Area

Answer: Option B
Solution:
Let’s define variables:

– Let the length of the rectangular park be 3x meters.
– Let the breadth of the rectangular park be 2x meters.

Step 1: Finding the Perimeter
The perimeter of the rectangle is given by:
\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})
\] \[
= 2 \times (3x + 2x) = 2 \times 5x = 10x
\]

Step 2: Finding the Total Distance Covered
– The man cycles along the boundary at a speed of 12 km/hr.
– Time taken to complete one round = 8 minutes = 8/60 hours
– Distance covered in this time = Speed × Time
\[
= 12 \times \frac{8}{60} = 12 \times \frac{2}{15} = \frac{24}{15} = 1.6 \text{ km} = 1600 \text{ m}
\]

Since this distance is the perimeter of the park:
\[
10x = 1600
\] \[
x = 160
\]

Step 3: Finding the Area
– Length = 3x = 3 × 160 = 480 m
– Breadth = 2x = 2 × 160 = 320 m
– Area = Length × Breadth
\[
= 480 \times 320 = 153600 \text{ sq. m}
\]

Final Answer:
\[
\mathbf{153600 \text{ sq. m}}
\]

Answer: Option D
Solution:
Let’s define variables:

– Let the actual side of the square be s.
– Due to the 2% excess error, the measured side becomes:
\[
s’ = s + 2\% \text{ of } s = s + \frac{2}{100} s = 1.02s
\]

Step 1: Finding the Actual and Measured Areas

-Actual area of the square:
\[
A = s^2
\] – Measured area with the incorrect side:
\[
A’ = (1.02s)^2 = 1.02^2 \times s^2
\]

Step 2: Calculating the Percentage Error in Area

\[
1.02^2 = 1.0404
\]

So the incorrect area is 4.04% greater than the actual area.

Final Answer:
\[
\mathbf{4.04\%}
\]

Answer: Option B
Solution:
Let the length of the rectangle be L and the breadth be B.

Step 1: Use the given ratio
The perimeter of a rectangle is given by:
\[
P = 2(L + B)
\]

According to the problem:
\[
\frac{P}{B} = \frac{5}{1}
\]

Substituting \(P = 2(L + B)\):

\[
\frac{2(L + B)}{B} = 5
\]

Multiplying both sides by \(B\):

\[
2(L + B) = 5B
\]

Step 2: Express L in terms of B
\[
2L + 2B = 5B
\]

\[
2L = 3B
\]

\[
L = \frac{3B}{2}
\]

Step 3: Use the given area
The area of a rectangle is given by:

\[
A = L \times B
\]

Given \( A = 216 \):

\[
\frac{3B}{2} \times B = 216
\]

\[
\frac{3B^2}{2} = 216
\]

Multiply both sides by 2:

\[
3B^2 = 432
\]

\[
B^2 = 144
\]

\[
B = 12 \text{ cm}
\]

Step 4: Find L
\[
L = \frac{3(12)}{2} = \frac{36}{2} = 18 \text{ cm}
\]

Answer:
The length of the rectangle is 18 cm.

Answer: Option C
Solution:
Let’s assume the original length of the rectangle is L and the original breadth is B.

Step 1: Calculate the Original Area
The original area of the rectangle is:

\[
\text{Original Area} = L \times B
\]

Step 2: Increase the Dimensions by 20%
Since both the length and breadth are increased by 20%, the new dimensions become:

\[
\text{New Length} = 1.2L
\] \[
\text{New Breadth} = 1.2B
\]

Step 3: Calculate the New Area
\[
\text{New Area} = (1.2L) \times (1.2B) = 1.44 (L \times B)
\]

Step 4: Find the Percentage Increase
\[
\text{Percentage Increase} = \left( \frac{\text{New Area} – \text{Original Area}}{\text{Original Area}} \right) \times 100
\]

\[
= \left( \frac{1.44LB – LB}{LB} \right) \times 100
\]

\[
= \left( \frac{0.44LB}{LB} \right) \times 100
\]

\[
= 44\%
\]

Final Answer:
The percentage increase in the area is 44%.

