = 3 \times 5 = 15$ cm. Therefore, the assertion is true.
The reason states that for any regular polygon, the perimeter can be calculated by multiplying the number of sides by the length of one side. This is correct as it is the general formula for the perimeter of a regular polygon. Additionally, this formula correctly explains why the perimeter of the given equilateral triangle is 15 cm (since it has 3 sides). Thus, the reason is true and correctly explains the assertion.

Key Concept: Perimeter of an equilateral triangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
An equilateral triangle has all three sides equal. Here, the side length is given as 4 cm. Therefore, the perimeter can be calculated as $3 \times \text{side length} = 3 \times 4 \, \text{cm} = 12 \, \text{cm}$. Thus, both Assertion and Reason are true, and Reason correctly explains why the perimeter is 12 cm.

Key Concept: Perimeter of an equilateral triangle

c) 12 cm
[Solution Description]
First, find the perimeter of the square. Perimeter of a square $= 4 \times \text{side} = 4 \times 9 = 36$ cm. This is equal to the perimeter of the equilateral triangle. Let the side of the equilateral triangle be $s$.
Using the perimeter formula for an equilateral triangle:
$3 \times s = 36$
Solve for $s$:
$s = \frac{36}{3} = 12 \text{ cm}$
So, the side length of the equilateral triangle is 12 cm.

Key Concept: Perimeter of equilateral triangle, conceptual complexity

d) 18 cm
[Solution Description]
First, find the perimeter of the first small triangle: $3 \times 7 = 21$ cm.
Next, find the perimeter of the second small triangle: $3 \times 11 = 33$ cm.
Add these perimeters: $21 + 33 = 54$ cm.
This is the perimeter of the larger triangle. Now, find its side length: $54 / 3 = 18$ cm.

Key Concept: Perimeter Change Due to Attachment

c) Increases by 2 units
[Solution Description]
When a new square is attached by fully aligning one of its sides with the figure, two sides of the square become internal (not part of the perimeter). The other two sides add to the perimeter. Therefore, the perimeter increases by $4 - 2 = 2$ units. New perimeter = $24 \text{ units} + 2 \text{ units} = 26 \text{ units}.$

Key Concept: Perimeter of split and rejoined figures

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The original perimeter of the 6 cm $\times$ 4 cm rectangle is $2(6 + 4) = 20$ cm. When it is split along the length (6 cm side), each piece has dimensions 3 cm $\times$ 4 cm. Rejoining these pieces can change the perimeter depending on how they are arranged. For example:
1. If rejoined side by side along the 4 cm side, the perimeter becomes $2(6 + 4) = 20$ cm (same as original).
2. If rejoined end to end along the 3 cm side, the perimeter becomes $2(12 + 2) = 28$ cm (greater than original).
Thus, the assertion is true as the perimeter can be equal or greater after rejoining, and the reason correctly explains why this happens due to the introduction of new boundary edges.

Key Concept: Split \& Rejoin: Perimeter Calculation

b) 22 cm
[Solution Description]
The original rectangle has dimensions 6 cm by 4 cm. When split into two equal parts along the longer side (6 cm), each part will be $3 \text{ cm} \times 4 \text{ cm}$. If they are rejoined along the longer side, the new figure will be a square with sides 4 cm + 4 cm = 8 cm length and 3 cm height. The perimeter is calculated as:
$2 \times (8 \text{ cm} + 3 \text{ cm}) = 22 \text{ cm}.$

Key Concept: Area of Rectangle and Square

d) Assertion is false, but Reason is true.
[Solution Description]
First, calculate the area of the rectangle using the formula $\text{length} \times \text{width}$:
Length = 6 m, Width = 4 m. So, Area of rectangle = $6 \, \text{m} \times 4 \, \text{m} = 24 \, \text{sq m}$.
Next, calculate the area of the square using the formula $\text{side} \times \text{side}$:
Side = 5 m. So, Area of square = $5 \, \text{m} \times 5 \, \text{m} = 25 \, \text{sq m}$.
Comparing both areas, we see that 24 sq m (rectangle) is not equal to 25 sq m (square). Hence, Assertion (A) is false.
Reason (R) states the correct formulas for calculating the areas of a rectangle and a square, so Reason (R) is true.
Therefore, the correct answer is: Assertion is false, but Reason is true.

