Key Concept: Perimeter of a Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To calculate the perimeter of the given rectangle, we use the formula for perimeter of a rectangle, which is $2 \times (\text{length} + \text{breadth})$. Here, length = 10 cm and breadth = 5 cm. Plugging the values into the formula:
Step 1: Add length and breadth: $10 \text{ cm} + 5 \text{ cm} = 15 \text{ cm}$.
Step 2: Multiply by 2: $2 \times 15 \text{ cm} = 30 \text{ cm}$.
So, the perimeter is indeed 30 cm. The reason provided is the correct formula for calculating the perimeter of a rectangle. Therefore, both assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Square Perimeter

c) 48 meters
[Solution Description]
The perimeter of a square is given by $4 \times side$. Substituting the given value:
$Perimeter = 4 \times 12\ m = 48\ m.$

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 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 composite shape

b) 90 m
[Solution Description]
To find the perimeter of the remaining part of the garden, first calculate the perimeter of the entire garden and then subtract the perimeter of the flower bed. However, since the flower bed is inside the garden, the boundary of the remaining garden includes both the outer boundary of the garden and the boundary around the flower bed.
Step 1: Perimeter of the rectangular garden:
$P_{\text{garden}} = 2 \times (20\,\text{m} + 15\,\text{m}) = 70\,\text{m}$
Step 2: Perimeter of the square flower bed:
$P_{\text{flower bed}} = 4 \times 5\,\text{m} = 20\,\text{m}$
Step 3: Total perimeter of the remaining garden:
Since the flower bed is removed, add the perimeters of the garden and the flower bed:
$P_{\text{total}} = P_{\text{garden}} + P_{\text{flower bed}} = 70\,\text{m} + 20\,\text{m} = 90\,\text{m}$

Key Concept: Perimeter of a rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion, we calculate the perimeter of the given rectangle using the formula. Given length = 10 cm and breadth = 6 cm, the perimeter is:
$2 \times (10\,\text{cm} + 6\,\text{cm}) = 2 \times 16\,\text{cm} = 32\,\text{cm}.$
The calculation matches the assertion, so the assertion is true. The reason provides the correct formula for calculating the perimeter of a rectangle, which justifies the assertion.
Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Perimeter of a Square and Equilateral Triangle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
First, calculate the perimeter of the square: $4 \times 5 = 20\,\text{cm}$. Next, calculate the perimeter of the equilateral triangle: $3 \times \frac{20}{3} = 20\,\text{cm}$. Both perimeters are equal. The Reason explains how the formulas for the perimeter of a square and an equilateral triangle differ.

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: 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: 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: 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: Perimeter of a Triangle

d) 24 cm
[Solution Description]
To find the perimeter of a triangle, add the lengths of all three sides. Given the sides are 6 cm, 8 cm, and 10 cm:
$\text{Perimeter} = 6\,\text{cm} + 8\,\text{cm} + 10\,\text{cm}$
Step 1: Add the first two sides:
$6\,\text{cm} + 8\,\text{cm} = 14\,\text{cm}$
Step 2: Add the third side to the result from Step 1:
$14\,\text{cm} + 10\,\text{cm} = 24\,\text{cm}$
The perimeter of the triangle is 24 cm.

Key Concept: Multi-step perimeter problem, Ratio applications

c) 80 cm
[Solution Description]
Perimeter of hexagon is 240 cm, so each side is $240/6 = 40$ cm. Same wire length is used for both shapes. Total wire length remains same. Triangle's perimeter = Hexagon's perimeter = 240 cm. Each side of triangle was $240/3 = 80$ cm.

Key Concept: Real-world application, Cost calculation

c) $800\sqrt{3}$ m$^2$
[Solution Description]
Total cost \$1,800 at \$15/m means perimeter is $1800/15 = 120$ m. Each side is $120/3 = 40$ m. Area is $\frac{\sqrt{3}}{4} \times 40^2 = 400\sqrt{3}$ m$^2$. Twice the area is $800\sqrt{3}$ m$^2$.