Answer: Option B
Solution:
Step 1: Define Given Information
– Length of the park = 60 m
– Width of the park = 40 m
– Total area of the park = \( 60 \times 40 = 2400 \) sq. m
– Area of the lawn = 2109 sq. m
– Area of the roads = Total park area – Lawn area
\[
2400 – 2109 = 291 \text{ sq. m}
\] – Two concrete crossroads (one along the length, one along the width) intersect at the center.

Step 2: Assume the Width of the Road
Let the width of each road be x meters.

1. First road along the length:
– Length = 60 m
– Width = x
– Area = \( 60x \)

2. Second road along the width:
– Length = 40 m
– Width = x
– Area = \( 40x \)

3. Overlapping (intersection) area (since it gets counted twice):
– A square of x × x
– Area = \( x^2 \)

Step 3: Form the Equation
\[
\text{Total area of roads} = (60x + 40x – x^2) = 291
\]

\[
100x – x^2 = 291
\]

\[
x^2 – 100x + 291 = 0
\]

Step 4: Solve for x
Solving the quadratic equation:

\[
x = \frac{100 \pm \sqrt{(-100)^2 – 4(1)(291)}}{2(1)}
\]

\[
x = \frac{100 \pm \sqrt{10000 – 1164}}{2}
\]

\[
x = \frac{100 \pm \sqrt{8836}}{2}
\]

\[
x = \frac{100 \pm 94}{2}
\]

Solving for two possible values of \(x\):

\[
x = \frac{100 + 94}{2} = \frac{194}{2} = 97 \quad \text{(Not possible, too large)}
\]

\[
x = \frac{100 – 94}{2} = \frac{6}{2} = 3
\]

Final Answer:
The width of the road is 3 meters.

Answer: Option C
Solution:
We are given that the diagonal of the rectangular closet is \( 7\frac{1}{2} \) feet and that the shorter side is \( 4\frac{1}{2} \) feet. We need to find the area of the closet.

Step 1: Convert Mixed Fractions to Improper Fractions
We first convert the given mixed fractions to improper fractions.

\[
7\frac{1}{2} = \frac{15}{2}, \quad 4\frac{1}{2} = \frac{9}{2}
\]

Step 2: Use the Pythagorean Theorem
Since the floor of the closet is a rectangle, we can use the Pythagorean theorem:

\[
a^2 + b^2 = c^2
\]

where:
– \( a = \frac{9}{2} \) (shorter side),
– \( b \) is the unknown longer side,
– \( c = \frac{15}{2} \) (diagonal).

Substituting the values:

\[
\left(\frac{9}{2}\right)^2 + b^2 = \left(\frac{15}{2}\right)^2
\]

\[
\frac{81}{4} + b^2 = \frac{225}{4}
\]

Step 3: Solve for \( b^2 \)
Rearrange the equation:

\[
b^2 = \frac{225}{4} – \frac{81}{4}
\]

\[
b^2 = \frac{144}{4} = 36
\]

\[
b = \sqrt{36} = 6
\]

Step 4: Find the Area of the Closet
The area of the rectangular closet is:

\[
A = a \times b = \left(\frac{9}{2}\right) \times 6
\]

\[
A = \frac{54}{2} = 27 \text{ square feet}
\]

Thus, the area of the closet is 27 square feet.

Answer: Option D
Solution:
We are given that a towel loses 20% of its length and 10% of its breadth due to bleaching. We need to determine the percentage decrease in area.

Step 1: Let Initial Dimensions Be \( L \) and \( B \)
Let the original length of the towel be \( L \) and the original breadth be \( B \).

The original area of the towel:

\[
A_{\text{original}} = L \times B
\]

Step 2: Compute the New Dimensions
– Since the length decreases by 20%, the new length is:

\[
L_{\text{new}} = L – 0.2L = 0.8L
\]

– Since the breadth decreases by 10%, the new breadth is:

\[
B_{\text{new}} = B – 0.1B = 0.9B
\]

The new area of the towel:

\[
A_{\text{new}} = (0.8L) \times (0.9B) = 0.72LB
\]

Step 3: Compute the Percentage Decrease in Area
The decrease in area is:

\[
\text{Decrease in area} = A_{\text{original}} – A_{\text{new}}
\]

\[
= LB – 0.72LB
\]

\[
= (1 – 0.72)LB = 0.28LB
\]

The **percentage decrease in area is:

\[
\frac{0.28LB}{LB} \times 100 = 28\%
\]

Final Answer:
The percentage decrease in area is 28%.