Key Concept: Combined areas of rectangles

c) 10 m $\times$ 8.3 m
[Solution Description]
First, calculate the area of the first field: $8 \, \text{m} \times 6 \, \text{m} = 48 \, \text{sq m}$. Then, calculate the area of the second field: $5 \, \text{m} \times 7 \, \text{m} = 35 \, \text{sq m}$. The total combined area is $48 \, \text{sq m} + 35 \, \text{sq m} = 83 \, \text{sq m}$. Now, check the given options to see which pair of dimensions gives an area of 83 sq m. For example, option c): $10 \, \text{m} \times 8.3 \, \text{m} = 83 \, \text{sq m}$.

Key Concept: Area application

c) 160 tiles
[Solution Description]
Effective tiled area dimensions after leaving 1m border on all sides:
Length: $10 \text{m} - 2 \text{m} = 8 \text{m}$, Width: $7 \text{m} - 2 \text{m} = 5 \text{m}$.
Number of tiles along length: $8 \text{m} / 0.5 \text{m} = 16$ tiles.
Number of tiles along width: $5 \text{m} / 0.5 \text{m} = 10$ tiles.
Total tiles needed: $16 \times 10 = 160$ tiles.

Key Concept: Area of a Rectangle

b) 24 square meters
[Solution Description]
The area of a rectangle is found by multiplying its length by its width. Given the length is 8 meters and the width is 3 meters, the area is $8 \times 3 = 24$ square meters.

Key Concept: Area of composite shapes, Area subtraction

c) 171.73 sq m
[Solution Description]
First calculate total garden area: $18 \times 12 = 216$ sq m. Fountain radius is $\frac{6}{2} = 3$ m, so area is $\pi (3)^2 = 9\pi$ sq m. Flower bed area is $4 \times 4 = 16$ sq m. Total covered area is $9\pi + 16 \approx 28.27 + 16 = 44.27$ sq m. Uncovered area is $216 - 44.27 = 171.73$ sq m.

Key Concept: Unit squares estimation, Non-rectilinear shapes

b) 121.25 sq m
[Solution Description]
Partial squares can be estimated as half coverage: $23 \div 2 \approx 11.5$. Total squares covered ≈ $37 + 11.5 = 48.5$. Each square represents $2.5$ sq m, so total area ≈ $48.5 \times 2.5 = 121.25$ sq m.

Key Concept: Calculating cost based on area

c) \$1500
[Solution Description]
Calculate the area of the plot: $length \times width = 20\,m \times 15\,m = 300\,sq\,m$. Multiply the area by the cost per sq m to get the total cost: $300\,sq\,m \times \$5/sq\,m = \$1500$.

Key Concept: Area of rectangle, percentage change

b) -16\%
[Solution Description]
Original area $A_1 = L \times W$.
New length = $1.2L$.
New width = $0.7W$.
New area $A_2 = 1.2L \times 0.7W = 0.84LW$.
Change in area = $A_2 - A_1 = 0.84LW - LW = -0.16LW$.
Percentage change = $-16\%$ (the negative sign indicates a decrease).

Key Concept: Real-life Application

c) 40 sq m
[Solution Description]
The area of a rectangle is given by $\text{length} \times \text{width}$. For the garden:
$8 \, \text{m} \times 5 \, \text{m} = 40 \, \text{sq m}.$

Key Concept: Area of Square

c) 49 square meters
[Solution Description]
The formula for the area of a square is $\text{Area} = \text{side} \times \text{side}$. Given the side length is 7 meters, we can calculate the area as:
$\text{Area} = 7 \, \text{meters} \times 7 \, \text{meters} = 49 \, \text{square meters}$.

Key Concept: Area of square, Conceptual Understanding

c) 10 m
[Solution Description]
First, find the area of the first smaller square: $6 \, \text{m} \times 6 \, \text{m} = 36 \, \text{sq m}$. Next, find the area of the second smaller square: $8 \, \text{m} \times 8 \, \text{m} = 64 \, \text{sq m}$. The total area of the larger square is the sum of these two areas: $36 \, \text{sq m} + 64 \, \text{sq m} = 100 \, \text{sq m}$. To find the side length of the larger square, take the square root of its area: $\sqrt{100 \, \text{sq m}} = 10 \, \text{m}$.