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: Regular Polygon Perimeter

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the perimeter of a regular pentagon, we use the formula: $Perimeter = n \times s$, where $n$ is the number of sides and $s$ is the length of one side. A pentagon has 5 sides. Substituting the given values: $Perimeter = 5 \times 4\text{ cm} = 20\text{ cm}$. This matches the Assertion (A). The Reason (R) correctly states the general formula for the perimeter of any regular polygon, which explains why the calculation in Assertion (A) is correct. Thus, both Assertion and Reason are true, and Reason correctly explains Assertion.

Key Concept: Perimeter of Regular Polygon

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A regular hexagon has 6 equal sides. Given that the length of one side is 5 cm, the perimeter can be calculated as:
$\text{Perimeter} = \text{Number of sides} \times \text{Length of one side} = 6 \times 5 \, \text{cm} = 30 \, \text{cm}$.
The Reason correctly states the formula for the perimeter of a regular polygon and directly explains why the Assertion is true.

Key Concept: Perimeter of Square

c) 12 inches
[Solution Description]
A square has 4 equal sides. The perimeter is given by:
$Perimeter = 4 \times \text{length of one side}$.
Given that perimeter = 48 inches, we can find the length of one side as:
$48 = 4 \times \text{length of one side}$.
Dividing both sides by 4:
$\text{length of one side} = \frac{48}{4} = 12 \text{ inches}$.

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: Perimeter of Connected Squares

b) 15.66 units
[Solution Description]
To find the total perimeter, we substitute the given values into the expression $10s + 4d$. Each straight side $s = 1$ unit, and each diagonal side $d = \sqrt{2} \approx 1.414$ units.
Calculation: $10s = 10 \times 1 = 10$ units.
$4d = 4 \times 1.414 \approx 5.656$ units.
Total perimeter $= 10 + 5.656 \approx 15.656$ units.
Rounded to two decimal places, the perimeter is approximately 15.66 units.

Key Concept: Perimeter using straight and diagonal lines

c) $4s$
[Solution Description]
A square has 4 sides. Since all sides are straight, the total perimeter is the sum of all straight units. Each side is $1s$, so:
$\text{Perimeter} = 4 \times s = 4s$

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 Pieces

d) Assertion is false, but Reason is true.
[Solution Description]
The original perimeter of the rectangle is $2(6 + 4) = 20 \text{ cm}$. When cut along its length, each piece will have dimensions $3 \text{ cm} \times 4 \text{ cm}$. If these pieces are rejoined side by side, they form a new rectangle with dimensions $6 \text{ cm} \times 4 \text{ cm}$, which has the same perimeter as the original ($20 \text{ cm}$). Thus, the assertion is false because the perimeter remains unchanged at $20 \text{ cm}$ if the rearrangement does not change the outer dimensions. However, the reason is true because the perimeter does depend on how the pieces are rearranged (e.g., different arrangements can lead to perimeters like $28 \text{ cm}$ or $22 \text{ cm}).

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: Total Distance Calculation

d) 900 meters
[Solution Description]
First, calculate the perimeter of the track: $2 \times (60 + 30) = 2 \times 90 = 180$ m per round. For 5 rounds, total distance = $5 \times 180 = 900$ m.

Key Concept: Perimeter of Rectangular Tracks

d) Assertion is false, but Reason is true.
[Solution Description]
First, calculate the perimeter of Akshi's outer track:
$2 \times (70 \, \text{m} + 40 \, \text{m}) = 220 \, \text{m}.$
Total distance covered by Akshi in 5 rounds:
$5 \times 220 \, \text{m} = 1100 \, \text{m}.$
Next, calculate the perimeter of Toshi's inner track (assuming dimensions 60 m $\times$ 30 m):
$2 \times (60 \, \text{m} + 30 \, \text{m}) = 180 \, \text{m}.$
Total distance covered by Toshi in 7 rounds:
$7 \times 180 \, \text{m} = 1260 \, \text{m}.$
Comparing the distances: $1260 \, \text{m} > 1100 \, \text{m}$. Thus, Toshi ran a longer distance, making the Assertion false. However, the Reason is true because the total distance depends on both the perimeter and the number of rounds.

