Chapter 6 Perimeter and Area

a. Given: Perimeter of a rectangle $= 14$ cm, breadth $(b) = 2$ cm.

To Find: Length $(l)$.

We know that the perimeter of a rectangle is given by the formula:

Substitute the given values into equation (i):

$14 = 2(l + 2)$

Divide both sides by 2:

$\frac{14}{2} = l + 2$

$7 = l + 2$

Subtract 2 from both sides:

$l = 7 - 2$

$l = 5$

Thus, the length of the rectangle is 5 cm.

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b. Given: Perimeter of a square $= 20$ cm.

To Find: Side $(s)$ of the square.

We know that the perimeter of a square is given by the formula:

Substitute the given value into equation (ii):

$20 = 4s$

Divide both sides by 4:

$s = \frac{20}{4}$

$s = 5$

Thus, the side of the square is 5 cm.

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c. Given: Perimeter of a rectangle $= 12$ m, length $(l) = 3$ m.

To Find: Breadth $(b)$ of the rectangle.

We use the formula for the perimeter of a rectangle (equation (i)):

Substitute the given values into equation (i):

$12 = 2(3 + b)$

Divide both sides by 2:

$\frac{12}{2} = 3 + b$

$6 = 3 + b$

Subtract 3 from both sides:

$b = 6 - 3$

$b = 3$

Thus, the breadth of the rectangle is 3 m.

Given:

Length of the rectangle $(l) = 5$ cm.

Breadth of the rectangle $(b) = 3$ cm.

The wire is used to form this rectangle and then a square.

To Find:

The length of a side of the square.

Solution:

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

The perimeter of a rectangle is given by the formula:

Substitute the given values into equation (i):

Perimeter of rectangle $= 2(5 \$cm + 3 \$cm)$

Perimeter of rectangle $= 2(8 \$cm)$

Perimeter of rectangle $= 16 \$cm$

So, the length of the wire is $16$ cm.

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When this wire is bent to form a square, the length of the wire becomes the perimeter of the square.

Perimeter of the square $= 16$ cm.

The perimeter of a square is given by the formula:

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

Substitute the perimeter value into equation (ii):

$16 = 4s$

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

$s = \frac{16}{4}$

$s = 4$

The length of a side of the square is 4 cm.

Given:

Perimeter of the triangle $= 55$ cm.

Length of the first side $(a) = 20$ cm.

Length of the second side $(b) = 14$ cm.

To Find:

The length of the third side $(c)$.

Solution:

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

Substitute the given values into equation (i):

$55 = 20 \$cm + 14 \$cm + c$

$55 = 34 \$cm + c$

To find the length of the third side $c$, subtract the sum of the other two sides from the perimeter:

$c = 55 - 34$

$c = 21$

The length of the third side of the triangle is 21 cm.

Given:

Length of the rectangular park $(l) = 150$ m.

Breadth of the rectangular park $(b) = 120$ m.

Cost of fencing per metre $= \textsf{₹}40$.

To Find:

The total cost of fencing the rectangular park.

Solution:

Fencing is done along the boundary of the park, which means we need to calculate the perimeter of the rectangle.

The perimeter of a rectangle is given by the formula:

Substitute the given length and breadth into equation (i):

Perimeter $= 2(150 \$\text{m} + 120 \$\text{m})$

Perimeter $= 2(270 \$\text{m})$

Perimeter $= 540 \$\text{m}$

The length of the fence required is 540 metres.

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The cost of fencing is $\textsf{₹}40$ per metre.

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

Total cost $= 540 \$\text{m} \times \textsf{₹}40/\text{m}$

Total cost $= 540 \times 40 \$\textsf{₹}$

Total cost $= 21600 \$\textsf{₹}$

The total cost of fencing the rectangular park is $\textsf{₹}21600$.

Given:

Length of the string $= 36$ cm.

This length represents the perimeter of the figure formed.

To Find:

The length of each side for different shapes.

Solution:

The perimeter of any polygon is the sum of the lengths of its sides. If the sides are of equal length, the perimeter is the number of sides multiplied by the length of one side.

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a. If the string is used to form a square:

A square has 4 sides of equal length.

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

The perimeter of a square is given by:

Substitute the perimeter (length of the string) into equation (i):

$36 = 4s$

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

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

$s = 9$

The length of each side of the square is 9 cm.

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b. If the string is used to form a triangle with all sides of equal length (equilateral triangle):

An equilateral triangle has 3 sides of equal length.

Let the side length of the equilateral triangle be $s$ cm.

The perimeter of an equilateral triangle is given by:

Substitute the perimeter (length of the string) into equation (ii):

$36 = 3s$

To find the side length $s$, divide the perimeter by 3:

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

$s = 12$

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

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c. If the string is used to form a hexagon with sides of equal length (regular hexagon):

A hexagon has 6 sides.

A regular hexagon has 6 sides of equal length.

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

The perimeter of a regular hexagon is given by:

Substitute the perimeter (length of the string) into equation (iii):

$36 = 6s$

To find the side length $s$, divide the perimeter by 6:

$s = \frac{36}{6}$

$s = 6$

The length of each side of the regular hexagon is 6 cm.

Given:

Length of the rectangular field $(l) = 230$ m.

Breadth of the rectangular field $(b) = 160$ m.

Number of rounds of fencing $= 3$.

To Find:

The total length of rope needed.

Solution:

The length of rope needed for one round of fencing is equal to the perimeter of the rectangular field.

The perimeter of a rectangle is given by the formula:

Substitute the given length and breadth into equation (i):

Perimeter $= 2(230 \$\text{m} + 160 \$\text{m})$

Perimeter $= 2(390 \$\text{m})$

Perimeter $= 780 \$\text{m}$

The length of rope needed for one round is 780 metres.

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The farmer wants to fence the field with 3 rounds of rope.

Total length of rope needed $=$ Number of rounds $\times$ Perimeter for one round

Total length of rope $= 3 \times 780 \$\text{m}$

Total length of rope $= 2340 \$\text{m}$

The total length of rope needed is 2340 m.

Given:

Length of the rectangular field $(l) = 20$ m.

Breadth of the rectangular field $(b) = 15$ m.

Number of rounds Akshi covered $= 5$.

To Find:

The total distance covered by Akshi in 5 rounds.

Solution:

The distance covered in one round is equal to the perimeter of the rectangular field.

The perimeter of a rectangle is given by the formula:

Substitute the given length and breadth into equation (i):

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

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

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

The distance covered in one round is 70 metres.

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To find the total distance covered in 5 rounds, multiply the distance of one round by the number of rounds.

Total distance covered $=$ Distance in one round $\times$ Number of rounds

Total distance covered $= 70 \$\text{m} \times 5$

Total distance covered $= 350 \$\text{m}$

The total distance Akshi has covered in 5 rounds is 350 m.

Given:

Length of the rectangular field $(l) = 20$ m.

Breadth of the rectangular field $(b) = 15$ m.

Number of rounds Toshi covered $= 7$.

Distance covered by Akshi in 5 rounds $= 350$ m (calculated in the previous question).

To Find:

The total distance covered by Toshi in 7 rounds.

Who ran a longer distance?

Solution:

The distance covered in one round is equal to the perimeter of the rectangular field.

The perimeter of a rectangle is given by the formula:

Substitute the given length and breadth into equation (i):

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

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

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

The distance covered in one round (perimeter) is 70 metres.

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To find the total distance covered by Toshi in 7 rounds, multiply the distance of one round by the number of rounds.

Total distance covered by Toshi $=$ Distance in one round $\times$ Number of rounds

Total distance covered by Toshi $= 70 \$\text{m} \times 7$

Total distance covered by Toshi $= 490 \$\text{m}$

The total distance Toshi has covered in 7 rounds is 490 m.

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Comparing the distances covered by Akshi and Toshi:

Distance covered by Akshi $= 350$ m

Distance covered by Toshi $= 490$ m

Since $490 \$\text{m} > 350 \$\text{m}$, Toshi ran a longer distance.

Toshi ran a longer distance.

Given:

Length of the rectangular track $(l) = 20$ m.

Breadth of the rectangular track $(b) = 15$ m.

Perimeter of the track (distance in one round) is given by:

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

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

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

One full round around the track is 70 m.

To determine the position after a certain distance, we find the number of full rounds completed and the remaining distance. Let's assume the runner starts at one corner and runs along the length side first.

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a. Akshi ran 250 m.

To find the number of full rounds, we divide the total distance by the perimeter:

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

Number of rounds $= \frac{250 \$\text{m}}{70 \$\text{m}}$

$250 = 3 \times 70 + 40$

Akshi completes 3 full rounds and has a remaining distance of 40 m.

After 3 full rounds, she is back at the starting point.

The remaining 40 m is covered as follows:

Along the first length: 20 m (Remaining distance $= 40 - 20 = 20$ m)

Along the first breadth: 15 m (Remaining distance $= 20 - 15 = 5$ m)

Along the second length: 5 m.

So, Akshi is on the second length side (opposite to the starting length), 5 m from the corner she just passed (after completing the first breadth).

Position 'A': On the top side, 5 m away from the corner reached after running one length and one breadth from the start.

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b. Akshi ran 500 m.

Number of rounds $= \frac{500 \$\text{m}}{70 \$\text{m}}$

$500 = 7 \times 70 + 10$

Akshi completes 7 full rounds and has a remaining distance of 10 m.

After 7 full rounds, she is back at the starting point.

The remaining 10 m is covered as follows:

Along the first length: 10 m.

So, Akshi is on the first length side (the starting side), 10 m from the starting corner.

Position 'B': On the bottom side, 10 m away from the starting corner.

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c. Akshi ran 1000 m.

Number of rounds $= \frac{1000 \$\text{m}}{70 \$\text{m}}$

$1000 = 14 \times 70 + 20$

Akshi completes 14 full rounds and has a remaining distance of 20 m.

Number of full rounds finished: 14 rounds.

After 14 full rounds, she is back at the starting point.

The remaining 20 m is covered as follows:

Along the first length: 20 m.

So, Akshi is at the end of the first length side.

Position 'C': At the corner reached after running one length from the starting corner.

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d. Toshi ran 250 m.

This is the same distance as Akshi in part (a).

$250 = 3 \times 70 + 40$

Toshi completes 3 full rounds and has a remaining distance of 40 m.

After 3 full rounds, she is back at the starting point.

The remaining 40 m is covered as follows: 20 m along the first length, 15 m along the first breadth, and 5 m along the second length.

Position 'X': On the top side, 5 m away from the corner reached after running one length and one breadth from the start.

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e. Toshi ran 500 m.

This is the same distance as Akshi in part (b).

$500 = 7 \times 70 + 10$

Toshi completes 7 full rounds and has a remaining distance of 10 m.

After 7 full rounds, she is back at the starting point.

The remaining 10 m is covered along the first length.

Position 'Y': On the bottom side, 10 m away from the starting corner.

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f. Toshi ran 1000 m.

This is the same distance as Akshi in part (c).

$1000 = 14 \times 70 + 20$

Toshi completes 14 full rounds and has a remaining distance of 20 m.

Number of full rounds finished: 14 rounds.

After 14 full rounds, she is back at the starting point.

The remaining 20 m is covered along the first length.

Position 'Z': At the corner reached after running one length from the starting corner.

Given:

Akshi says the perimeter of the triangle shape is 9 units.