Key Concept: Area of Square

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the area of a square with side length 6 m is 36 sq m. According to the given syllabus, the area of a square is $\text{side}^2$. Here, the side length is 6 m, so the area is $6 \times 6 = 36 \, \text{sq m}$. Therefore, the assertion is true. The reason states the correct formula for calculating the area of a square, which aligns with the syllabus. Moreover, the reason correctly explains why the assertion is true. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Area comparison

c) Two shapes can have different areas for given perimeters.
[Solution Description]
While Shape A has a longer perimeter but smaller area compared to Shape B, this implies that Shape A is likely more elongated (e.g., a long thin rectangle) while Shape B could be more compact (e.g., a square). Therefore, it must be true that two shapes can have different areas for given perimeters.

Key Concept: Area estimation using grid paper

a) 16 sq units
[Solution Description]
Full squares contribute $12 \times 1 = 12 \, \text{sq units}$. Half-squares contribute $8 \times \frac{1}{2} = 4 \, \text{sq units}$. The total estimated area is the sum of these contributions: $12 + 4 = 16 \, \text{sq units}$.

Key Concept: Area of Rectangle and Square

d) Assertion is false, but Reason is true.
[Solution Description]
First, calculate the area of the rectangle using its formula: $\text{Area}_{\text{rectangle}} = \text{length} \times \text{width} = 10 \, \text{m} \times 5 \, \text{m} = 50 \, \text{sq m}$. Next, calculate the area of the square: $\text{Area}_{\text{square}} = \text{side} \times \text{side} = 7 \, \text{m} \times 7 \, \text{m} = 49 \, \text{sq m}$. Comparing both areas: $50 \, \text{sq m} \neq 49 \, \text{sq m}$. Therefore, the Assertion is false. However, the Reason correctly states the formulas for calculating the area of a rectangle and a square, making it true.

Key Concept: Counting Partial Squares

b) 7 sq units
[Solution Description]
Full squares contribute 3 sq units. Squares more than half covered count as 1 sq unit each: $4 \times 1 = 4$ sq units. Total area: $3 + 4 = 7$ sq units.

Key Concept: Area comparison using grid paper

a) Shape X
[Solution Description]
Area of Shape X: 15 $\times$ 1 + 5 $\times$ 1 = 20 sq units.
Area of Shape Y: 10 $\times$ 1 + 8 $\times$ 1 + 2 $\times$ $\frac{1}{2}$ = 19 sq units.
Shape X has a larger area (20 sq units) than Shape Y (19 sq units).

Key Concept: Area of triangles on grid paper

b) 4.5 sq units
[Solution Description]
Base AB = $5 - 2 = 3$ units.
Height BC = $6 - 3 = 3$ units.
Area of triangle ABC = $\frac{1}{2} \times$ base $\times$ height = $\frac{1}{2} \times 3 \times 3 = 4.5$ sq units.

Key Concept: Area of Rectangle

b) 40 sq units
[Solution Description]
The area of a rectangle is calculated using the formula $length \times width$. Here, $8 \times 5 = 40$ sq units.

Key Concept: Counting Unit Squares, Area Estimation

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion claims that the area calculated by counting only full unit squares is accurate, which is false because partial squares also contribute to the total area. The reason explains the convention for counting partial squares (ignoring if less than half, counting as 1 if more than half), which is true. However, this method provides an approximate area, not an exact one. Thus, the assertion is false, but the reason is true.

Key Concept: Perimeter and Area

a) $3 \times 8$ and $4 \times 6$
[Solution Description]
Possible dimensions for area 24 sq units: $(1 \times 24)$, $(2 \times 12)$, $(3 \times 8)$, $(4 \times 6)$. The perimeters would be $50$, $28$, $22$, and $20$ units respectively.

Key Concept: Combined area calculation, non-overlapping regions

b) 73 sq m
[Solution Description]
Calculate the area of the rectangular section: $8 \text{ m} \times 6 \text{ m} = 48 \text{ sq m}$.
Calculate the area of the square section: $5 \text{ m} \times 5 \text{ m} = 25 \text{ sq m}$.
Add both areas to get the total area: $48 \text{ sq m} + 25 \text{ sq m} = 73 \text{ sq m}$.