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: Cost of fencing

b) Rs.8400
[Solution Description]
First, find the perimeter of the garden, then multiply by the cost per meter to find the total cost.
Step 1: Calculate the perimeter.
$2 \times (80\, \text{m} + 60\, \text{m}) = 2 \times 140\, \text{m} = 280\, \text{m}$
Step 2: Multiply the perimeter by the cost per meter.
$280\, \text{m} \times Rs.30/\text{m} = Rs.8400$
Therefore, the total cost of fencing the garden is Rs.8400.

Key Concept: Perimeter transformation, Rectangle to square

b) 100 cm$^2$
[Solution Description]
Perimeter of rectangle: $2 \times (12\,cm + 8\,cm) = 40\,cm$.
Side of square: $\frac{40\,cm}{4} = 10\,cm$.
Area of square: $10\,cm \times 10\,cm = 100\,cm^2$.

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: Multiple Rounds Fencing

c) 840 m
[Solution Description]
Calculate single-round perimeter: $2 \times (80m + 60m) = 280m$.
For 3 rounds: $3 \times 280m = 840m$.

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: Perimeter to Cost Conversion

c) \Rs.3,750
[Solution Description]
First find perimeter: $2 \times (length + breadth) = 2 \times (45m + 30m) = 2 \times 75m = 150m$.
Then calculate cost: $150m \times \Rs.25/m = \Rs.3750$.

Key Concept: Perimeter changes, multi-step folding

c) $32\sqrt{2} \text{ cm}$
[Solution Description]
The square has side length $8 \text{ cm}$. When folded along the diagonal, the diagonal length is calculated using the Pythagorean theorem:
$\text{Diagonal} = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} \text{ cm}$
After cutting, we get two right-angled isosceles triangles with legs of $8 \text{ cm}$ and hypotenuse $8\sqrt{2} \text{ cm}$.
Rearranging these triangles so their right angles meet forms a rhombus with each side equal to the hypotenuse:
Side length = $8\sqrt{2} \text{ cm}$
Perimeter of the rhombus:
$P = 4 \times 8\sqrt{2} = 32\sqrt{2} \text{ cm}$

Key Concept: Paper Folding and Perimeter Changes

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Original perimeter of the rectangle is $2(6 + 4) = 20 \text{ cm}$. When cut along the length (6 cm), each piece has dimensions $3 \text{ cm} \times 4 \text{ cm}$. Rejoining them side by side forms a new rectangle of dimensions $6 \text{ cm} \times 4 \text{ cm}$, same as original. However, if joined end to end, forming an L-shape, the perimeter becomes $6 + 4 + 3 + 3 + 4 + 6 = 26 \text{ cm}$, which is indeed greater than the original perimeter. The Reason correctly explains why the perimeter increases in this specific arrangement.

Key Concept: Paper folding, perimeter changes

c) 20 cm
[Solution Description]
The original rectangle is cut into two pieces of size $3 \text{ cm} \times 4 \text{ cm}$.
When the longer sides ($4 \text{ cm}$) are joined, the new figure will have dimensions:
Length = $3 \text{ cm} + 3 \text{ cm} = 6 \text{ cm}$
Width = $4 \text{ cm}$
The perimeter of this new rectangle is calculated as:
$P = 2 \times (\text{Length} + \text{Width}) = 2 \times (6 \text{ cm} + 4 \text{ cm}) = 20 \text{ cm}$

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 rectangle, Real-world application

b) 10 meters
[Solution Description]
Let the breadth be $b$ meters. Then the length is $2b$ meters.
Formula for perimeter: $P = 2 \times (length + breadth) = 2 \times (2b + b) = 6b$ meters.
Total cost of fencing is \$300 at \$5 per meter, so the perimeter must be $\frac{300}{5} = 60$ meters.
Set the perimeter equal to 60 meters: $6b = 60$.
Divide both sides by 6: $b = 10$ meters.
So, the breadth of the field is 10 meters.

Key Concept: Perimeter of a rectangle

c) 26 m
[Solution Description]
To find the total distance around the garden, we calculate the perimeter using $2 \times (length + breadth)$. Here, length = 9 m and breadth = 4 m.
$2 \times (9\,m + 4\,m) = 2 \times 13\,m = 26\,m$.
Hence, the total distance around the garden is 26 m.