Toshi says the perimeter will be more than 9 units.

To Determine:

Who is correct (Akshi or Toshi).

Solution:

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

When a figure is drawn on a grid or dotted paper, the lengths of horizontal and vertical segments can be found by counting the grid units or the distance between adjacent dots.

However, the length of a diagonal segment on a grid is calculated using the Pythagorean theorem, based on the horizontal and vertical distances it covers.

For example, if a diagonal segment goes from point (0,0) to point (a,b) on a grid, its actual length is $\sqrt{a^2 + b^2}$. This length is always greater than or equal to the larger of $|a|$ and $|b|$, and strictly greater than $a+b$ (unless $a$ or $b$ is zero, meaning it's a horizontal or vertical line).

If the triangle shape in the image includes any diagonal sides, their lengths calculated correctly using the distance formula or Pythagorean theorem will typically result in values that are not simple integers and are greater than a casual count along the grid lines might suggest.

It is very likely that Akshi's calculation of 9 units involves incorrectly adding up lengths, possibly by counting diagonal steps on the grid as single units or by making a visual estimate that doesn't account for the true length of diagonal segments.

Toshi's statement that the perimeter is more than 9 units suggests that the actual calculated lengths of the sides, when summed up, exceed 9 units. This is common for triangles with diagonal sides on a grid, where the calculated perimeter will be $\geq 9$ units.

Given that Toshi says the perimeter will be more than 9 units, and assuming the triangle's dimensions on the grid lead to side lengths (especially diagonal ones) whose sum exceeds 9 when calculated correctly, Toshi is likely correct.

The perimeter of a triangle formed by points on a grid with diagonal sides is generally greater than an estimate based on simply counting grid segments or "blocks" without using the distance formula.

Therefore, Toshi is correct.

What I think: Toshi is correct. The perimeter of the triangle, calculated by finding the true lengths of its sides (using the distance formula for diagonal sides), will be greater than 9 units if the diagram implies sides leading to such a value.

Given:

Inner square track side length $= 100$ m.

Outer square track side length $= 150$ m.

Common finishing line: Midpoint of one side (let's assume the bottom side).

Total race distance $= 350$ m.

Assume runners run counter-clockwise (standard track direction).

To Find:

The starting positions 'A' (inner track) and 'B' (outer track) such that running 350 m counter-clockwise reaches the finish line.

Solution:

First, calculate the perimeter of each track.

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

Perimeter of inner track $= 4 \times 100 \$\text{m} = 400 \$\text{m}$.

Perimeter of outer track $= 4 \times 150 \$\text{m} = 600 \$\text{m}$.

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A runner starting at point S and running a distance D counter-clockwise to reach a finish line F means that the distance along the track from S to F in the counter-clockwise direction is D.

Alternatively, the distance along the track from F to S in the counter-clockwise direction is (Perimeter - D).

Let's find the starting position for the runner on the inner track ('A').

Perimeter of inner track $= 400$ m.

Distance to run $= 350$ m.

The starting point A is $(400 \$\text{m} - 350 \$\text{m}) = 50 \$\text{m}$ counter-clockwise from the finish line.

Assume the finish line is at the midpoint of the bottom side. The distance from the midpoint of the bottom side to the bottom-right corner, moving counter-clockwise, is half the side length, which is $\frac{100}{2} = 50$ m.

Tracing 50 m counter-clockwise from the midpoint of the bottom side leads directly to the bottom-right corner.

Therefore, the starting position 'A' for the inner track is the bottom-right corner.

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Let's find the starting position for the runner on the outer track ('B').

Perimeter of outer track $= 600$ m.

Distance to run $= 350$ m.

The starting point B is $(600 \$\text{m} - 350 \$\text{m}) = 250 \$\text{m}$ counter-clockwise from the finish line.

Assume the finish line is at the midpoint of the bottom side. Side length is 150 m. Midpoint is $\frac{150}{2} = 75$ m from each bottom corner.

Tracing 250 m counter-clockwise from the midpoint of the bottom side:

Distance from mid-bottom to bottom-right corner (CCW) $= 75$ m. (Remaining distance $= 250 - 75 = 175$ m)

Distance from bottom-right corner to top-right corner (CCW) $= 150$ m. (Remaining distance $= 175 - 150 = 25$ m)

The remaining 25 m is along the top side, starting from the top-right corner and moving towards the top-left corner (CCW direction).

Therefore, the starting position 'B' for the outer track is on the top side, 25 m away from the top-right corner (or 125 m away from the top-left corner).

To find the perimeter of the figures in terms of straight and diagonal units, we count the number of segments along the boundary. A straight unit corresponds to a horizontal or vertical segment connecting adjacent grid points. A diagonal unit corresponds to a diagonal segment connecting adjacent grid points (i.e., a segment with a horizontal displacement of 1 and a vertical displacement of 1).

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Figure 1:

This figure is a rectangle.

We count the segments along its perimeter.

There are 3 straight units along the top side and 3 straight units along the bottom side.

There are 2 straight units along the left side and 2 straight units along the right side.

Total straight units $= 3 + 2 + 3 + 2 = 10$.

There are no diagonal segments.

Total diagonal units $= 0$.

Perimeter of Figure 1: 10 straight units, 0 diagonal units.

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Figure 2:

We count the segments along its perimeter.

There is 1 straight unit on the top, 1 on the right, 1 on the bottom, and 1 on the left.

Total straight units $= 1 + 1 + 1 + 1 = 4$.

There is 1 diagonal unit on the top-right, 1 on the bottom-right, 1 on the bottom-left, and 1 on the top-left.

Total diagonal units $= 1 + 1 + 1 + 1 = 4$.

Perimeter of Figure 2: 4 straight units, 4 diagonal units.

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Figure 3:

This figure is an octagon.

We count the segments along its perimeter.

There is 1 straight (horizontal) unit on the top, 1 straight (vertical) unit on the right, 1 straight (horizontal) unit on the bottom, and 1 straight (vertical) unit on the left.

Total straight units $= 1 + 1 + 1 + 1 = 4$.

There is 1 diagonal unit on the top-right, 1 on the bottom-right, 1 on the bottom-left, and 1 on the top-left.

Total diagonal units $= 1 + 1 + 1 + 1 = 4$.

Perimeter of Figure 3: 4 straight units, 4 diagonal units.

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Figure 4:

This figure is a rectangle.

We count the segments along its perimeter.

There are 3 straight units along the top side and 3 straight units along the bottom side.

There are 2 straight units along the left side and 2 straight units along the right side.

Total straight units $= 3 + 2 + 3 + 2 = 10$.

There are no diagonal segments.

Total diagonal units $= 0$.

Perimeter of Figure 4: 10 straight units, 0 diagonal units.

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Figure 5:

This figure is a cross shape.

We count the segments along its perimeter.

All segments along the boundary are horizontal or vertical unit lengths.

Counting the segments: There are 4 segments forming the ends of the arms (1 on top, 1 on right, 1 on bottom, 1 on left). There are 8 segments forming the inner corners (2 segments at each of the four inner corners).

Total straight units $= 4 + 8 = 12$.

There are no diagonal segments.

Total diagonal units $= 0$.

Perimeter of Figure 5: 12 straight units, 0 diagonal units.

A regular polygon is a polygon where all sides are of equal length and all interior angles are equal.

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Examples of objects with regular shapes from our surroundings and finding their perimeters:

1. Shape: Square tile or a square photograph frame.

A square is a regular polygon with 4 equal sides.

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

Perimeter of the square $= \text{sum of the lengths of its 4 sides}$.

Since all sides are equal, Perimeter $= s + s + s + s = 4 \times s$.

If the side length is 15 cm, the perimeter is $4 \times 15 \$\text{cm} = 60$ cm.

2. Shape: Traffic yield sign (typically an equilateral triangle).

An equilateral triangle is a regular polygon with 3 equal sides.

Let the side length of the equilateral triangle be $s$.

Perimeter of the equilateral triangle $= \text{sum of the lengths of its 3 sides}$.

Since all sides are equal, Perimeter $= s + s + s = 3 \times s$.

If the side length is 30 cm, the perimeter is $3 \times 30 \$\text{cm} = 90$ cm.

3. Shape: Stop sign (typically a regular octagon).

A regular octagon is a regular polygon with 8 equal sides.

Let the side length of the regular octagon be $s$.

Perimeter of the regular octagon $= \text{sum of the lengths of its 8 sides}$.

Since all sides are equal, Perimeter $= s + s + s + s + s + s + s + s = 8 \times s$.

If the side length is 25 cm, the perimeter is $8 \times 25 \$\text{cm} = 200$ cm.

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Generalisation for the perimeter of other regular polygons:

Based on the examples, we observe a pattern:

For a square (4 sides), Perimeter $= 4 \times \text{side length}$.

For an equilateral triangle (3 sides), Perimeter $= 3 \times \text{side length}$.

For a regular octagon (8 sides), Perimeter $= 8 \times \text{side length}$.

This pattern holds true for any regular polygon.

If a regular polygon has $n$ sides, and each side has a length $s$, then the perimeter is the sum of the lengths of these $n$ equal sides.

Perimeter $= s + s + ... + s$ ($n$ times)

Perimeter $= n \times s$

So, the general formula for the perimeter of a regular polygon with $n$ sides, each of length $s$, is:

Here, $n$ is the number of sides and $s$ is the length of each side.

Given:

Original rectangular paper chit dimension = 6 cm $\times$ 4 cm.

The chit is cut into two equal pieces as shown.

These two pieces are joined in different ways (arrangements a, b, c, d).

The perimeter of arrangement a is given as 28 cm.

To Find:

The perimeter of arrangements b, c, and d.

Solution:

The original rectangle has a length of 6 cm and a breadth of 4 cm. Its perimeter is $2(6+4) = 2(10) = 20$ cm.

The cut shown in the first image divides the 6 cm $\times$ 4 cm rectangle into two equal pieces. The cut is a zig-zag line across the rectangle.

When pieces are cut and then rejoined, the perimeter of the new shape is the sum of the perimeters of the individual pieces minus twice the length of the boundary along which they are joined.

Since the two pieces are equal, their perimeters are the same. Let $P_{piece}$ be the perimeter of one piece.

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We are given that the perimeter of arrangement a is 28 cm.

Looking at the image for arrangements (6.jpg), arrangement a appears to be formed by joining the two pieces side-by-side along their shorter straight edges. The original rectangle is 6 cm $\times$ 4 cm. The cut shown in the first image appears to divide the 4 cm width into two parts, resulting in pieces that are roughly 6 cm long and 2 cm wide, but with a zig-zag cut edge.

If we assume the pieces are joined along a straight edge corresponding to the 2 cm dimension, the joined boundary length is 2 cm.

Using equation (ii) for arrangement a:

$28 = 2 \times P_{piece} - 2 \times 2$

$28 = 2 \times P_{piece} - 4$

$28 + 4 = 2 \times P_{piece}$

$32 = 2 \times P_{piece}$

$P_{piece} = \frac{32}{2} = 16$ cm.

So, the perimeter of each piece is 16 cm.

This implies that the boundaries of each piece consist of some straight edges and the zig-zag cut edge, and their total length is 16 cm.

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Now we can find the perimeters of the other arrangements using $P_{piece} = 16$ cm and the length of the joined boundary.

b. Arrangement b:

Looking at the image, arrangement b appears to be formed by joining the two pieces along their shorter straight edges (corresponding to the 2 cm dimension), similar to arrangement a, but possibly with an offset.