Key Concept: Comparison of Shapes for Area Measurement

b) Squares can cover a region without leaving gaps
[Solution Description]
Squares can be packed tightly without gaps, making it easier to cover any enclosed region completely. Circles leave gaps when packed together, leading to inaccurate area measurement.

Key Concept: Area Estimation Rules

b) 23.5 sq units
[Solution Description]
Full squares contribute 15 square units. Squares more than half covered contribute 7 square units. Exactly half covered squares contribute $3 \times \frac{1}{2} = 1.5$ square units. Total area = $15 + 7 + 1.5 = 23.5$ square units.

Key Concept: Area of Triangle, Diagonal Division

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let’s analyze the assertion and reason step-by-step:
1. **Understanding the Rectangle and Diagonal**:
- A rectangle has four sides with opposite sides equal and all angles as right angles.
- A diagonal connects two non-adjacent vertices, splitting the rectangle into two right-angled triangles.
2. **Area of Rectangle**:
- If the rectangle has length $l$ and width $w$, its area is $A = l \times w$.
3. **Area of Each Triangle**:
- The diagonal splits the rectangle into two identical (congruent) right-angled triangles.
- Each triangle will have legs of lengths $l$ and $w$.
- The area of each right-angled triangle is $\frac{1}{2} \times l \times w = \frac{A}{2}$.
- Thus, the area of each triangle is half of the rectangle's area.
4. **Congruency of Triangles**:
- Both triangles are congruent because their corresponding sides and angles are equal (by Side-Angle-Side congruency).
- Congruent implies they have the same shape and size, hence equal areas.
5. **Relationship Between Assertion and Reason**:
- The assertion states the area relationship between the triangle and the rectangle.
- The reason explains why this relationship exists (due to congruency and equal division of area).
- Therefore, the reason correctly explains the assertion.

Key Concept: Basic Concept

c) 6 units
[Solution Description]
The formula for the area of a triangle is $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. Given: Area = 30 square units, base = 10 units.
Step 1: Rearrange the formula to solve for height:
$\text{height} = \frac{2 \times \text{Area}}{\text{base}}$.
Step 2: Substitute the given values:
$\text{height} = \frac{2 \times 30}{10} = \frac{60}{10} = 6$ units.
Therefore, the height of the triangle is 6 units.

Key Concept: Area of Triangle via Rectangle

c) 30 cm$^2$
[Solution Description]
First, find the area of the rectangle: $Area_{rectangle} = length \times width = 10\ cm \times 6\ cm = 60\ cm^2$. The diagonal divides the rectangle into two equal triangles, so each triangle's area is half of the rectangle's area: $Area_{triangle} = \frac{1}{2} \times Area_{rectangle} = \frac{1}{2} \times 60\ cm^2 = 30\ cm^2$.

Key Concept: Relationship between Triangle and Rectangle Areas

c) 15 cm$^2$
[Solution Description]
The area of each triangle formed by the diagonal of a rectangle is half the area of the rectangle: $Area_{triangle} = \frac{1}{2} \times Area_{rectangle} = \frac{1}{2} \times 30 \, cm^2 = 15 \, cm^2$.

Key Concept: Multi-concept: Diagonal division, Combined areas

c) 80 cm$^2$
[Solution Description]
First, find the area of square PQRS: $10 \times 10 = 100$ cm$^2$. The diagonal divides it into two triangles, each with area $\frac{100}{2} = 50$ cm$^2$.
Next, find the area of rectangle LMNO: $10 \times 6 = 60$ cm$^2$. The diagonal divides it into two triangles, each with area $\frac{60}{2} = 30$ cm$^2$.
The problem asks for the combined area of triangle PRS (from the square) and triangle LMO (from the rectangle). Thus, total area $= 50 + 30 = 80$ cm$^2$.