Key Concept: Perimeter of a square

d) 200 meters
[Solution Description]
First, calculate the perimeter of the square field: $4 \times 25\,\text{m} = 100\,\text{m}$. Since Priya runs around the field twice, the total distance covered is $2 \times 100\,\text{m} = 200\,\text{m}$.

Key Concept: Perimeter of a square

c) 16 cm
[Solution Description]
The perimeter of a square is four times the length of one side. Given the perimeter is 64 cm, we can find the side length by dividing the perimeter by 4. Side length = $64\,\text{cm} \div 4 = 16\,\text{cm}$. So, each side of the square is 16 cm long.

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 general triangle

c) 24 cm
[Solution Description]
The perimeter of a triangle is the sum of all its sides. Given the sides are 6 cm, 8 cm, and 10 cm, we add them together to find the perimeter.
Step 1: Add the first two sides: $6 \text{ cm} + 8 \text{ cm} = 14 \text{ cm}$.
Step 2: Add the third side to the previous result: $14 \text{ cm} + 10 \text{ cm} = 24 \text{ cm}$.
The perimeter of the triangle is 24 cm.

Key Concept: Perimeter of a Triangle, Equilateral Triangle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
First, check Assertion (A): The perimeter of an equilateral triangle with each side 6 cm can be calculated using the formula for the perimeter of an equilateral triangle, which is $3 \times \text{length of one side}$. So, $3 \times 6 = 18$ cm. Thus, Assertion (A) is true.
Next, check Reason (R): The statement in Reason (R) is true because, by definition, the perimeter of any triangle is the sum of its three sides.
Now, evaluate if Reason (R) correctly explains Assertion (A): Although Reason (R) is true, it is not the correct explanation for Assertion (A) since the assertion specifically refers to an equilateral triangle, and the reason is a general property applicable to all triangles. Therefore, the correct option is b).

Key Concept: Perimeter of a Triangle

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 side length 5 cm is 15 cm. Using the formula for the perimeter of an equilateral triangle, we calculate: $Perimeter = 3 \times side \\ length = 3 \times 5 \\ cm = 15 \\ cm$. This matches the Assertion. The Reason correctly explains why the perimeter is 15 cm, as it links to the formula specific to equilateral triangles. Therefore, both Assertion and Reason are true, and Reason provides the correct explanation.

Key Concept: Perimeter of an equilateral triangle

c) 9 cm
[Solution Description]
The perimeter of an equilateral triangle is three times the length of one side. Let the length of one side be $s$. Then,
$\text{Perimeter} = 3 \times s$.
Given that the perimeter is 27 cm,
$3s = 27 \text{ cm}$.
Solving for $s$,
$s = \frac{27 \text{ cm}}{3} = 9 \text{ cm}$.

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: Perimeter of regular polygon

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the perimeter of a regular pentagon, we multiply the number of sides (5) by the length of one side (4 cm). So, $5 \times 4\,\text{cm} = 20\,\text{cm}$. Thus, the Assertion is true. The Reason states a general formula for the perimeter of any regular polygon, which is also correct. The Reason correctly explains why the Assertion holds true because it provides the underlying rule applied in the Assertion.

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 of an equilateral triangle

c) 7 cm
[Solution Description]
We know that the perimeter of an equilateral triangle is given by $3 \times \text{side}$. Let the length of one side be $s$. Given perimeter $P = 21$ cm. Using the formula:
$3 \times s = 21$
To find $s$, divide both sides by 3:
$s = \frac{21}{3} = 7 \text{ cm}$
So, the length of one side is 7 cm.

Key Concept: Perimeter of an equilateral triangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The perimeter of any polygon is the sum of all its sides. An equilateral triangle has three sides that are equal in length by definition. So, if each side has a length $l$, then the total perimeter would be $l + l + l = 3l$. This confirms that the Assertion is true. The Reason states the defining property of an equilateral triangle (all sides are equal), which is also true and correctly explains why the perimeter formula ($3 \times \text{side length}$) holds for an equilateral triangle.