Answer: Option C
Solution:
Let’s break it down again and calculate the percentage saved.

Step 1: Define the side length
Let the side length of the square lot be \( s \).

Step 2: Compare the two distances
– If the man walks along the edges, the total distance traveled is simply the sum of the two sides of the square, which is \( 2s \).
– If the man walks diagonally across the square, we use the Pythagorean theorem to calculate the diagonal distance:

\[
\text{Diagonal} = \sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}
\]

Step 3: Calculate the distance saved
The distance saved by walking diagonally instead of along the edges is:

\[
\text{Distance saved} = 2s – s\sqrt{2}
\]

Step 4: Calculate the percentage saved
To find the percentage saved, divide the distance saved by the total distance along the edges and multiply by 100:

\[
\text{Percentage saved} = \frac{2s – s\sqrt{2}}{2s} \times 100
\]

Simplifying:

\[
\text{Percentage saved} = \frac{2 – \sqrt{2}}{2} \times 100
\]

Approximating \( \sqrt{2} \approx 1.414 \):

\[
\text{Percentage saved} = \frac{2 – 1.414}{2} \times 100
\]

\[
\text{Percentage saved} \approx \frac{0.586}{2} \times 100
\]

\[
\text{Percentage saved} \approx 29.3\%
\]

Final Answer:
The man saved approximately 29.3% by walking diagonally across the square lot.

Answer: Option B
Solution:
We are given the following information about a rectangle:
– The diagonal of the rectangle is \( \sqrt{41} \) cm,
– The area of the rectangle is 20 square cm.

We need to find the perimeter of the rectangle.

Step 1: Use the Pythagorean Theorem
Let the length and breadth of the rectangle be \( l \) and \( b \), respectively. The diagonal \( d \) is given by the Pythagorean theorem:

\[
d^2 = l^2 + b^2
\]

Since \( d = \sqrt{41} \), we have:

\[
(\sqrt{41})^2 = l^2 + b^2
\]

\[
41 = l^2 + b^2
\]

Step 2: Use the Area of the Rectangle
The area of the rectangle is given by:

\[
\text{Area} = l \times b = 20
\]

Thus, we have two equations:
1. \( l^2 + b^2 = 41 \)
2. \( l \times b = 20 \)

Step 3: Solve the System of Equations
To solve this system, we can use the identity \( (l + b)^2 = l^2 + b^2 + 2lb \):

\[
(l + b)^2 = 41 + 2(20) = 41 + 40 = 81
\]

\[
l + b = \sqrt{81} = 9
\]

Step 4: Find the Perimeter
The perimeter \( P \) of the rectangle is given by:

\[
P = 2(l + b) = 2 \times 9 = 18 \text{ cm}
\]

Final Answer:
The perimeter of the rectangle is 18 cm.

Answer: Option A
Solution:
We are given the dimensions of a room as:
– Length = 15 meters 17 centimeters
– Breadth = 9 meters 2 centimeters

We need to find the least number of square tiles required to completely cover the floor.

Step 1: Convert Dimensions to a Consistent Unit
To simplify the calculation, let’s convert the dimensions into centimeters.

– Length: \( 15 \, \text{m} 17 \, \text{cm} = 15 \times 100 + 17 = 1517 \, \text{cm} \)
– Breadth: \( 9 \, \text{m} 2 \, \text{cm} = 9 \times 100 + 2 = 902 \, \text{cm} \)

Thus, the dimensions of the floor are:
– Length = 1517 cm
– Breadth = 902 cm

Step 2: Find the Greatest Common Divisor (GCD)
The least number of square tiles required will correspond to the largest possible square tile that can fit perfectly along both dimensions of the floor. To find this, we need to determine the GCD of 1517 and 902.