Length of the joined boundary $= 2$ cm.

Using equation (ii):

Perimeter(b) $= 2 \times P_{piece} - 2 \times \text{Length(Joined Boundary)}$

Perimeter(b) $= 2 \times 16 \text{ cm} - 2 \times 2 \text{ cm}$

Perimeter(b) $= 32 \text{ cm} - 4 \text{ cm}$

Perimeter(b) $= 28$ cm.

The perimeter of arrangement b is 28 cm.

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c. Arrangement c:

Looking at the image, arrangement c appears to be formed by joining the two pieces along their longer straight edges (corresponding to the 6 cm dimension).

Length of the joined boundary $= 6$ cm.

Using equation (ii):

Perimeter(c) $= 2 \times P_{piece} - 2 \times \text{Length(Joined Boundary)}$

Perimeter(c) $= 2 \times 16 \text{ cm} - 2 \times 6 \text{ cm}$

Perimeter(c) $= 32 \text{ cm} - 12 \text{ cm}$

Perimeter(c) $= 20$ cm.

This arrangement forms a rectangle of 6 cm $\times$ 4 cm, which is the original shape, and its perimeter is indeed 20 cm. This supports our calculated piece perimeter of 16 cm and the interpretation of joined edge lengths.

The perimeter of arrangement c is 20 cm.

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d. Arrangement d:

Looking at the image, arrangement d appears to be formed by joining the two pieces to create an L-shape. This is typically done by joining a shorter straight edge of one piece to a part of the longer straight edge of the other piece. The length of the joined boundary is limited by the shorter edge.

Assuming the pieces are joined along a straight edge corresponding to the 2 cm dimension (joined to a 6 cm dimension edge), the length of the joined boundary $= 2$ cm.

Using equation (ii):

Perimeter(d) $= 2 \times P_{piece} - 2 \times \text{Length(Joined Boundary)}$

Perimeter(d) $= 2 \times 16 \text{ cm} - 2 \times 2 \text{ cm}$

Perimeter(d) $= 32 \text{ cm} - 4 \text{ cm}$

Perimeter(d) $= 28$ cm.

The perimeter of arrangement d is 28 cm.

Given:

Original rectangular paper chit dimension = 6 cm $\times$ 4 cm.

The chit is cut into two equal pieces by a zig-zag line.

Perimeter of arrangement a = 28 cm.

Perimeter of arrangement c (original rectangle) = 20 cm.

To Find:

An arrangement of the two pieces that forms a figure with a perimeter of 22 cm.

Solution:

Let $P_{piece}$ be the perimeter of one of the two equal pieces, and let $L_j$ be the length of the boundary along which the two pieces are joined in an arrangement. The perimeter of the combined shape is given by:

Since the pieces are equal, $P_{Piece 1} = P_{Piece 2} = P_{piece}$.

From arrangement c, which forms the original 6 cm $\times$ 4 cm rectangle with a perimeter of 20 cm, the pieces are joined along the entire length of the zig-zag cut. Let the length of the zig-zag cut be $L_{cut}$. So, for arrangement c, $L_j = L_{cut}$ and Perimeter(c) $= 20$ cm.

$20 = 2 \times P_{piece} - 2 \times L_{cut}$

$10 = P_{piece} - L_{cut}$

From arrangement a, with a perimeter of 28 cm, the pieces are joined along a straight edge. Looking at the figure of the cut (4.jpg), the rectangle is 6 cm wide and 4 cm high, and the cut goes horizontally across the 6 cm dimension, starting and ending at the 4 cm side edges. The cut appears to start and end at the midpoint of the 4 cm side edges (2 cm from the top/bottom). Thus, each piece retains a straight edge of length 6 cm (from the original top/bottom) and two straight edge segments of length 2 cm (from the original left/right sides).

Arrangement a appears to be formed by joining the two pieces along their shorter straight edges (length 2 cm). So, for arrangement a, $L_j = 2$ cm and Perimeter(a) $= 28$ cm.

$28 = 2 \times P_{piece} - 2 \times 2$

$28 = 2 \times P_{piece} - 4$

$32 = 2 \times P_{piece}$

Now we have the perimeter of each piece, $P_{piece} = 16$ cm.

Using equation (ii), we can find the length of the zig-zag cut:

$16 = 10 + L_{cut}$

So, the length of the zig-zag cut is 6 cm.

The boundaries of each piece consist of a 6 cm straight edge, two 2 cm straight edges, and the 6 cm zig-zag cut edge. The perimeter of a piece is $6 + 2 + 2 + 6 = 16$ cm, which is consistent with $P_{piece}=16$ cm.

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We want to form a figure with a perimeter of 22 cm. Using equation (i):

$22 = 2 \times P_{piece} - 2 \times L_j$

Substitute $P_{piece} = 16$ cm:

$22 = 2 \times 16 - 2 \times L_j$

$22 = 32 - 2 \times L_j$

$2 \times L_j = 32 - 22$

$2 \times L_j = 10$

We need to join the two pieces along a boundary length of 5 cm.

The available edge lengths on each piece for joining are the straight edges (6 cm and 2 cm) and the zig-zag cut edge (6 cm).

To achieve a joined length of 5 cm, we must join along a segment of an edge that is at least 5 cm long. The edges that are long enough are the 6 cm straight edge and the 6 cm zig-zag edge.

A simple way to achieve a joined length of 5 cm is to align and join a 5 cm segment of the 6 cm straight edge of one piece with a 5 cm segment of the 6 cm straight edge of the other piece.

Arrangement:

Take the two identical pieces. Place the 6 cm straight edge of one piece next to the 6 cm straight edge of the other piece. Slide them together until they are aligned and joined along a length of 5 cm. The remaining 1 cm of the 6 cm edge on each piece becomes part of the external perimeter of the new shape.

The resulting shape will have a total length of $(6 - 5 + 6) = 7$ cm along the direction of the joined edge. The perimeter will consist of the unjoined portions of the 6 cm straight edges ($1+1=2$ cm), the two 2 cm straight edges from each piece ($2+2+2+2=8$ cm), and the 6 cm zig-zag edge from each piece ($6+6=12$ cm).

Calculated Perimeter $= 2 \text{ cm} + 8 \text{ cm} + 12 \text{ cm} = 22$ cm. This matches the requirement.

The arrangement is formed by joining the two pieces along a 5 cm segment of their 6 cm straight edges.

Given:

Area of the rectangular garden $= 300$ sq m ($300 \text{ m}^2$).

Length of the rectangular garden $(l) = 25$ m.

To Find:

The width (breadth) of the garden $(b)$.

Solution:

The area of a rectangle is given by the formula:

Substitute the given values into equation (i):

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

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

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

$b = \frac{300}{25} \$\text{m}$

We can simplify the fraction:

$b = \frac{\cancel{300}^{12}}{\cancel{25}_{1}} \$\text{m}$

$b = 12 \$\text{m}$

The width of the garden is 12 m.

Given:

Length of the rectangular plot $(l) = 500$ m.

Breadth of the rectangular plot $(b) = 200$ m.

Rate of tiling $= \textsf{₹}8$ per hundred sq m.

To Find:

The total cost of tiling the rectangular plot.

Solution:

First, calculate the area of the rectangular plot.

The area of a rectangle is given by the formula:

Substitute the given length and breadth into equation (i):

Area $= 500 \$\text{m} \times 200 \$\text{m}$

Area $= (500 \times 200) \$\text{m}^2$

Area $= 100000 \$\text{m}^2$

The area of the rectangular plot is 100000 square meters.

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The cost of tiling is $\textsf{₹}8$ per hundred sq m. This means the cost for every $100 \$\text{m}^2$ is $\textsf{₹}8$.

Cost per sq m $= \frac{\textsf{₹}8}{100 \$\text{m}^2}$

To find the total cost, multiply the total area by the cost per square meter.

Total cost $= \text{Area} \times \text{Cost per sq m}$

Total cost $= 100000 \$\text{m}^2 \times \frac{\textsf{₹}8}{100 \$\text{m}^2}$

Total cost $= \frac{100000}{100} \times 8 \$\textsf{₹}$

Total cost $= 1000 \times 8 \$\textsf{₹}$

Total cost $= 8000 \$\textsf{₹}$

The total cost of tiling the rectangular plot is $\textsf{₹}8000$.

Given:

Length of the rectangular coconut grove $(l) = 100$ m.

Breadth of the rectangular coconut grove $(b) = 50$ m.

Area required for each coconut tree $= 25$ sq m ($25 \text{ m}^2$).

To Find:

The maximum number of trees that can be planted in the grove.

Solution:

First, calculate the total area of the rectangular coconut grove.

The area of a rectangle is given by the formula:

Substitute the given length and breadth into equation (i):

Area $= 100 \$\text{m} \times 50 \$\text{m}$

Area $= (100 \times 50) \$\text{m}^2$

Area $= 5000 \$\text{m}^2$

The total area of the coconut grove is 5000 square meters.

---

Each coconut tree requires an area of $25 \$\text{m}^2$.

To find the maximum number of trees that can be planted, divide the total area of the grove by the area required per tree.

Maximum number of trees $= \frac{\text{Total area of the grove}}{\text{Area required per tree}}$

Maximum number of trees $= \frac{5000 \$\text{m}^2}{25 \$\text{m}^2}$

Maximum number of trees $= \frac{5000}{25}$

We can simplify the fraction:

Maximum number of trees $= \frac{\cancel{5000}^{200}}{\cancel{25}_{1}}$

Maximum number of trees $= 200$

The maximum number of coconut trees that can be planted in the grove is 200.

To find the area of the given figures, we can split them into smaller rectangles whose areas are easy to calculate. The total area of the figure is the sum of the areas of the rectangles it is split into.

The area of a rectangle is given by the formula: Area $= \text{length} \times \text{breadth}$.

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Figure 1:

Let's split this figure into two rectangles by extending the horizontal line.

Rectangle 1 (Top): This is the upper rectangle with length 6 m and breadth 3 m.

Rectangle 2 (Bottom): This is the lower left rectangle with length 4 m and breadth 2 m.

The total area of Figure 1 is the sum of the areas of Rectangle 1 and Rectangle 2.

Total Area of Figure 1 $= 18 \$\text{m}^2 + 8 \$\text{m}^2$

Total Area of Figure 1 $= 26 \$\text{m}^2$

The area of Figure 1 is 26 sq m.

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Figure 2:

Let's split this figure into three vertical rectangles based on the given widths.

Rectangle 1 (Left): This is the leftmost rectangle with width 1 m and height 5 m.

Rectangle 2 (Middle): This is the middle rectangle with width 2 m and height 3 m.

Rectangle 3 (Right): This is the rightmost rectangle with width 3 m and height 2 m.

The total area of Figure 2 is the sum of the areas of Rectangle 1, Rectangle 2, and Rectangle 3.

Total Area of Figure 2 $= 5 \$\text{m}^2 + 6 \$\text{m}^2 + 6 \$\text{m}^2$

Total Area of Figure 2 $= 17 \$\text{m}^2$

The area of Figure 2 is 17 sq m.

The tangram puzzle consists of 7 pieces: two large right triangles, one medium right triangle, two small right triangles, one square, and one parallelogram.