Key Concept: Perimeter after Split and Rejoin

b) 20 cm
[Solution Description]
The original rectangle is split into two smaller rectangles of size 3 cm $\times$ 4 cm. When joined along the 3 cm side, the new figure becomes a rectangle of size 6 cm $\times$ 4 cm again, but with an overlapping side removed. The perimeter is calculated as $2 \times (6 + 4) = 20$ cm.

Key Concept: Perimeter after Split and Rejoin

b) 8 cm
[Solution Description]
The original rectangle is split into two smaller rectangles of size 6 cm $\times$ 2 cm. When joined along the 6 cm side, the new figure measures 6 cm $\times$ 4 cm again but has shared edges. The perimeter is $2 \times (6 + 4) - 2 \times 6 = 8$ cm.

Key Concept: Regular polygons, Perimeter generalization

b) 42 cm
[Solution Description]
A regular hexagon has 6 equal sides. Therefore, the perimeter is simply the sum of all its sides: $6 \times 7 \text{ cm} = 42 \text{ cm}$.

Key Concept: Subtraction of areas

c) 20 sq m
[Solution Description]
First, find the area of the floor:
$6 \, \text{m} \times 4 \, \text{m} = 24 \, \text{sq m}.$
Next, find the area of the carpet:
$2 \, \text{m} \times 2 \, \text{m} = 4 \, \text{sq m}.$
Subtract the carpet's area from the floor's area to find the uncarpeted area:
$24 \, \text{sq m} - 4 \, \text{sq m} = 20 \, \text{sq m}.$

Key Concept: Area of composite shapes

b) 78 sq m
[Solution Description]
First calculate area of the garden: $12 \text{m} \times 8 \text{m} = 96 \text{sq m}$.
Area of one flower bed: $3 \text{m} \times 3 \text{m} = 9 \text{sq m}$.
Total area of two flower beds: $9 \text{sq m} \times 2 = 18 \text{sq m}$.
Remaining area: $96 \text{sq m} - 18 \text{sq m} = 78 \text{sq m}$.

Key Concept: Area of a square

b) 25 sq m
[Solution Description]
The formula for the area of a square is $\text{side} \times \text{side}$. Given the side length is 5 m, the area is calculated as:
$5 \, \text{m} \times 5 \, \text{m} = 25 \, \text{sq m}$

Key Concept: Area of a Square

c) 25 square centimeters
[Solution Description]
The area of a square is calculated by multiplying the length of its side by itself. Here, the side length is given as 5 cm. So, the area is $5 \times 5 = 25$ square centimeters.

Key Concept: Area of Square

c) 49 square cm
[Solution Description]
The area of a square is calculated using the formula $Area = side \times side$. Here, the side length is 7 cm.
Step 1: Multiply the side by itself.
$Area = 7 \times 7 = 49$ square cm.
Thus, the area of the square is 49 square cm.

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: 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: 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 rectangle

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description] To check the assertion, first calculate the area of the rectangle: $length \times width = 8 m \times 5 m = 40 sq m$. Next, calculate the area of the square: $side \times side = 7 m \times 7 m = 49 sq m$. Since 40 sq m < 49 sq m, the Assertion is true. The Reason correctly states the formulas for the area of a rectangle and a square but does not explain why the rectangle's area is less than the square's. Therefore, both are true, but the Reason does not correctly explain the Assertion.

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

c) 8 cm
[Solution Description]
The area of a square is given by $\text{Area} = \text{side}^2$. We are given that the area is 64 square centimeters. To find the side length, we take the square root of the area:
$\text{Side} = \sqrt{64} \, \text{cm} = 8 \, \text{cm}$.

Key Concept: Side Length from Area

c) 7 m
[Solution Description]
To find the side length of a square when the area is known, we take the square root of the area. Here, the area is given as 49 sq m.
Step 1: Given the area = 49 sq m
Step 2: Use the formula relating area and side length:
$\text{Area} = \text{side} \times \text{side}$
Step 3: Rearrange to find the side length:
$\text{side} = \sqrt{\text{Area}} = \sqrt{49 \, \text{sq m}} = 7 \, \text{m}$
Thus, the length of each side of the garden is 7 meters.