Let’s calculate the GCD of 1517 and 902.

Step 3: Calculate the Area of the Floor
The area of the floor is:

\[
\text{Area of the floor} = 1517 \times 902 \, \text{cm}^2
\]

Step 4: Find the Area of the Square Tile
The area of the square tile is the square of the side length, which is the GCD of 1517 and 902.

Step 5: Calculate the Number of Tiles
Finally, we divide the area of the floor by the area of the tile to get the least number of square tiles required.

Let me calculate that for you.

The least number of square tiles required to pave the floor of the room is 814 tiles.

Answer: Option D
Solution:
We are given the following information about a rectangle:
– The difference between the length and breadth is 23 m,
– The perimeter of the rectangle is 206 m.

We need to find the area of the rectangle.

Step 1: Set up equations
Let the length of the rectangle be \( l \) and the breadth be \( b \). We are given:
1. \( l – b = 23 \) (The difference between length and breadth is 23 m)
2. The perimeter formula for a rectangle is:

\[
\text{Perimeter} = 2(l + b) = 206
\]

So,

\[
l + b = \frac{206}{2} = 103
\]

Step 2: Solve the system of equations
We have the following system of equations:
1. \( l – b = 23 \)
2. \( l + b = 103 \)

We can solve this system by adding the two equations:

\[
(l – b) + (l + b) = 23 + 103
\]

\[
2l = 126
\]

\[
l = \frac{126}{2} = 63
\]

Now, substitute \( l = 63 \) into the second equation \( l + b = 103 \):

\[
63 + b = 103
\]

\[
b = 103 – 63 = 40
\]

Step 3: Find the Area
The area \( A \) of the rectangle is given by:

\[
A = l \times b = 63 \times 40 = 2520 \, \text{m}^2
\] Final Answer:
The area of the rectangle is 2520 square meters.

Answer: Option B
Solution:
We are given that the length of a rectangle is halved, and its breadth is tripled. We need to determine the percentage change in the area of the rectangle.

Step 1: Define the original dimensions
Let the original **length** of the rectangle be \( l \) and the **breadth** be \( b \). The original area \( A_{\text{original}} \) is:

\[
A_{\text{original}} = l \times b
\]

Step 2: New dimensions after the changes
– The new length is \( \frac{l}{2} \).
– The new breadth is \( 3b \).

The new area \( A_{\text{new}} \) is:

\[
A_{\text{new}} = \left( \frac{l}{2} \right) \times (3b) = \frac{3}{2} \times l \times b = \frac{3}{2} A_{\text{original}}
\]

Step 3: Find the percentage change in area
The percentage change in area is:

\[
\text{Percentage change} = \frac{A_{\text{new}} – A_{\text{original}}}{A_{\text{original}}} \times 100
\]

Substitute \( A_{\text{new}} = \frac{3}{2} A_{\text{original}} \):

\[
\text{Percentage change} = \frac{\frac{3}{2} A_{\text{original}} – A_{\text{original}}}{A_{\text{original}}} \times 100
\]

\[
= \frac{\frac{3}{2} A_{\text{original}} – \frac{2}{2} A_{\text{original}}}{A_{\text{original}}} \times 100
\]

\[
= \frac{\frac{1}{2} A_{\text{original}}}{A_{\text{original}}} \times 100
\]

\[
= \frac{1}{2} \times 100 = 50\%
\]

Final Answer:
The area of the rectangle increases by 50%.

Answer: Option E
Solution:
We are given that:
– The length of the rectangular plot is 20 meters more than its breadth,
– The cost of fencing the plot is Rs. 5300, with the cost of fencing being Rs. 26.50 per meter.

We need to find the length of the rectangular plot.

Step 1: Define variables
Let the breadth of the rectangle be \( b \) meters. Then, the length of the rectangle will be \( l = b + 20 \) meters.