Key Concept: Area of Triangle from Rectangle Diagonal

c) 24 cm$^2$
[Solution Description]
First, find the area of the rectangle: $Area_{rectangle} = length \times width = 8 \, cm \times 6 \, cm = 48 \, cm^2$. The area of each triangle formed by the diagonal is half the area of the rectangle: $Area_{triangle} = \frac{1}{2} \times Area_{rectangle} = \frac{1}{2} \times 48 \, cm^2 = 24 \, cm^2$.

Key Concept: Area of triangle, Rectangle decomposition

d) 30 cm$^2$
[Solution Description]
First, find the area of rectangle ABCD: $10 \text{ cm} \times 6 \text{ cm} = 60 \text{ cm}^2$. The area of triangle ABE is half the area of the rectangle formed by its base and height. The base of triangle ABE is AB = 10 cm, and the height is the width of the rectangle = 6 cm. So, the area of triangle ABE is $\frac{1}{2} \times 60 = 30 \text{ cm}^2$.

Key Concept: Area of Triangle from Rectangle

d) 30 cm$^2$
[Solution Description]
To find the area of the triangle, first calculate the area of the rectangle. The formula for the area of a rectangle is:
$\text{Area of Rectangle} = \text{Length} \times \text{Width} = 10\,\text{cm} \times 6\,\text{cm} = 60\,\text{cm}^2$
Since the diagonal divides the rectangle into two equal triangles, the area of each triangle is half the area of the rectangle:
$\text{Area of Triangle} = \frac{1}{2} \times 60\,\text{cm}^2 = 30\,\text{cm}^2$
Therefore, the area of one triangle is 30 cm$^2$.

Key Concept: Area of Triangle, Congruent halves

b) 36 cm$^2$
[Solution Description] First, calculate the area of rectangle ABCD: $12 \times 8 = 96$ cm$^2$. The area of triangle BAD is half of this, so $48$ cm$^2$. Now, triangle BED shares vertex B with triangle BAD and has base ED = AD - AE = 9 cm (since AD = 12 cm and AE = 3 cm). Its height remains the same as triangle BAD, which is 8 cm. Thus, the area of triangle BED is $\frac{1}{2} \times 9 \times 8 = 36$ cm$^2$.

Key Concept: Area of Triangle, Congruent halves

c) 4:1
[Solution Description] Area of rectangle EFGH is $16 \times 6 = 96$ cm$^2$. Area of triangle EFG is half of this, so 48 cm$^2$. For triangle EKG, base EK = 4 cm, and height FG = 6 cm. Thus, its area is $\frac{1}{2} \times 4 \times 6 = 12$ cm$^2$. The ratio of the area of triangle EFG to triangle EKG is $\frac{48}{12} = 4:1$.

Key Concept: Area of Triangle from Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the area of a triangle formed by splitting a rectangle along its diagonal is half the area of the rectangle. From the syllabus, this is confirmed as a correct relationship. The reason states that the two triangles obtained are congruent (exactly overlapping) and have equal areas, which is also true from the syllabus. Moreover, the reason correctly explains why the assertion is true: since both triangles are congruent and together form the entire rectangle, each must have half the area.

Key Concept: Partial split, Combined area

c) 35 cm$^2$
[Solution Description]
The area of rectangle EFGH = $14 \times 7 = 98$ cm$^2$.
Triangle EFK has base FK = 5 cm (part of FG) and height EF = 14 cm.
So, the area of triangle EFK = $\frac{1}{2} \times 5 \times 14 = 35$ cm$^2$.

Key Concept: Comparing Areas

b) The triangle's area is half the rectangle's area.
[Solution Description]
When a triangle and a rectangle have the same base and height, the area of the triangle is half the area of the rectangle.
Step 1: Let the area of the rectangle = base $\times$ height.
Step 2: Area of the triangle = $\frac{1}{2} \times$ base $\times$ height.
Step 3: Thus, the triangle's area is half of the rectangle's area.

Key Concept: Area of Triangle via Rectangle Diagonal

c) 30 cm$^2$
[Solution Description]
First, find the area of the rectangle: $10 \text{ cm} \times 6 \text{ cm} = 60 \text{ cm}^2$. Since the diagonal divides the rectangle into two equal triangles, the area of one triangle is half of the rectangle's area: $\frac{1}{2} \times 60 \text{ cm}^2 = 30 \text{ cm}^2$.