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 estimation using grid paper

c) 13 sq units
[Solution Description]
Add the area contributed by full squares and half-covered squares. Full squares give 12 sq units, and half-covered squares contribute $2 \times \frac{1}{2} = 1$ sq unit. Total estimated area is $12 + 1 = 13$ sq units.

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 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 estimation using grid paper

b) 20 sq units
[Solution Description]
Full squares: 12 $\times$ 1 = 12 sq units.
More than half squares: 6 $\times$ 1 = 6 sq units.
Exactly half squares: 4 $\times$ $\frac{1}{2}$ = 2 sq units.
Total area = 12 + 6 + 2 = 20 sq units.

Key Concept: Area using Grid Paper

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the area of a triangle drawn on grid paper can be exactly calculated by counting the number of full squares it covers. However, this is not entirely accurate because the area calculation on grid paper follows specific conventions: only full squares or more-than-half squares are counted as 1 square unit, while less-than-half squares are ignored. Exact halves are counted as $\frac{1}{2}$ square unit. Thus, the assertion is false.
The reason states that a triangle's area is always half the area of a rectangle that shares the same base and height. This is true for any triangle, as derived from the formula for the area of a triangle ($\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$), which confirms that the area is indeed half that of a corresponding rectangle with the same dimensions.
Therefore, the assertion is false, but the reason is true.

Key Concept: Counting Unit Squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a rectangle with dimensions 4 units by 6 units has an area of 24 square units. This is correct because the area of a rectangle is calculated as length multiplied by width, which in this case gives $4 \times 6 = 24$ square units. The reason explains that the area of any rectangle can be found by counting the number of unit squares fitting inside it. This is also true and directly explains why the assertion is correct.
Both the assertion and reason are individually true, and the reason provides a correct explanation for the assertion.

Key Concept: Counting unit squares

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To find the area of the rectangle, multiply its length by its width: $6 \times 4 = 24$ square units. This matches the assertion. The reason states that counting unit squares gives an accurate measurement for rectangles, which is true because rectangles can be perfectly tiled with unit squares without gaps or overlaps. The reason correctly explains the assertion because the formula for the area of a rectangle is derived from counting these unit squares.

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: Area estimation rules, Counting squares on graph paper

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion claims that the area of a circle drawn on a graph sheet can be accurately determined by counting full squares only. However, this is false because circles have curved boundaries, and relying solely on full squares leads to significant inaccuracies.
The reason states that circles cannot be packed without gaps, making exact area measurement using squares impossible. This is true because squares cannot perfectly cover the curved region of a circle due to their straight edges, resulting in either overcounting or undercounting areas.
Therefore, the assertion is false, but the reason is true.

Key Concept: Area Calculation

d) 71 sq m
[Solution Description]
Area of the field = $length \times width$ = $12 \, \text{m} \times 8 \, \text{m} = 96 \, \text{sq m}$
Area of the square garden = $side \times side$ = $5 \, \text{m} \times 5 \, \text{m} = 25 \, \text{sq m}$
Remaining area = Area of field - Area of garden = 96 - 25 = 71 sq m

Key Concept: Unit Square Area

b) 1 sq unit
[Solution Description]
The area of a square is calculated by multiplying its length and width. Here, both are 1 unit. Thus, area = $1 \text{ unit} \times 1 \text{ unit} = 1 \text{ square unit}$.

Key Concept: Basic Concept

c) 8 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 = 20 square units, height = 5 units.
Step 1: Rearrange the formula to solve for base:
$\text{base} = \frac{2 \times \text{Area}}{\text{height}}$.
Step 2: Substitute the given values:
$\text{base} = \frac{2 \times 20}{5} = \frac{40}{5} = 8$ units.
Therefore, the base of the triangle is 8 units.

Key Concept: Area of Triangle formed by Diagonal, Division into Rectangles

b) 9 cm$^2$
[Solution Description]
First, find the area of the rectangle: $Area = 12 \text{ cm} \times 6 \text{ cm} = 72 \text{ cm}^2$. The diagonal divides it into two equal triangles, each with area $\frac{1}{2} \times 72 \text{ cm}^2 = 36 \text{ cm}^2$. Now, the new triangle is formed by joining the midpoints of two adjacent sides (say 6 cm and 12 cm sides) and one vertex. This creates a smaller triangle within one of the larger triangles. Its base is half of 12 cm (6 cm), and height is half of 6 cm (3 cm). So, its area is $\frac{1}{2} \times 6 \text{ cm} \times 3 \text{ cm} = 9 \text{ cm}^2$.