Step 2: Use the perimeter formula
The perimeter of a rectangle is given by:

\[
P = 2(l + b)
\]

Substituting \( l = b + 20 \):

\[
P = 2((b + 20) + b) = 2(2b + 20)
\]

\[
P = 4b + 40
\]

Step 3: Use the cost of fencing
The total cost of fencing is Rs. 5300, and the cost per meter of fencing is Rs. 26.50. The total length of fencing is the perimeter of the rectangle. Therefore, the perimeter can be calculated by:

\[
\text{Cost} = \text{Perimeter} \times \text{Cost per meter}
\]

Substitute the values:

\[
5300 = (4b + 40) \times 26.50
\]

Step 4: Solve for \( b \)
Solve for \( b \):

\[
5300 = (4b + 40) \times 26.50
\]

\[
5300 = 26.50(4b + 40)
\]

Divide both sides by 26.50:

\[
\frac{5300}{26.50} = 4b + 40
\]

\[
200 = 4b + 40
\]

\[
4b = 200 – 40 = 160
\]

\[
b = \frac{160}{4} = 40
\]

Step 5: Find the length
Now that we know the breadth is \( b = 40 \) meters, the length is:

\[
l = b + 20 = 40 + 20 = 60 \, \text{meters}
\]

Final Answer:
The length of the plot is 60 meters.

Answer: Option D
Solution:
We are given:
– The area of a rectangular field is 680 square feet.
– The field is to be fenced on three sides, leaving one side (the longer side) uncovered.
– The side that is uncovered has a length of 20 feet.

We need to determine how many feet of fencing will be required.

Step 1: Define Variables
Let the length of the rectangular field be \( l \) and the breadth be \( b \). Since the uncovered side is the longer side, we assume that \( l = 20 \) feet.

The area of the rectangle is given by:

\[
\text{Area} = l \times b
\]

Substituting \( l = 20 \):

\[
680 = 20 \times b
\]

Solving for \( b \):

\[
b = \frac{680}{20} = 34 \, \text{feet}
\]

Step 2: Calculate the Amount of Fencing Required
Since the field is to be fenced on three sides (two sides of the breadth and one side of the length), the total length of fencing required is:

\[
\text{Fencing required} = 2b + l = 2 \times 34 + 20 = 68 + 20 = 88 \, \text{feet}
\]

Final Answer:
The amount of fencing required is 88 feet.

Answer: Option C
Solution:
We are given the dimensions of the tank:
– Length = 25 m,
– Width = 12 m,
– Depth = 6 m.

We need to find the cost of plastering the walls and bottom of the tank, at a rate of 75 paise per square meter.

Step 1: Calculate the Area to Be Plastered
The area to be plastered includes the walls and the bottom of the tank. The tank has:
– Two walls of dimensions 25 m by 6 m (length × depth),
– Two walls of dimensions 12 m by 6 m (width × depth),
– The bottom of the tank, which is 25 m by 12 m (length × width).

Area of the Walls:
1. Two walls of 25 m by 6 m:
\[
\text{Area of two walls} = 2 \times (25 \times 6) = 2 \times 150 = 300 \, \text{sq. m}
\]

2. Two walls of 12 m by 6 m:
\[
\text{Area of two walls} = 2 \times (12 \times 6) = 2 \times 72 = 144 \, \text{sq. m}
\]

Area of the Bottom:
3.Bottom of the tank (25 m by 12 m):
\[
\text{Area of the bottom} = 25 \times 12 = 300 \, \text{sq. m}
\]

Step 2: Total Area to Be Plastered
The total area to be plastered is the sum of the areas of the walls and the bottom:

\[
\text{Total area} = 300 + 144 + 300 = 744 \, \text{sq. m}
\]

Step 3: Calculate the Cost of Plastering
The cost of plastering is given as 75 paise per square meter. Since 1 rupee = 100 paise, the cost per square meter in rupees is:

\[
\text{Cost per sq. m} = \frac{75}{100} = 0.75 \, \text{rupees}
\]

Therefore, the total cost is:

\[
\text{Total cost} = 744 \times 0.75 = 558 \, \text{rupees}
\]

Final Answer:
The cost of plastering the walls and bottom of the tank is 558 rupees.