Key Concept: Perimeter Comparison

c) Greater than P
[Solution Description]
The perimeter of shapes changes based on their arrangement but the area remains the same. Since a square minimizes perimeter for a given area, rearranging the pieces into a rectangle will result in a larger perimeter than the square's perimeter P.

Key Concept: Tangram Area Relationships

d) Shape D has twice the area of Shape F
[Solution Description]
Start with the given relationships:
1. Shape D can be covered exactly by Shapes C and E ($D = C + E$). Since $C = E$, then $D = 2C$.
2. Shape A and B have the same area, and Shape G has the same area as Shape A ($G = A$).
3. Shape F has half the area of Shape G ($F = \frac{1}{2}G$), so $F = \frac{1}{2}A$.
Now, derive the relationship between Shape D and F. From step 1, $D = 2C$. The syllabus hints that $A$ is twice as big as $C$ or $E$ ($A = 2C$). Therefore, $D = 2C = A = G$. Then, $F = \frac{1}{2}A = \frac{1}{2}D$. Thus, the area of Shape D is twice the area of Shape F.

Key Concept: Perimeter and Area

b) Same as the area of Shape C
[Solution Description]
From the syllabus hint, Shape D can be covered using Shapes C and E, meaning Shape D = Shape C + Shape E. Since Shape C = Shape E, Shape D = 2 × Shape C. The syllabus does not explicitly mention Shape F's relationship with Shape C, but based on typical tangram configurations, Shape F has the same area as Shape C.

Key Concept: Tangram shapes and comparison, Area relationships

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the syllabus, we know that Shape D can be exactly covered using Shapes C and E. This means the area of Shape D is equal to the sum of the areas of Shapes C and E. Since Shapes C and E have the same area, the area of Shape D is twice that of Shape C or Shape E.
Now, consider the big square formed with all seven tangram pieces. According to the given relationships:
- The area of Shape A + B (both equal) = $4 \times$ area of Shape C.
- Shape D = $2 \times$ area of Shape C.
- Shapes F and G combined also contribute proportionally based on their relationships (as derived from overlapping).
Adding up all these contributions:
Total area of the big square = $(4 + 4 + 2 + 1 + 1 + 2 + 2) \times$ area of Shape C = $16 \times$ area of Shape C.
Thus, the Assertion is true. The Reason correctly explains this by stating how Shape D relates to Shape C and how the big square can be divided into regions each equal to Shape C, totaling 16 such regions.

Key Concept: Area Comparison

b) 3 square units
[Solution Description]
From the syllabus, we know that the area of Shape D is twice the area of Shape C ($\text{Area of D} = 2 \times \text{Area of C}$). Also, Shapes C and E have the same area ($\text{Area of E} = \text{Area of C}$).
Given:
$\text{Area of C} = 1 \text{ square unit}$
Step 1: Calculate the area of Shape D.
$\text{Area of D} = 2 \times \text{Area of C} = 2 \times 1 = 2 \text{ square units}$
Step 2: Calculate the area of Shape E.
$\text{Area of E} = \text{Area of C} = 1 \text{ square unit}$
Step 3: Add the areas of Shapes D and E.
$\text{Total Area} = \text{Area of D} + \text{Area of E} = 2 + 1 = 3 \text{ square units}$

Key Concept: Area comparison among Shapes C, D, and E

a) Shape D is equal to Shapes C and E combined, and Shapes C and E are equal
[Solution Description]
From the syllabus, we know that Shape D has twice the area of Shape C or Shape E ($\text{Area of D} = 2 \times \text{Area of C}$), and Shapes C and E have the same area.

Key Concept: Area comparison

b) 2 square units
[Solution Description]
From the key observations, we know that $\text{Area of D} = 2 \times \text{Area of C}$. Given $\text{Area of C} = 1$, then $\text{Area of D} = 2 \times 1 = 2$ square units.

Key Concept: Tangram and Puzzle-based Area Exploration

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the syllabus hints, we know that Shape D can be exactly covered by two Shapes C. This implies the area of Shape D is twice that of Shape C. Therefore, both Assertion (A) and Reason (R) are true. Additionally, the reason correctly explains why the assertion is true.