---

To figure out which pieces have the same area, we can compare their sizes. A common way to do this with a tangram is to consider the area of the smallest triangle as a unit area.

By observation and fitting the pieces together, we can see how the larger pieces are composed of the smallest triangles:

• Each small triangle has an area of 1 unit. (There are 2 such pieces).
• The square can be formed by combining two small triangles. So its area is $1 + 1 = 2$ units. (There is 1 such piece).
• The parallelogram can be formed by combining two small triangles. So its area is $1 + 1 = 2$ units. (There is 1 such piece).
• The medium triangle can be formed by combining two small triangles. So its area is $1 + 1 = 2$ units. (There is 1 such piece).
• Each large triangle can be formed by combining four small triangles (or two medium triangles). So its area is $1 + 1 + 1 + 1 = 4$ units. (There are 2 such pieces).

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Based on these areas:

The pieces with the same area are:

• The two small triangles have the same area (1 unit each). There are 2 pieces in this group.
• The medium triangle, the square, and the parallelogram have the same area (2 units each). There are 3 pieces in this group.
• The two large triangles have the same area (4 units each). There are 2 pieces in this group.

So, pieces have the same area in groups of 2, 3, and 2.

Based on the analysis from the previous question, we consider the area of the smallest triangle (Shape C) as 1 unit.

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Shape C is one of the small triangles.

Area of Shape C $= 1$ unit.

Shape D is the medium triangle.

Area of Shape D $= 2$ units (equivalent to 2 small triangles).

Shape E is one of the large triangles.

Area of Shape E $= 4$ units (equivalent to 4 small triangles).

---

To find how many times bigger Shape D is compared to Shape C, we compare their areas:

So, Shape D is 2 times bigger than Shape C in terms of area.

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The relationship between Shapes C, D, and E based on their areas is:

• Area of Shape D is twice the Area of Shape C ($2 \times 1 = 2$).
• Area of Shape E is twice the Area of Shape D ($2 \times 2 = 4$).
• Area of Shape E is four times the Area of Shape C ($4 \times 1 = 4$).

There is a proportional relationship where the area of each successive shape (C to D, and D to E) doubles.

Relationship: Area(D) = 2 $\times$ Area(C), Area(E) = 2 $\times$ Area(D), Area(E) = 4 $\times$ Area(C).

Based on the analysis of the tangram pieces, we can compare the areas of Shape D (medium triangle) and Shape F (square) by considering how many times the area of the smallest triangle (Shape C) fits into them.

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Shape D is the medium triangle.

Shape F is the square.

From observation of the tangram pieces and how they fit together:

• The medium triangle (Shape D) can be formed by combining two small triangles (Shape C). So, the area of Shape D is 2 times the area of the smallest triangle.
• The square (Shape F) can also be formed by combining two small triangles (Shape C). So, the area of Shape F is 2 times the area of the smallest triangle.

---

Comparison of Areas:

Since both Shape D and Shape F have an area equivalent to two smallest triangles (Shape C), their areas are equal.

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Conclusion and Reason:

Neither Shape D nor Shape F has more area. They have the same area.

Reason: Both the medium triangle (Shape D) and the square (Shape F) can be decomposed into, or are equivalent in area to, two of the smallest triangles (Shape C) of the tangram puzzle. Therefore, their areas are equal.

Based on the analysis of the tangram pieces, we can compare the areas of Shape F (square) and Shape G (parallelogram) by considering how many times the area of the smallest triangle (Shape C) fits into them.

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Shape F is the square.

Shape G is the parallelogram.

From observation of the tangram pieces and how they fit together:

• The square (Shape F) can be formed by combining two small triangles (Shape C). So, the area of Shape F is 2 times the area of the smallest triangle.
• The parallelogram (Shape G) can also be formed by combining two small triangles (Shape C). So, the area of Shape G is 2 times the area of the smallest triangle.

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Comparison of Areas:

Since both Shape F and Shape G have an area equivalent to two smallest triangles (Shape C), their areas are equal.

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Conclusion and Reason:

Neither Shape F nor Shape G has more area. They have the same area.

Reason: Both the square (Shape F) and the parallelogram (Shape G) can be decomposed into, or are equivalent in area to, two of the smallest triangles (Shape C) of the tangram puzzle. Therefore, their areas are equal.

To compare the areas of Shape A and Shape G, we can use the relative areas of the tangram pieces based on the smallest triangle as a unit of area (e.g., Shape C or Shape E).

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From the properties of the tangram pieces:

• The smallest triangles (Shape C and Shape E) have an area of 1 unit each.
• The medium triangle (Shape D), the square (Shape F), and the parallelogram (Shape G) each have an area equivalent to 2 of the smallest triangles. So, Area(D) = Area(F) = Area(G) = 2 units.
• The large triangles (Shape A and Shape B) each have an area equivalent to 4 of the smallest triangles. So, Area(A) = Area(B) = 4 units.

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We want to compare the area of Shape A with the area of Shape G.

To find out how many times bigger Shape A is compared to Shape G, we divide the area of Shape A by the area of Shape G:

So, Shape A is 2 times as big as Shape G in terms of area.

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Conclusion:

Shape A has more area than Shape G.

Specifically, Shape A is twice as big as Shape G.

The large square formed by all seven tangram pieces has a total area equal to the sum of the areas of all the individual pieces.

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Let's denote the area of Shape C (one of the smallest triangles) as 1 unit of area.

Based on the properties of the tangram pieces (as determined in previous questions or by observation):

• Area of Shape C (Small Triangle) = 1 unit
• Area of Shape E (Small Triangle) = 1 unit
• Area of Shape D (Medium Triangle) = 2 units (equivalent to 2 small triangles)
• Area of Shape F (Square) = 2 units (equivalent to 2 small triangles)
• Area of Shape G (Parallelogram) = 2 units (equivalent to 2 small triangles)
• Area of Shape A (Large Triangle) = 4 units (equivalent to 4 small triangles)
• Area of Shape B (Large Triangle) = 4 units (equivalent to 4 small triangles)

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The area of the big square is the sum of the areas of all 7 pieces:

Area (Big Square) = Area(A) + Area(B) + Area(C) + Area(D) + Area(E) + Area(F) + Area(G)

Area (Big Square) $= 4 \text{ units} + 4 \text{ units} + 1 \text{ unit} + 2 \text{ units} + 1 \text{ unit} + 2 \text{ units} + 2 \text{ units}$

Area (Big Square) $= (4 + 4 + 1 + 2 + 1 + 2 + 2)$ units

Area (Big Square) $= 16$ units

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Since we defined the area of Shape C as 1 unit, the area of the big square is 16 times the area of Shape C.

The area of the big square formed with all seven pieces is 16 times the area of Shape C.

When the 7 tangram pieces are arranged to form a rectangle (or any other shape without overlapping or gaps), the total area of the resulting figure is the sum of the areas of the individual pieces.

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Let the area of Shape C (one of the smallest triangles) be our unit of area. So, Area(C) = 1 unit.

From previous observations about the tangram pieces:

• Area of Shape C = 1 unit
• Area of Shape E = 1 unit
• Area of Shape D (medium triangle) = 2 units (equivalent to 2 small triangles)
• Area of Shape F (square) = 2 units (equivalent to 2 small triangles)
• Area of Shape G (parallelogram) = 2 units (equivalent to 2 small triangles)
• Area of Shape A (large triangle) = 4 units (equivalent to 4 small triangles)
• Area of Shape B (large triangle) = 4 units (equivalent to 4 small triangles)

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The area of the rectangle formed by all seven pieces is the sum of their individual areas:

Area (Rectangle) = Area(A) + Area(B) + Area(C) + Area(D) + Area(E) + Area(F) + Area(G)

Area (Rectangle) $= 4 \text{ units} + 4 \text{ units} + 1 \text{ unit} + 2 \text{ units} + 1 \text{ unit} + 2 \text{ units} + 2 \text{ units}$

Area (Rectangle) $= (4 + 4 + 1 + 2 + 1 + 2 + 2)$ units

Area (Rectangle) $= 16$ units

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

The area of the rectangle formed is 16 times the area of Shape C.

Reason:

The total area of a figure constructed from multiple pieces is the sum of the areas of those pieces. Rearranging the pieces only changes the shape and its perimeter, not the total area occupied by the pieces.

When the 7 tangram pieces are arranged to form different shapes (like a large square or a rectangle), the total area covered by the pieces remains the same, as the area is the sum of the areas of the individual pieces.

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However, the perimeters of the shapes formed are generally different.

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Explanation:

The perimeter of a shape is the total length of its outer boundary. When you arrange the tangram pieces, some edges of the individual pieces are joined together. These joined edges become internal to the new combined shape and are no longer part of the outer boundary.

The set of edges that make up the external boundary is different for different arrangements of the pieces, even though the pieces themselves are the same.

For example, when forming the large square, certain edges of the triangles, square, and parallelogram form the four sides of the square. When forming a rectangle, a different combination of edges forms the four sides of the rectangle.

Since the specific edges contributing to the outer boundary change with the arrangement, the total length of the outer boundary (the perimeter) changes.

Therefore, the perimeters of the square and the rectangle formed from these 7 pieces will be different.

To find the area of the figures drawn on a grid, we can count the number of complete unit squares covered by each figure.

Let's assume each square in the grid is a unit square with an area of 1 square unit.

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Figure 1:

We count the number of full squares the figure covers:

• Row 1 (from top): 3 squares
• Row 2: 1 square (on the left)
• Row 3: 1 square (on the left)
• Row 4 (bottom): 1 square (on the left)

Total number of squares covered $= 3 + 1 + 1 + 1 = 6$ squares.

Alternatively, we can split the figure into two rectangles.

Split into a top rectangle (3 units $\times$ 1 unit) and a vertical rectangle below it (1 unit $\times$ 3 units).

Area of top rectangle $= 3 \times 1 = 3$ square units.

Area of vertical rectangle $= 1 \times 3 = 3$ square units.

Total Area $= 3 + 3 = 6$ square units.

The area of Figure 1 is 6 square units.

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Figure 2:

We count the number of full squares the figure covers:

• Top horizontal part: 3 squares
• Bottom vertical part: 3 squares (below the middle top square)

Total number of squares covered $= 3 + 3 = 6$ squares.

Alternatively, we can split the figure into two rectangles.

Split into a top horizontal rectangle (3 units $\times$ 1 unit) and a bottom vertical rectangle (1 unit $\times$ 3 units).

Area of top rectangle $= 3 \times 1 = 3$ square units.

Area of bottom rectangle $= 1 \times 3 = 3$ square units.

Total Area $= 3 + 3 = 6$ square units.

The area of Figure 2 is 6 square units.

To make rectangles with an area of 24 square units and whole number lengths and widths, we need to find pairs of whole numbers whose product is 24.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

The possible pairs of length and width $(l, b)$ such that $l \times b = 24$ and $l \geq b$ are:

• (24, 1)
• (12, 2)
• (8, 3)
• (6, 4)

Now, let's calculate the perimeter for each rectangle using the formula Perimeter $= 2(l + b)$.

Length (m)	Width (m)	Area ($m^2$)	Perimeter (m) ($2(l+b)$)
24	1	$24 \times 1 = 24$	$2(24+1) = 2(25) = 50$
12	2	$12 \times 2 = 24$	$2(12+2) = 2(14) = 28$
8	3	$8 \times 3 = 24$	$2(8+3) = 2(11) = 22$
6	4	$6 \times 4 = 24$	$2(6+4) = 2(10) = 20$

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a. Which rectangle has the greatest perimeter?