Answer: Option D
Solution:
We are given the following:
– The length of the room is 5.5 meters,
– The width of the room is 3.75 meters,
– The cost of paving the floor with slabs is Rs. 800 per square meter.

We need to find the cost of paving the floor.

Step 1: Calculate the Area of the Floor
The area of the floor is given by the formula:

\[
\text{Area} = \text{Length} \times \text{Width}
\]

Substitute the given values:

\[
\text{Area} = 5.5 \times 3.75 = 20.625 \, \text{square meters}
\]

Step 2: Calculate the Cost of Paving
The cost of paving is Rs. 800 per square meter. Therefore, the total cost is:

\[
\text{Total cost} = \text{Area} \times \text{Cost per square meter}
\]

\[
\text{Total cost} = 20.625 \times 800 = 16,500 \, \text{rupees}
\]

Final Answer:
The cost of paving the floor is Rs. 16,500.

Answer: Option C
Solution:
We are given the following information:
– The artist has completed one-fourth of a rectangular oil painting,
– When the artist paints an additional 100 square centimeters, the total area completed will be three-quarters of the painting,
– The height of the oil painting is 10 cm.

We need to find the length of the oil painting.

Step 1: Define the Variables
Let the length of the painting be \( l \) cm, and the height is given as 10 cm.

The area of the rectangle is:

\[
\text{Area of the painting} = \text{Length} \times \text{Height} = l \times 10
\]

Step 2: Set Up the Equation for the Areas
The artist has completed one-fourth of the total area, and after painting an additional 100 square centimeters, he will have completed three-quarters of the painting.

Let’s express this in terms of the area:
– The area completed after painting the additional 100 cm² is \( \frac{3}{4} \) of the total area.
– The area already completed is \( \frac{1}{4} \) of the total area.

Thus, the additional 100 cm² completed is the difference between \( \frac{3}{4} \) and \( \frac{1}{4} \), which is:

\[
\frac{3}{4} \times (\text{Total area}) – \frac{1}{4} \times (\text{Total area}) = 100
\]

This simplifies to:

\[
\frac{2}{4} \times (\text{Total area}) = 100
\]

\[
\frac{1}{2} \times (\text{Total area}) = 100
\]

Thus, the total area of the painting is:

\[
\text{Total area} = 100 \times 2 = 200 \, \text{cm}^2
\]

Step 3: Find the Length
We know the total area is 200 cm², and the height of the painting is 10 cm. Using the area formula:

\[
\text{Area} = \text{Length} \times \text{Height}
\]

Substitute the known values:

\[
200 = l \times 10
\]

Solve for \( l \):

\[
l = \frac{200}{10} = 20 \, \text{cm}
\]

Final Answer:
The length of the oil painting is 20 cm.

Answer: Option A
Solution:
We are given that the length of a rectangle is increased by 60%, and we need to find by what percent the width should be decreased to maintain the same area.

Step 1: Define Variables
Let the original length of the rectangle be \( l \) and the original width be \( w \). The original area of the rectangle is:

\[
\text{Original Area} = l \times w
\]

Step 2: Increase in Length
The length is increased by 60%, so the new length becomes:

\[
\text{New Length} = l + 0.60l = 1.60l
\]

Step 3: Maintain the Same Area
We want to maintain the same area, so the new area should be equal to the original area. The new area is:

\[
\text{New Area} = 1.60l \times w_{\text{new}}
\]

Since the area remains the same, we have:

\[
1.60l \times w_{\text{new}} = l \times w
\]

Step 4: Solve for the New Width
Cancel out \( l \) from both sides:

\[
1.60 \times w_{\text{new}} = w
\]

Solve for \( w_{\text{new}} \):

\[
w_{\text{new}} = \frac{w}{1.60}
\]

\[
w_{\text{new}} = 0.625w
\]

Step 5: Calculate the Percentage Decrease in Width
The width decreases from \( w \) to \( 0.625w \), so the decrease in width is:

\[
\text{Decrease in width} = w – 0.625w = 0.375w
\]

The percentage decrease is:

\[
\text{Percentage decrease} = \frac{0.375w}{w} \times 100 = 37.5\%
\]

Final Answer:
The width must be decreased by 37.5% to maintain the same area.