Key Concept: Comparing Areas

c) The triangle's area is half the rectangle's area.
[Solution Description]
The yellow triangle is formed by the diagonal of the rectangle, so its area is exactly half of the rectangle's area. Thus, $Area_{triangle} = \frac{1}{2} \times 48 = 24$ square units.

Key Concept: Area comparison between rectangle and triangle, Non-uniform subdivision

d) 40 sq. units
[Solution Description]
First, calculate the total area of rectangle ABCD: $AB = AF + FB = 4 + 6 = 10$ units. Total area $= AB \times AD = 10 \times 8 = 80$ sq. units.
Next, divide rectangle ABCD into AFED and BFEC. Areas are: AFED $= 4 \times 8 = 32$ sq. units, BFEC $= 6 \times 8 = 48$ sq. units.
Triangle ABE includes parts from both small rectangles. For any point E on DC, the area of triangle ABE will be $\frac{1}{2}$ of AFED plus $\frac{1}{2}$ of BFEC because it spans the entire height.
So, $\text{Area of triangle ABE} = \frac{32}{2} + \frac{48}{2} = 16 + 24 = 40$ sq. units.

Key Concept: Area of a Triangle formed by Diagonal of 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 the diagonal of a rectangle is half the area of the rectangle. This is mathematically correct because the formula for the area of such a triangle is $\frac{1}{2} \times \text{length} \times \text{breadth}$, which is indeed half the area of the rectangle ($\text{length} \times \text{breadth}$).
The reason given is that the diagonal divides the rectangle into two congruent triangles with the same base and height as the rectangle. This is also true because both triangles share the diagonal as their common base and have heights equal to the other dimension of the rectangle.
Furthermore, the reason correctly explains the assertion since congruent triangles imply equal areas, and the area of each triangle being half the rectangle's area is directly derived from this property.

Key Concept: Area of Triangle via Diagonal of Rectangle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The syllabus specifies that when a diagonal is drawn in a rectangle, it divides the rectangle into two triangles. These two triangles are congruent (exactly overlap each other if cut and placed one over the other). Since both triangles are identical in shape and size, their areas must be equal. The total area of the rectangle is the sum of the areas of these two triangles. Therefore, the area of each triangle is exactly half of the area of the rectangle.
The assertion states that the area of a triangle formed by the diagonal of a rectangle is half the area of the rectangle, which is correct based on the syllabus. The reason provided is also correct because a diagonal indeed divides the rectangle into two congruent triangles with equal areas. Moreover, the reason correctly explains why the assertion is true, as the equality of areas arises from the congruence of the triangles created by the diagonal.

Key Concept: Grid Paper Calculation

c) 40 square units
[Solution Description]
The area of triangle BAD is half the area of rectangle ABCD since it is formed by one side of the rectangle and its diagonal.
Given:
$\text{Area of Rectangle ABCD} = 80\,\text{square units}$
Thus:
$\text{Area of Triangle BAD} = \frac{1}{2} \times 80\,\text{square units} = 40\,\text{square units}$
The area of triangle BAD is 40 square units.

Key Concept: Area of a Triangle

c) 20 square cm
[Solution Description]
First, find the area of the rectangle:
$8 \, \text{cm} \times 5 \, \text{cm} = 40 \, \text{square cm}$
Then, the area of the triangle formed by the diagonal is half of this:
$\frac{1}{2} \times 40 = 20 \, \text{square cm}$

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: Rectangle to Triangle

c) 18 cm\textsuperscript{2}
[Solution Description]
The blue rectangle has an area of 36 cm\textsuperscript{2}. The yellow triangle is formed by taking half of the rectangle, so its area is half of the rectangle's area.
Calculation: Area of yellow triangle = $\frac{1}{2} \times 36 = 18 \text{ cm}^2$.