Comparing the perimeters (50 m, 28 m, 22 m, 20 m), the greatest perimeter is 50 m.

This corresponds to the rectangle with dimensions 24 m $\times$ 1 m.

The rectangle with the greatest perimeter is the one with dimensions 24 units $\times$ 1 unit, and its perimeter is 50 units.

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b. Which rectangle has the least perimeter?

Comparing the perimeters (50 m, 28 m, 22 m, 20 m), the least perimeter is 20 m.

This corresponds to the rectangle with dimensions 6 m $\times$ 4 m.

The rectangle with the least perimeter is the one with dimensions 6 units $\times$ 4 units, and its perimeter is 20 units.

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c. If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.

Let's find the possible rectangles with an area of 32 sq cm and whole number sides. We find pairs of whole numbers whose product is 32.

The factors of 32 are 1, 2, 4, 8, 16, and 32.

The possible pairs of length and width $(l, b)$ such that $l \times b = 32$ and $l \geq b$ are:

• (32, 1)
• (16, 2)
• (8, 4)

Now, let's calculate the perimeter for each rectangle (in cm).

Length (cm)	Width (cm)	Area ($cm^2$)	Perimeter (cm) ($2(l+b)$)
32	1	$32 \times 1 = 32$	$2(32+1) = 2(33) = 66$
16	2	$16 \times 2 = 32$	$2(16+2) = 2(18) = 36$
8	4	$8 \times 4 = 32$	$2(8+4) = 2(12) = 24$

For an area of 32 sq cm:

The greatest perimeter is 66 cm, corresponding to the rectangle with dimensions 32 cm $\times$ 1 cm.

The least perimeter is 24 cm, corresponding to the rectangle with dimensions 8 cm $\times$ 4 cm.

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Generalization:

Yes, given any area, it is possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter (assuming whole number sides).

Observations from Area 24 and Area 32:

• The rectangle with the **greatest perimeter** for a fixed area is the one where the length and width are **farthest apart** (i.e., a very long and narrow rectangle). This corresponds to using the factors of the area that are 1 and the number itself (e.g., 24 $\times$ 1, 32 $\times$ 1).
• The rectangle with the **least perimeter** for a fixed area is the one where the length and width are **closest together** (i.e., a shape closest to a square). This corresponds to using the factors of the area that are closest to the square root of the area (e.g., for 24, $\sqrt{24} \approx 4.9$, factors 6 and 4 are closest; for 32, $\sqrt{32} \approx 5.6$, factors 8 and 4 are closest). If the area is a perfect square, the square shape will have the absolute least perimeter among all rectangles with that area.

Reasons:

Let the area of a rectangle be $A$, and the length and width be $l$ and $b$. So, $A = l \times b$. The perimeter is $P = 2(l + b)$.

As the dimensions $l$ and $b$ get farther apart for a fixed area $A$, the sum $(l+b)$ increases, leading to a larger perimeter $P$. The extreme case is when $b=1$ and $l=A$, where $l+b = A+1$, giving the maximum perimeter $2(A+1)$ (for whole number sides).

As the dimensions $l$ and $b$ get closer together for a fixed area $A$, the sum $(l+b)$ decreases, leading to a smaller perimeter $P$. The minimum value of $(l+b)$ for a fixed product $l \times b$ occurs when $l$ and $b$ are equal (i.e., a square), or as close as possible to being equal if limited to whole numbers.

Example for Area 100 sq units (perfect square):

Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100.

Possible $(l, b)$ pairs ($l \geq b$): (100, 1), (50, 2), (25, 4), (20, 5), (10, 10).

Perimeters:

• 100 $\times$ 1: $2(100+1) = 202$
• 50 $\times$ 2: $2(50+2) = 104$
• 25 $\times$ 4: $2(25+4) = 58$
• 20 $\times$ 5: $2(20+5) = 50$
• 10 $\times$ 10: $2(10+10) = 40$

Greatest perimeter: 202 (dimensions 100 $\times$ 1 - longest and narrowest).

Least perimeter: 40 (dimensions 10 $\times$ 10 - a square, sides are equal/closest).

The question asks about whether two specific triangles overlap exactly and if they have the same area. Without the image of the two triangles, it is not possible to definitively say whether those particular triangles overlap exactly.

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However, we can discuss the mathematical principle behind this type of question.

If two figures overlap each other exactly, it means that they are congruent. Congruent figures have the same size and the same shape. When two figures are congruent, they can be placed one upon the other so that they perfectly cover each other.

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A fundamental property of geometric shapes is that congruent figures have the same area.

Therefore, if the two triangles mentioned in the question are indeed identical (congruent) such that they can overlap each other exactly, then they must necessarily have the same area.

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The hint suggests trying this with more rectangles and squares. Let's consider a rectangle or a square. If you draw a diagonal line across a rectangle or a square, you divide it into two triangles.

For example, if you take a rectangle and cut it along one of its diagonals, you get two right-angled triangles. These two triangles are congruent. You can verify this by placing one on top of the other; they will overlap exactly.

Since these two triangles are congruent, they will have the same area.

The area of the original rectangle (or square) is the sum of the areas of these two congruent triangles. Since their areas are equal, the area of each triangle is half the area of the rectangle (or square).

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Conclusion:

In general, if two figures can overlap each other exactly, they are congruent, and thus they have the same area. If the specific triangles referred to in the question are congruent, they will overlap exactly and have equal areas.

The exercise involving cutting a rectangle (or square) along a diagonal into two triangles and observing that they overlap exactly allows us to draw the following inferences:

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Inference 1:

If two figures can be placed one upon the other such that they perfectly cover each other, then these two figures are congruent.

This is the definition of congruence in geometry – shapes that are identical in size and shape.

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Inference 2:

Any two congruent figures have the same area.

If two shapes are identical, they must cover the same amount of space, thus having equal areas. The exercise demonstrates this by showing the triangles overlap exactly, implying they have equal area.

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Inference 3 (Specific to the Rectangle/Square example):

A diagonal divides a rectangle (or a square) into two triangles of equal area.

Since cutting a rectangle along its diagonal results in two congruent triangles, and congruent triangles have equal area, the area of each triangle is exactly half the area of the original rectangle.

These inferences highlight the relationship between congruence and area, and provide a method to find the area of a triangle by relating it to the area of a rectangle (specifically, a right-angled triangle formed by the sides and diagonal of a rectangle).

The figures are drawn on a squared grid paper. We can calculate the area by counting the unit squares. From the image, we can determine the dimensions of the rectangle ABCD.

The length of the rectangle (e.g., AB or CD) covers 6 units on the grid.

The breadth of the rectangle (e.g., AD or BC) covers 4 units on the grid.

Area of a rectangle = length $\times$ breadth.

Area of rectangle ABCD $= 6 \text{ units} \times 4 \text{ units} = 24$ square units.

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Now let's find the areas of the triangles.

The blue triangle is $\triangle$BAD. This is a right-angled triangle formed by the diagonal BD and sides AB and AD of the rectangle ABCD.

The base of $\triangle$BAD can be taken as AD = 4 units, and the height can be taken as AB = 6 units (since it's a right-angled triangle at A).

Alternatively, $\triangle$BAD is one of the two congruent triangles formed by cutting rectangle ABCD along its diagonal BD. The other triangle is $\triangle$BCD.

The area of each of these triangles is half the area of the rectangle.

Area of $\triangle$BAD $= \frac{1}{2} \times 24 \text{ square units}$

Area of $\triangle$BAD $= 12 \text{ square units}$

1. Find the area of blue triangle BAD. 12 square units.

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The red triangle is $\triangle$ABE. This triangle is formed within the rectangle ABCD. The base of this triangle can be taken as the side AB of the rectangle, and the height is the perpendicular distance from point E to the base AB. Point E is on the side CD of the rectangle, so the perpendicular distance from E to AB is equal to the breadth of the rectangle, which is AD or BC = 4 units.

Base of $\triangle$ABE $= \text{AB} = 6$ units.

Height of $\triangle$ABE $= \text{AD} = 4$ units.

The area of a triangle is given by the formula: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

Area of $\triangle$ABE $= \frac{1}{2} \times 6 \text{ units} \times 4 \text{ units}$

Area of $\triangle$ABE $= \frac{1}{2} \times 24 \text{ square units}$

Area of $\triangle$ABE $= 12 \text{ square units}$

2. Find the area of red triangle ABE. 12 square units.

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Area of rectangle ABCD = 24 square units.

To find the area of the figures, we can split them into smaller, simpler shapes like rectangles and triangles, calculate the area of each part, and then add them together.

Assume each small square in the grid has a side length of 1 unit, so the area of each small square is $1 \times 1 = 1$ square unit.

The area of a rectangle is given by: Area $= \text{length} \times \text{breadth}$.

The area of a triangle is given by: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

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Figure 1:

We can split Figure 1 into a rectangle and a triangle.

Let's split it into a bottom rectangle and a top triangle.

Rectangle: The bottom part is a rectangle with length 3 units and breadth 2 units.

Triangle: The top part is a triangle. Its base is the top side of the rectangle, which is 3 units. Its height is the vertical distance from the top side of the rectangle to the peak, which is 2 units.

Area of Triangle $= \frac{1}{2} \times 6 \$\text{square units} = 3 \$\text{square units}$.

The total area of Figure 1 is the sum of the areas of the rectangle and the triangle.

Total Area of Figure 1 $= \text{Area of Rectangle} + \text{Area of Triangle}$

Total Area of Figure 1 $= 6 \$\text{square units} + 3 \$\text{square units}$

Total Area of Figure 1 $= 9 \$\text{square units}$.

The area of Figure 1 is 9 square units.

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Figure 2:

We can split Figure 2 into a parallelogram and a triangle.

Let's split it into a bottom triangle and a top parallelogram.

Triangle: The bottom part is a triangle with a horizontal base of 4 units and a vertical height of 1 unit.

Area of Triangle $= \frac{1}{2} \times 4 \$\text{square units} = 2 \$\text{square units}$.

Parallelogram: The top part is a parallelogram. Its base can be taken as the top side of the triangle, which is 4 units. Its height is the perpendicular vertical distance from this base to the top edge of the figure, which is 3 units.

Area of Parallelogram $= 12 \$\text{square units}$.

The total area of Figure 2 is the sum of the areas of the triangle and the parallelogram.

Total Area of Figure 2 $= \text{Area of Triangle} + \text{Area of Parallelogram}$

Total Area of Figure 2 $= 2 \$\text{square units} + 12 \$\text{square units}$

Total Area of Figure 2 $= 14 \$\text{square units}$.

The area of Figure 2 is 14 square units.

Here are the solutions to the questions based on forming shapes using 9 unit squares.

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1. What is the smallest perimeter possible?

To find the smallest perimeter, we need to arrange the 9 unit squares to form the most compact shape possible. This configuration maximizes the number of shared edges between squares, thus minimizing the length of the exposed boundary, which is the perimeter.

The most compact shape using 9 unit squares is a 3x3 square.

An illustration of the shape:

    S S S
    S S S
    S S S
        

Each side of this square is 3 units long.

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

Perimeter = $3 + 3 + 3 + 3 = 4 \times 3 = 12$ units.

The smallest perimeter possible using 9 unit squares is 12 units.