Answer: Option B
Solution:
We are given the following:

– The area of a rectangular plot is \( 45 \, \text{m} \times 40 \, \text{m} \),
– We need to find the length of the diagonal of a square plot whose area is equal to the area of the rectangular plot.

Step 1: Calculate the Area of the Rectangular Plot
The area of the rectangular plot is:

\[
\text{Area of the rectangle} = \text{Length} \times \text{Breadth} = 45 \times 40 = 1800 \, \text{sq. meters}
\]

Step 2: Area of the Square Plot
The area of the square plot is also 1800 square meters. The area of a square is given by:

\[
\text{Area of the square} = \text{Side}^2
\]

Let the side of the square be \( s \). Thus:

\[
s^2 = 1800
\]

Solve for \( s \):

\[
s = \sqrt{1800} \approx 42.43 \, \text{meters}
\]

Step 3: Calculate the Diagonal of the Square
The diagonal \( d \) of a square is related to the side length \( s \) by the Pythagorean theorem:

\[
d = \sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}
\]

Substitute \( s = 42.43 \):

\[
d = 42.43 \times \sqrt{2} \approx 42.43 \times 1.414 \approx 59.99 \, \text{meters}
\]

Final Answer:
The length of the diagonal of the square plot is approximately 60 meters.

Answer: Option D
Solution:
We are given two square fields:
– The area of the first square field is 1 hectare.
– The second square field is broader by 1% compared to the first one.

We need to find the difference in their areas.

Step 1: Convert 1 hectare to square meters
1 hectare is equal to 10,000 square meters. So, the area of the first square field is:

\[
\text{Area of first field} = 10,000 \, \text{sq. meters}
\]

Step 2: Side length of the first field
The area of a square is given by \( \text{Area} = \text{Side}^2 \), so the side length of the first field is:

\[
\text{Side of first field} = \sqrt{10,000} = 100 \, \text{meters}
\]

Step 3: Side length of the second field
The second field is 1% broader than the first one, which means its side length is 1% greater than 100 meters. So, the side length of the second field is:

\[
\text{Side of second field} = 100 \times (1 + 0.01) = 100 \times 1.01 = 101 \, \text{meters}
\]

Step 4: Area of the second field
The area of the second field is:

\[
\text{Area of second field} = 101^2 = 10201 \, \text{sq. meters}
\]

Step 5: Difference in areas
The difference in their areas is:

\[
\text{Difference} = 10201 – 10,000 = 201 \, \text{sq. meters}
\]

Final Answer:
The difference in the areas of the two square fields is 201 square meters.

Answer: Option C
Solution:
We are given two squares with the following conditions:

1.Twenty-nine times the area of the first square is one square meter less than six times the area of the second square.
2. Nine times the side of the first square exceeds the perimeter of the second square by 1 meter.

We need to find the **difference in the sides of these squares.

Step 1: Define Variables
Let:
– \( s_1 \) be the side of the first square,
– \( s_2 \) be the side of the second square.

Step 2: Translate the Conditions into Equations
Condition 1:
“Twenty-nine times the area of the first square is one square meter less than six times the area of the second square.”

The area of the first square is \( s_1^2 \), and the area of the second square is \( s_2^2 \). This condition can be written as:

\[
29 \times s_1^2 = 6 \times s_2^2 – 1 \quad \text{(Equation 1)}
\]

Condition 2:
“Nine times the side of the first square exceeds the perimeter of the second square by 1 meter.”