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: Relationship Between Rectangle and Triangle Areas

b) 12 square units
[Solution Description]
Since the area of each triangle formed by the diagonal of a rectangle is half the area of the rectangle, we calculate it as follows:
$\text{Area of triangle} = \frac{1}{2} \times \text{Area of rectangle} = \frac{1}{2} \times 24 = 12 \text{ square units}.$
Therefore, the area of the triangle is 12 square units.

Key Concept: Composite Triangle Area

c) 18 cm$^2$
[Solution Description]
Calculate the area of triangle AEF as half of rectangle AFED: $\frac{1}{2} \times 20 \text{ cm}^2 = 10 \text{ cm}^2$. Similarly, the area of triangle BEF is half of rectangle BFEC: $\frac{1}{2} \times 16 \text{ cm}^2 = 8 \text{ cm}^2$. Add these to find the area of triangle ABE: $10 \text{ cm}^2 + 8 \text{ cm}^2 = 18 \text{ cm}^2$.

Key Concept: Variable position point, proportional areas

c) 3.6 units
[Solution Description]
Area of rectangle = $8 \times 6 = 48$ square units.
Area of triangle PQT = $\frac{1}{2} \times PQ \times ST = \frac{1}{2} \times 8 \times y = 4y$.
Area of triangle QRT = $\frac{1}{2} \times QR \times RT = \frac{1}{2} \times 6 \times (6-y) = 3(6-y) = 18-3y$.
Given condition: $4y = 2 \times (18-3y)$
Solving: $4y = 36-6y$ → $10y = 36$ → $y = 3.6$ units

Key Concept: Tangram Area Relationships

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that Shape D has twice the area of Shape C because it can be covered using Shapes C and E. From the syllabus hint, we know that Shapes C and E have the same area, and Shape D can be exactly covered using both C and E. This confirms that Shape D's area is indeed the sum of the areas of Shapes C and E, which means it is twice the area of Shape C. The reason explains that the area relationships in a tangram are determined by how shapes can be combined or overlaid. Since the assertion correctly describes the relationship based on this principle, and the reason provides the correct explanation for the assertion, both are true, and the reason correctly explains the assertion.

Key Concept: Perimeter Comparison

c) 172 sq m
[Solution Description]
Calculate the area of the garden: $15 \text{ m} \times 12 \text{ m} = 180 \text{ sq m}$. Area of one small rectangle: $2 \text{ m} \times 1 \text{ m} = 2 \text{ sq m}$. Total area occupied by four such rectangles: $4 \times 2 = 8 \text{ sq m}$. Remaining area for lawn: $180 - 8 = 172 \text{ sq m}$.

Key Concept: Tangram Area Comparison

b) 4 times the area of Shape C
[Solution Description]
Given that Shape D has twice the area of Shape C, and Shape E has the same area as Shape C. Let the area of Shape C be $x$. Then, the area of Shape D is $2x$ and Shape E is $x$. The total area of Shapes C, D, and E combined is $x + 2x + x = 4x$.

Key Concept: Tangram shapes and comparison

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the given syllabus, it is mentioned that Shape D can be exactly covered using Shapes C and E, meaning the area of Shape D is equal to the combined area of Shapes C and E. Since Shapes C and E have the same area, Shape D must have twice the area of Shape C. Therefore, the Assertion is true. The Reason correctly explains why the Assertion is true, as it provides the logical basis for the area relationship between Shapes D, C, and E.

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 Relationships in Tangram

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the syllabus, Shape D can be completely covered by placing Shapes C and E on it. The areas of Shapes C and E are equal as per the syllabus hint. Therefore, if $\text{Area of C} = x$, then $\text{Area of E} = x$. Since Shape D is covered by both C and E, its area must be $x + x = 2x$. Thus, the area of Shape D is indeed twice that of Shape C or E. The reason correctly explains why the assertion is true.

Key Concept: Tangram Area Comparison

b) 2 square units
[Solution Description]
According to the hint provided, Shape D can be exactly covered using Shapes C and E, each having an area of 1 square unit. Therefore, the area of Shape D is twice the area of Shape C. So, the area of Shape 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.