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2. What is the largest perimeter possible?

To find the largest perimeter, we need to arrange the 9 unit squares to form the least compact, or most elongated, shape possible. This configuration minimizes the number of shared edges between squares, thus maximizing the length of the exposed boundary.

The most elongated shape using 9 unit squares is a 1x9 rectangle (a straight line of 9 squares).

An illustration of the shape:

    S S S S S S S S S
        

This rectangle has two short sides of length 1 unit each and two long sides of length 9 units each.

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

Perimeter = $1 + 9 + 1 + 9 = 2 \times (1 + 9) = 2 \times 10 = 20$ units.

The largest perimeter possible using 9 unit squares is 20 units.

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3. Make a figure with a perimeter of 18 units.

A perimeter of 18 units is an intermediate value between the minimum (12) and maximum (20) perimeters. We need to arrange the 9 unit squares in a shape that is neither the most compact square nor the most elongated line.

One possible figure with a perimeter of 18 units is an L-shaped figure formed by joining a 5x1 strip of squares and a 4x1 strip of squares at a corner.

An illustration of one such L-shaped figure:

    S S S S S
    S
    S
    S
        

Let's count the number of exposed edges around the boundary of this shape:

Top edge: 5 units

Right edge: 5 units (from the vertical part)

Bottom edge: 4 units (from the vertical part)

Left edge: 1 unit (from the horizontal part) + 3 units (from the vertical part) = 4 units

The total perimeter is the sum of these exposed edge lengths.

Perimeter = $5 + 5 + 4 + 4 = 18$ units.

There are multiple distinct shapes using 9 unit squares that have a perimeter of 18 units.

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4. Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning?

Let's examine the possibility of making other shaped figures for each perimeter:

Perimeter 12 (Smallest):

There is only one shape using 9 unit squares that has a perimeter of 12 units. This unique shape is the 3x3 square.

Reasoning: The perimeter of a shape formed by unit squares is minimized when the squares are arranged as compactly as possible, maximizing the number of internal shared edges. The 3x3 square is the most compact arrangement for 9 squares, leaving the fewest number of edges exposed on the boundary. Any other arrangement will involve breaking some of the internal connections present in the 3x3 square, which will expose more edges and increase the perimeter.

Perimeter 20 (Largest):

There is more than one shape using 9 unit squares that has a perimeter of 20 units.

Reasoning: The perimeter is maximized when the shape is as elongated or as 'notched' as possible, minimizing the number of internal shared edges. The 1x9 rectangle (a straight line) is one shape with this maximum perimeter, as it minimizes internal connections by arranging squares in a single line. However, other shapes that are not straight lines can also have a large number of exposed edges, potentially resulting in the same perimeter. For example, shapes with many 'indentations' or a spread-out form can also achieve a perimeter of 20 units. There are several distinct arrangements of 9 squares (nonominoes) that have a perimeter of 20 units.

Perimeter 18:

There is more than one shape using 9 unit squares that has a perimeter of 18 units.

Reasoning: A perimeter of 18 is an intermediate value. Shapes with intermediate perimeters are less restricted in form than the extremely compact (minimum perimeter) or extremely elongated (maximum perimeter) shapes. There are many ways to arrange 9 squares to have a number of internal connections that results in a perimeter of 18. As illustrated in point 3, an L-shape is one such example, and there are numerous other distinct arrangements of 9 squares (nonominoes) that are neither the 3x3 square nor the 1x9 rectangle but happen to have a perimeter of 18 units. Studies on polyominoes confirm the existence of multiple distinct shapes for this perimeter.

Let the original figure be $F$ and its perimeter be $P_F = 24$ units.

We are adding a new unit square, let's call it $S$. The new square $S$ has a perimeter of $4$ units on its own.

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When we attach the new square $S$ to the figure $F$, they will share one or more edges. Let $n$ be the number of edges that the new square $S$ shares with the figure $F$.

Each shared edge was previously part of the perimeter of either $F$ or $S$ (or both, if we think about them separately before joining). When the edges are joined, they are no longer part of the external perimeter. So, each shared edge reduces the total perimeter by 2 units (1 unit from $F$'s boundary and 1 unit from $S$'s boundary).

The new square $S$ has 4 edges. If $n$ edges are shared, then $4-n$ edges of $S$ will become part of the new figure's perimeter.

The original figure $F$ loses $n$ edges from its perimeter where the new square is attached.

The change in perimeter is the number of new edges added minus the number of old edges lost:

Change in Perimeter $= (4 - n) - n = 4 - 2n$

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Answering the first part: What will be the change in the perimeter if a new square is attached as shown on the right (in the provided image)?

Looking at the image, the new square is attached to one of the exposed edges of the original figure. This means the new square shares exactly one edge with the original figure.

In this case, the number of shared edges is $n=1$.

The change in perimeter is calculated using the formula: Change $= 4 - 2n$

Change $= 4 - 2(1) = 4 - 2 = 2$ units.

So, the perimeter will increase by 2 units.

The new perimeter will be $24 + 2 = 26$ units.

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Answering the second part: Experiment placing this new square at different places and think what the change in perimeter will be. Can you place the square so that the perimeter: a) increases; b) decreases; c) stays the same?

Yes, the change in perimeter depends entirely on how many edges ($n$) the new square shares with the existing figure when it is attached. We can achieve different outcomes by placing the square at different locations on the boundary of the original figure.

a) Perimeter increases:

The perimeter increases if the new square shares only one edge ($n=1$) with the original figure. The change is $4 - 2(1) = 2$ units. This happens when you attach the square to a straight outer edge of the figure, not in a corner or indentation where it could touch more than one edge.

Example: Attaching the square to the side of one of the 'arms' of the cross shape shown in the original image. This was the placement shown in the image itself.

b) Perimeter decreases:

The perimeter decreases if the new square shares three edges ($n=3$) with the original figure. The change is $4 - 2(3) = 4 - 6 = -2$ units. This happens when you attach the square into a 'pocket' or a concave corner of the figure, where it fits snugly against three existing edges.

Example: The cross shape has several inward-facing corners. If you place the new square in one of these corners such that it touches edges on three sides of the corner, the perimeter will decrease by 2 units.

c) Perimeter stays the same:

The perimeter stays the same if the new square shares exactly two edges ($n=2$) with the original figure. The change is $4 - 2(2) = 4 - 4 = 0$ units. This happens when you attach the square to a corner of the figure, or along an edge such that it also touches an adjacent edge of the figure.

Example: Attaching the square into an 'L-shaped' corner of the figure (a concave angle formed by two adjacent sides). The cross shape has many such corners where an added square could touch two existing edges, resulting in no change to the perimeter.

Given:

A rectangular plot with dimensions $15 \text{ m} \times 10 \text{ m}$.

The house plan within the plot shows several rooms with some dimensions provided:

- Kitchen: $6 \text{ m} \times 4 \text{ m}$

- Bedroom 1: $6 \text{ m} \times 5 \text{ m}$

- Bedroom 2: $4 \text{ m} \times 5 \text{ m}$

- Living Room: $8 \text{ m} \times ?$

- Toilets (combined): $4 \text{ m} \times ?$

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To Find:

a. The missing measurements of the rooms.

b. The area of the house.

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Solution:

a. Finding the missing measurements.

Based on the plan, let's assume the plot is $15 \text{ m}$ wide (horizontal) and $10 \text{ m}$ deep (vertical).

Total Plot Width = $15 \text{ m}$

Total Plot Depth = $10 \text{ m}$

By observing the layout, we can infer the presence of internal passages or wall thicknesses not explicitly dimensioned, which make the sum of room dimensions less than the total plot dimensions in certain directions.

Considering the vertical dimensions on the left side:

Depth of Kitchen + Depth of Vertical Passage + Depth of Bedroom 1 = Total Plot Depth

$4 \text{ m} + \text{Vertical Passage Depth} + 5 \text{ m} = 10 \text{ m}$

Vertical Passage Depth = $10 \text{ m} - 4 \text{ m} - 5 \text{ m} = 1 \text{ m}$

There is a $1 \text{ m}$ deep vertical passage that runs horizontally, separating the top row of rooms (Kitchen and Living Room) from the bottom row of rooms (Bedroom 1, Bedroom 2, and Toilets).

---

Considering the horizontal dimensions along the top edge:

Width of Kitchen + Width of Horizontal Passage + Width of Living Room = Total Plot Width

$6 \text{ m} + \text{Horizontal Passage Width} + 8 \text{ m} = 15 \text{ m}$

Horizontal Passage Width = $15 \text{ m} - 6 \text{ m} - 8 \text{ m} = 1 \text{ m}$

There is a $1 \text{ m}$ wide horizontal passage that runs vertically, separating the left column of rooms (Kitchen and Bedroom 1) from the right column of rooms (Living Room, Bedroom 2, and Toilets).

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Now, let's determine the missing dimensions for the Living Room and the combined Toilets area.

The Living Room is in the top-right section of the house. This top section, containing the Kitchen and Living Room, is bounded below by the $1 \text{ m}$ vertical passage. Since the Kitchen has a depth of $4 \text{ m}$, and it occupies the full depth of this top section, the Living Room must also have the same depth.

Missing measurement for Living Room (Depth) = $4 \text{ m}$.

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The combined Toilets area is in the bottom-right section of the house, next to Bedroom 2. This bottom section, containing Bedroom 1, Bedroom 2, and the Toilets, is bounded above by the $1 \text{ m}$ vertical passage. Since Bedroom 1 and Bedroom 2 both have a depth of $5 \text{ m}$, and they occupy the full depth of this bottom section, the combined Toilets area must also have the same depth.

Missing measurement for Combined Toilets (Depth) = $5 \text{ m}$.

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The missing measurements are the depth of the Living Room ($4 \text{ m}$) and the depth of the combined Toilets area ($5 \text{ m}$).

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b. Finding the area of the house.

The area of the house is the sum of the areas of all the rooms. The area of a rectangle is calculated as Width $\times$ Depth.

Area of Kitchen = $6 \text{ m} \times 4 \text{ m} = 24 \text{ m}^2$

Area of Bedroom 1 = $6 \text{ m} \times 5 \text{ m} = 30 \text{ m}^2$

Area of Bedroom 2 = $4 \text{ m} \times 5 \text{ m} = 20 \text{ m}^2$

Area of Living Room = $8 \text{ m} \times 4 \text{ m} = 32 \text{ m}^2$ (Using the depth found in part a)

Area of Combined Toilets = $4 \text{ m} \times 5 \text{ m} = 20 \text{ m}^2$ (Using the depth found in part a)

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Total Area of House = Area(Kitchen) + Area(Bedroom 1) + Area(Bedroom 2) + Area(Living Room) + Area(Combined Toilets)

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The area of the house is $126 \text{ m}^2$.

Given:

The house plan with the following room dimensions:

- Kitchen: $3 \text{ m} \times 4 \text{ m}$

- Bedroom: $4 \text{ m} \times 5 \text{ m}$

- Toilet: $3 \text{ m} \times 2 \text{ m}$

- Hall: $? \times ?$

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To Find:

a. The missing measurements (dimensions) of the Hall.

b. The area of the house.

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Solution:

a. Finding the missing measurements.

Let's analyze the layout of the rooms to determine the missing dimensions of the Hall.

Observing the horizontal arrangement, the Kitchen and Hall are placed side-by-side in the top section, and the Bedroom and Toilet are placed side-by-side in the bottom section. The total width of the house is the sum of the widths of the rooms in each row.

Total Width = Width of Kitchen + Width of Hall

Total Width = Width of Bedroom + Width of Toilet

From the given dimensions:

Using the first equation for the top row:

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Observing the vertical arrangement, the Kitchen is placed above the Bedroom, and the Hall is placed above the Toilet. The total depth of the house is the sum of the depths of the rooms in each column.

Total Depth = Depth of Kitchen + Depth of Bedroom

Total Depth = Depth of Hall + Depth of Toilet

From the given dimensions:

Using the first equation for the right column:

The missing measurements for the Hall are a width of $4 \text{ m}$ and a depth of $7 \text{ m}$.

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b. Finding the area of the house.

The area of the house is the sum of the areas of all the rooms. The area of each room is calculated by multiplying its width and depth.

Area of Kitchen = Width $\times$ Depth = $3 \text{ m} \times 4 \text{ m} = 12 \text{ m}^2$

Area of Bedroom = Width $\times$ Depth = $4 \text{ m} \times 5 \text{ m} = 20 \text{ m}^2$

Area of Toilet = Width $\times$ Depth = $3 \text{ m} \times 2 \text{ m} = 6 \text{ m}^2$

Area of Hall = Width $\times$ Depth = $4 \text{ m} \times 7 \text{ m} = 28 \text{ m}^2$ (Using the dimensions found in part a)

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Total Area of House = Area(Kitchen) + Area(Bedroom) + Area(Toilet) + Area(Hall)

The area of the house is $66 \text{ m}^2$.

We need to find the missing value for each of the four figures provided.

The figures are either rectangles or squares, and the missing value is either the area or a side length.

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Figure 1:

This is a rectangle.

Given: Length ($l$) = $8$ units, Width ($w$) = $6$ units.

To find: Area ($A$).

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

The missing value in Figure 1 is the Area, which is $\mathbf{48}$ square units.

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Figure 2:

This is a rectangle.

Given: Area ($A$) = $30$ square units, Length ($l$) = $6$ units.

To find: Width ($w$).

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

We can rearrange the formula to find the width: $w = \frac{A}{l}$.

The missing value in Figure 2 is the Width, which is $\mathbf{5}$ units.

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Figure 3:

This is a square.

Given: Side length ($s$) = $10$ units.

To find: Area ($A$).

The formula for the area of a square is $A = s \times s = s^2$.

The missing value in Figure 3 is the Area, which is $\mathbf{100}$ square units.

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Figure 4:

This is a rectangle.

Given: Area ($A$) = $56$ square units, Width ($w$) = $7$ units.

To find: Length ($l$).

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

We can rearrange the formula to find the length: $l = \frac{A}{w}$.

The missing value in Figure 4 is the Length, which is $\mathbf{8}$ units.

Given:

Rectangle 1 dimensions: $5 \text{ m} \times 10 \text{ m}$.

Rectangle 2 dimensions: $2 \text{ m} \times 7 \text{ m}$.

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To Find:

The dimensions of a rectangle whose area is the sum of the areas of the two given rectangles.

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Solution:

First, we calculate the area of each given rectangle.

The formula for the area of a rectangle is Area = Length $\times$ Width.

Area of Rectangle 1 ($A_1$):

Area of Rectangle 2 ($A_2$):

The area of the third rectangle ($A_3$) is the sum of the areas of the two given rectangles.

We need to find the dimensions (Length $\times$ Width) of a rectangle whose area is $64 \text{ m}^2$.

Let the dimensions of the third rectangle be $l \times w$. Then, $l \times w = 64 \text{ m}^2$.

There are multiple possible pairs of positive numbers for length and width that multiply to give 64.

Some possible pairs of integer dimensions (Length $\times$ Width) are:

• $64 \text{ m} \times 1 \text{ m}$
• $32 \text{ m} \times 2 \text{ m}$
• $16 \text{ m} \times 4 \text{ m}$
• $8 \text{ m} \times 8 \text{ m}$ (This is a square, which is a special type of rectangle)

Any pair of positive numbers $(l, w)$ such that $l \times w = 64$ can be the dimensions of the required rectangle.

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Answer:

The area of the required rectangle is $64 \text{ m}^2$.

The dimensions of such a rectangle can be any pair of positive numbers $l$ and $w$ such that $l \times w = 64$.

For example, the dimensions could be $\mathbf{16 \text{ m} \times 4 \text{ m}}$ or $\mathbf{8 \text{ m} \times 8 \text{ m}}$.

Given:

Length of the rectangular garden ($l$) = $50 \text{ m}$.

Area of the rectangular garden ($A$) = $1000 \text{ m}^2$.

---

To Find:

The width of the garden ($w$).

---

Solution:

The formula for the area of a rectangle is:

Substitute the given values into the formula:

To find the width ($w$), divide the area by the length:

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Answer:

The width of the garden is $\mathbf{20 \text{ m}}$.

Given:

Dimensions of the rectangular room: Length ($l_{room}$) = $5 \text{ m}$, Width ($w_{room}$) = $4 \text{ m}$.

Dimensions of the square carpet: Side ($s_{carpet}$) = $3 \text{ m}$.

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To Find:

The area of the floor that is not carpeted.

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Solution:

First, we calculate the area of the rectangular room.

The formula for the area of a rectangle is Area = Length $\times$ Width.

Next, we calculate the area of the square carpet.

The formula for the area of a square is Area = Side $\times$ Side = Side$^2$.

The area that is not carpeted is the difference between the area of the room and the area of the carpet.

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Answer:

The area that is not carpeted is $\mathbf{11 \text{ m}^2}$.

Given:

Dimensions of the rectangular garden: Length ($l_{garden}$) = $15 \text{ m}$, Width ($w_{garden}$) = $12 \text{ m}$.

Dimensions of each rectangular flower bed: Length ($l_{bed}$) = $2 \text{ m}$, Width ($w_{bed}$) = $1 \text{ m}$.

Number of flower beds = $4$, located at the four corners.

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To Find:

The area of the garden available for laying down a lawn.

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Solution:

First, we calculate the total area of the rectangular garden.

The formula for the area of a rectangle is Area = Length $\times$ Width.

Next, we calculate the area of one rectangular flower bed.

Since there are 4 identical flower beds, the total area occupied by the flower beds is:

The area available for laying down a lawn is the area of the garden minus the total area of the flower beds.

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Answer:

The area now available for laying down a lawn is $\mathbf{172 \text{ m}^2}$.

Given:

Shape A is a rectangle with Area = $18$ square units.

Shape B is a rectangle with Area = $20$ square units.

Perimeter of Shape A is greater than the Perimeter of Shape B.

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To Find:

Draw (or describe how to draw) two such shapes satisfying the given conditions.

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Solution:

We need to find dimensions (Length and Width) for two rectangles, one for Shape A and one for Shape B, such that their areas match the given values and the perimeter of Shape A is greater than the perimeter of Shape B.

The formula for the area of a rectangle with length $l$ and width $w$ is $A = l \times w$.

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

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For Shape A, Area $= 18$ square units. We need to find pairs of positive integers $(l_A, w_A)$ such that $l_A \times w_A = 18$. Possible integer pairs are:

• $1 \times 18$: Perimeter $= 2(1+18) = 2(19) = 38$ units.
• $2 \times 9$: Perimeter $= 2(2+9) = 2(11) = 22$ units.
• $3 \times 6$: Perimeter $= 2(3+6) = 2(9) = 18$ units.

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For Shape B, Area $= 20$ square units. We need to find pairs of positive integers $(l_B, w_B)$ such that $l_B \times w_B = 20$. Possible integer pairs are:

• $1 \times 20$: Perimeter $= 2(1+20) = 2(21) = 42$ units.
• $2 \times 10$: Perimeter $= 2(2+10) = 2(12) = 24$ units.
• $4 \times 5$: Perimeter $= 2(4+5) = 2(9) = 18$ units.

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Now we need to select a pair of dimensions for Shape A and a pair for Shape B such that Perimeter(A) $>$ Perimeter(B).

Let's compare the perimeters:

• If we choose A as $1 \times 18$ (Perimeter 38): We need B with Perimeter less than 38. Possible B dimensions: $2 \times 10$ (Perimeter 24), $4 \times 5$ (Perimeter 18). Both satisfy $38 > 24$ and $38 > 18$.
• If we choose A as $2 \times 9$ (Perimeter 22): We need B with Perimeter less than 22. Possible B dimensions: $4 \times 5$ (Perimeter 18). This satisfies $22 > 18$.
• If we choose A as $3 \times 6$ (Perimeter 18): We need B with Perimeter less than 18. None of the integer perimeter values for B (42, 24, 18) are less than 18.

We can choose any valid combination from the first two cases. Let's choose the second case:

• Shape A: Dimensions $2$ units $\times 9$ units. Area = $2 \times 9 = 18$ sq units. Perimeter = $2(2+9) = 22$ units.
• Shape B: Dimensions $4$ units $\times 5$ units. Area = $4 \times 5 = 20$ sq units. Perimeter = $2(4+5) = 18$ units.

Here, Area(A) $= 18$, Area(B) $= 20$, and Perimeter(A) $= 22$, Perimeter(B) $= 18$. The condition Perimeter(A) $>$ Perimeter(B) ($22 > 18$) is satisfied.

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How to draw the shapes:

Assuming we are drawing on a grid where each square represents one square unit:

Shape A: Draw a rectangle that is 2 units wide and 9 units long (or 9 units wide and 2 units long). This rectangle will cover $2 \times 9 = 18$ grid squares.

Shape B: Draw a rectangle that is 4 units wide and 5 units long (or 5 units wide and 4 units long). This rectangle will cover $4 \times 5 = 20$ grid squares.

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These two rectangles satisfy the given conditions.

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Alternate Solution Example:

Another valid combination is:

• Shape A: Dimensions $1$ unit $\times 18$ units. Area = $1 \times 18 = 18$ sq units. Perimeter = $2(1+18) = 38$ units.
• Shape B: Dimensions $2$ units $\times 10$ units. Area = $2 \times 10 = 20$ sq units. Perimeter = $2(2+10) = 24$ units.

Here, Area(A) $= 18$, Area(B) $= 20$, and Perimeter(A) $= 38$, Perimeter(B) $= 24$. The condition Perimeter(A) $>$ Perimeter(B) ($38 > 24$) is satisfied.

How to draw these shapes:

Shape A: Draw a rectangle that is 1 unit wide and 18 units long.

Shape B: Draw a rectangle that is 2 units wide and 10 units long.

Given:

Distance of the rectangular border from the top edge of the page = $1 \text{ cm}$.

Distance of the rectangular border from the bottom edge of the page = $1 \text{ cm}$.

Distance of the rectangular border from the left edge of the page = $1.5 \text{ cm}$.

Distance of the rectangular border from the right edge of the page = $1.5 \text{ cm}$.

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To Find:

The perimeter of the border.

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Solution:

Let the length of the book page (from top to bottom) be $L \text{ cm}$.

Let the width of the book page (from left to right) be $W \text{ cm}$.

The border is $1 \text{ cm}$ from the top and $1 \text{ cm}$ from the bottom. This means the length of the border is the length of the page reduced by $1 \text{ cm}$ at the top and $1 \text{ cm}$ at the bottom.

The border is $1.5 \text{ cm}$ from the left and $1.5 \text{ cm}$ from the right. This means the width of the border is the width of the page reduced by $1.5 \text{ cm}$ on the left and $1.5 \text{ cm}$ on the right.

The formula for the perimeter of a rectangle is Perimeter = $2 \times (\text{Length} + \text{Width})$.

Substitute the expressions for $l_{border}$ and $w_{border}$ from equations (2) and (4) into equation (5):

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Answer:

The question asks for "the perimeter", implying a specific numerical value. However, the dimensions of the book page are not provided. The perimeter of the border depends on the dimensions of the page.

Assuming the length of the page is $L \text{ cm}$ and the width of the page is $W \text{ cm}$, the perimeter of the border is given by the formula:

To find a numerical answer, the length and width of the specific book page must be measured.

Given:

Outer rectangle dimensions: 12 units $\times$ 8 units.

Inner rectangle is drawn inside the outer rectangle without touching it.

Area of the inner rectangle is exactly half the area of the outer rectangle.

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To Find:

Describe or draw the two rectangles satisfying the conditions.

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Solution:

First, let's find the area of the outer rectangle.

The formula for the area of a rectangle is Area = Length $\times$ Width.

The area of the inner rectangle is half the area of the outer rectangle.

We need to find dimensions (Length and Width) for the inner rectangle such that their product is 48 square units. Let the dimensions of the inner rectangle be $l_{inner} \times w_{inner}$. Then, $l_{inner} \times w_{inner} = 48$.

Possible pairs of positive integer dimensions for the inner rectangle are (1, 48), (2, 24), (3, 16), (4, 12), (6, 8), and their swapped pairs.

To draw the inner rectangle inside the outer rectangle without touching it, the dimensions of the inner rectangle must be strictly less than the dimensions of the outer rectangle ($l_{inner} < 12$ and $w_{inner} < 8$, or $l_{inner} < 8$ and $w_{inner} < 12$).

Let's check the possible dimensions for the inner rectangle:

• $1 \times 48$: $1 < 8$ (or 12), $48$ is not less than 8 or 12. Cannot fit.
• $2 \times 24$: $2 < 8$ (or 12), $24$ is not less than 8 or 12. Cannot fit.
• $3 \times 16$: $3 < 8$ (or 12), $16$ is not less than 8 or 12. Cannot fit.
• $4 \times 12$: $4 < 8$, but $12$ is not less than 8. If we swap, $12$ is not less than 12. Cannot fit without touching (unless aligned exactly).
• $6 \times 8$: $6 < 8$, but $8$ is not less than 8. Cannot fit without touching (unless aligned exactly).

It appears there might be a misunderstanding in interpreting "without touching". It likely means there must be some space between the boundaries of the two rectangles on all four sides. If we require a minimum gap, say 1 unit, then the maximum dimensions of the inner rectangle must be $(12-2 \times \text{gap}) \times (8-2 \times \text{gap})$. For a 1 unit gap, this would be $10 \times 6$. The area is $60$, which is not 48.

Let's reconsider the interpretation. The question asks to draw *a* rectangle occupying exactly half the area, *without touching* the outer rectangle. This implies there must be some positive distance between the boundaries. If using integer coordinates on a grid, this means the inner rectangle's coordinates $(x_1, y_1)$ to $(x_2, y_2)$ must satisfy $0 < x_1 < x_2 < 12$ and $0 < y_1 < y_2 < 8$ (assuming the outer rectangle is from $(0,0)$ to $(12,8)$). The length of the inner rectangle would be $x_2 - x_1$ and the width $y_2 - y_1$.

However, if we are allowed to use non-integer dimensions for the inner rectangle, any dimensions $l_{inner}, w_{inner}$ such that $l_{inner} \times w_{inner} = 48$ and $l_{inner} < 12$, $w_{inner} < 8$ would work. For example, if we take $l_{inner} = 9.6$ and $w_{inner} = 5$, then $9.6 \times 5 = 48$. Since $9.6 < 12$ and $5 < 8$, a $9.6 \times 5$ rectangle can be placed inside a $12 \times 8$ rectangle with a gap. For instance, centered with a gap of $(12-9.6)/2 = 1.2$ on the sides and $(8-5)/2 = 1.5$ on the top/bottom.

Given the context is likely elementary (Page 149 suggests basic geometry), integer dimensions are usually preferred. Let's assume the question implies a non-integer solution or has a slight ambiguity in "without touching". If "without touching" means no shared boundary points, then $l_{inner} < 12$ and $w_{inner} < 8$ (or vice versa) is sufficient.

Let's use the $6 \times 8$ dimensions for the inner rectangle. Area $= 6 \times 8 = 48$ square units. These dimensions are 6 and 8.

We have outer dimensions 12 and 8.

Can a $6 \times 8$ rectangle fit inside a $12 \times 8$ rectangle without touching? Yes, if we place the $6 \times 8$ rectangle such that its 6-unit side is along the 12-unit side of the outer rectangle, there will be $12 - 6 = 6$ units of space left horizontally. We can position it with 3 units of space on the left and 3 units on the right. Vertically, the 8-unit side of the inner rectangle matches the 8-unit side of the outer rectangle. If "without touching" means there must be a positive gap on all four sides, then the $6 \times 8$ cannot fit. If it means they cannot share any boundary points, then the $6 \times 8$ can fit if its dimensions are strictly less than 12 and 8. The pair $(6, 8)$ has 6 < 12 but 8 is not < 8. So this pair strictly touches on two sides if aligned fully on the vertical.

Perhaps the dimension pairs for the inner rectangle should be strictly less than BOTH outer dimensions when considered in order. $l_{inner} < 12$ AND $w_{inner} < 8$. Let's recheck the pairs for 48:

• (1, 48): $1 < 12$ but $48 \not< 8$. No.
• (2, 24): $2 < 12$ but $24 \not< 8$. No.
• (3, 16): $3 < 12$ but $16 \not< 8$. No.
• (4, 12): $4 < 12$ but $12 \not< 8$. No.
• (6, 8): $6 < 12$ but $8 \not< 8$. No.

This suggests integer dimensions might not work unless the meaning of "without touching" allows sharing edges (which contradicts "without touching"). A likely interpretation for this level is that we need $l_{inner} \le 12$ and $w_{inner} \le 8$, but there should be a visible gap. This forces us to use non-integer dimensions or assume the drawing can be slightly conceptual.

Let's proceed with non-integer dimensions that work mathematically and explain how to draw them with a gap.

Choose inner rectangle dimensions $l_{inner} = 9.6$ units and $w_{inner} = 5$ units. Area $= 9.6 \times 5 = 48$. $9.6 < 12$ and $5 < 8$.

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How to draw the shapes:

1. Draw the outer rectangle with length 12 units and width 8 units. On a grid, this would be a rectangle covering $12 \times 8 = 96$ grid squares.

2. Calculate the required area for the inner rectangle: $96 / 2 = 48$ square units.

3. Choose dimensions for the inner rectangle whose area is 48. For example, choose length $9.6$ units and width $5$ units.

4. Draw the inner rectangle inside the outer rectangle. To ensure it doesn't touch, leave a gap around it. For instance, place the inner rectangle such that it is centered within the outer rectangle.

Gap on the left/right = $(12 - 9.6) / 2 = 2.4 / 2 = 1.2$ units.

Gap on the top/bottom = $(8 - 5) / 2 = 3 / 2 = 1.5$ units.

So, the inner rectangle can be drawn $1.2$ units from the left edge, $1.2$ units from the right edge, $1.5$ units from the top edge, and $1.5$ units from the bottom edge of the outer rectangle.

Alternatively, using simpler numbers but potentially implying non-integer steps in drawing or assuming a different interpretation:

Choose inner rectangle dimensions $l_{inner} = 8$ units and $w_{inner} = 6$ units. Area $= 8 \times 6 = 48$. Here $8 < 12$ but $6 < 8$. We can fit an $8 \times 6$ rectangle inside a $12 \times 8$ rectangle with a gap.

Place the $8 \times 6$ rectangle inside the $12 \times 8$ rectangle. For example, horizontally centered: gap = $(12 - 8) / 2 = 4 / 2 = 2$ units on left/right. Vertically centered: gap = $(8 - 6) / 2 = 2 / 2 = 1$ unit on top/bottom.

This seems a more likely intended solution using simpler numbers, accepting that $l_{inner} < L_{outer}$ and $w_{inner} < W_{outer}$ is the fitting condition.

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Answer (using 8x6 inner rectangle):

1. Draw a rectangle with length 12 units and width 8 units.

2. Calculate the area of this rectangle: $12 \times 8 = 96$ square units.

3. The required area for the inner rectangle is half of this: $96 / 2 = 48$ square units.

4. Choose dimensions for the inner rectangle that multiply to 48, and are smaller than 12 and 8. A suitable choice is length 8 units and width 6 units (since $8 \times 6 = 48$, $8 < 12$, and $6 < 8$).

5. Draw the 8 units $\times$ 6 units rectangle inside the 12 units $\times$ 8 units rectangle, leaving a space around it. For instance, place it 2 units from the left edge, 2 units from the right edge, 1 unit from the top edge, and 1 unit from the bottom edge of the outer rectangle.

Given:

A square piece of paper is folded in half and cut along the fold, resulting in two rectangles.

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To Find:

Which of the given statements is always true, regardless of the size of the square.

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Solution:

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

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

The perimeter of the square is $P_{square} = 4 \times s = 4s$ units.

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When the square is folded in half and cut along the fold, it creates two identical rectangles. The fold must be parallel to one of the sides. This cut divides one dimension of the square ($s$) into two equal parts ($s/2$). The other dimension remains $s$.

So, the dimensions of each of the two rectangles are Length = $s$ units and Width = $\frac{s}{2}$ units.

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Now, let's calculate the area and perimeter of each rectangle.

Area of each rectangle ($A_{rect}$):

Perimeter of each rectangle ($P_{rect}$):

Total area of the two rectangles added together = $A_{rect} + A_{rect} = \frac{s^2}{2} + \frac{s^2}{2} = s^2$ square units.

Total perimeter of the two rectangles added together ($P_{total\_rect}$) = $P_{rect} + P_{rect} = 3s + 3s = 6s$ units.

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Now let's examine each statement:

a. The area of each rectangle is larger than the area of the square.

Is $\frac{s^2}{2} > s^2$? This is false, as $\frac{s^2}{2}$ is half of $s^2$.

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b. The perimeter of the square is greater than the perimeters of both the rectangles added together.

Is $4s > 6s$? This is false (assuming $s>0$).

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c. The perimeters of both the rectangles added together is always $1\frac{1}{2}$ times the perimeter of the square.

Total perimeter of the two rectangles = $6s$.

$1\frac{1}{2}$ times the perimeter of the square = $1.5 \times 4s = \frac{3}{2} \times 4s = 6s$.

Yes, this statement is true for any positive value of $s$.

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d. The area of the square is always three times as large as the areas of both rectangles added together.

Area of the square = $s^2$.

Total area of both rectangles added together = $s^2$.

Is $s^2 = 3 \times s^2$? This is false (assuming $s>0$).

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Based on the analysis, statement (c) is the only one that is always true.

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Answer:

The statement that is always true is:

c. The perimeters of both the rectangles added together is always $1\frac{1}{2}$ times the perimeter of the square.