The perimeter of a square is four times the side length. Therefore, the perimeter of the second square is \( 4s_2 \). This condition can be written as:

\[
9 \times s_1 = 4 \times s_2 + 1 \quad \text{(Equation 2)}
\]

Step 3: Solve the System of Equations
We now have the system of two equations:
1. \( 29s_1^2 = 6s_2^2 – 1 \)
2. \( 9s_1 = 4s_2 + 1 \)

Solve Equation 2 for \( s_2 \):
From Equation 2:

\[
9s_1 = 4s_2 + 1
\]

Solve for \( s_2 \):

\[
4s_2 = 9s_1 – 1
\]

\[
s_2 = \frac{9s_1 – 1}{4} \quad \text{(Equation 3)}
\]

Substitute Equation 3 into Equation 1:
Now substitute \( s_2 = \frac{9s_1 – 1}{4} \) into Equation 1:

\[
29s_1^2 = 6\left( \frac{9s_1 – 1}{4} \right)^2 – 1
\]

First, expand \( \left( \frac{9s_1 – 1}{4} \right)^2 \):

\[
\left( \frac{9s_1 – 1}{4} \right)^2 = \frac{(9s_1 – 1)^2}{16}
\]

Expand \( (9s_1 – 1)^2 \):

\[
(9s_1 – 1)^2 = 81s_1^2 – 18s_1 + 1
\]

Thus:

\[
\left( \frac{9s_1 – 1}{4} \right)^2 = \frac{81s_1^2 – 18s_1 + 1}{16}
\]

Substitute this into Equation 1:

\[
29s_1^2 = 6 \times \frac{81s_1^2 – 18s_1 + 1}{16} – 1
\]

Simplify the right side:

\[
29s_1^2 = \frac{6(81s_1^2 – 18s_1 + 1)}{16} – 1
\]

\[
29s_1^2 = \frac{486s_1^2 – 108s_1 + 6}{16} – 1
\]

Multiply through by 16 to eliminate the fraction:

\[
464s_1^2 = 486s_1^2 – 108s_1 + 6 – 16
\]

\[
464s_1^2 = 486s_1^2 – 108s_1 – 10
\]

Move all terms to one side:

\[
464s_1^2 – 486s_1^2 + 108s_1 + 10 = 0
\]

\[
-22s_1^2 + 108s_1 + 10 = 0
\]

### Step 4: Solve the Quadratic Equation
Now, solve the quadratic equation:

\[
22s_1^2 – 108s_1 – 10 = 0
\]

Using the quadratic formula:

\[
s_1 = \frac{-(-108) \pm \sqrt{(-108)^2 – 4(22)(-10)}}{2(22)}
\]

\[
s_1 = \frac{108 \pm \sqrt{11664 + 880}}{44}
\]

\[
s_1 = \frac{108 \pm \sqrt{12544}}{44}
\]

\[
s_1 = \frac{108 \pm 112}{44}
\]

Thus, we have two possible solutions:

\[
s_1 = \frac{108 + 112}{44} = \frac{220}{44} = 5
\]

or

\[
s_1 = \frac{108 – 112}{44} = \frac{-4}{44} = -\frac{1}{11} \quad (\text{which is not a valid solution since side length cannot be negative})
\]

Thus, \( s_1 = 5 \) meters.

Step 5: Find \( s_2 \)
Substitute \( s_1 = 5 \) into Equation 3:

\[
s_2 = \frac{9(5) – 1}{4} = \frac{45 – 1}{4} = \frac{44}{4} = 11 \, \text{meters}
\]

Step 6: Find the Difference in Sides
The difference in the sides of the squares is:

\[
|s_2 – s_1| = |11 – 5| = 6 \, \text{meters}
\]

Final Answer:
The difference in the sides of the squares is 6 meters.

Answer: Option C
Solution:
We are given an isosceles triangle with the following properties:

– The measure of each of the equal sides is \( 10 \, \text{cm} \),
– The angle between the two equal sides is \( 45^\circ \).

We need to find the area of the triangle.

Step 1: Use the Formula for the Area of a Triangle
The area of a triangle can be found using the following formula when two sides and the included angle are known:

\[
\text{Area} = \frac{1}{2} \times a \times b \times \sin(\theta)
\]

Where:
– \( a \) and \( b \) are the two equal sides (in this case, both are \( 10 \, \text{cm} \)),
– \( \theta \) is the included angle (in this case, \( 45^\circ \)).

Step 2: Substitute the Known Values
Substitute \( a = 10 \), \( b = 10 \), and \( \theta = 45^\circ \) into the formula:

\[
\text{Area} = \frac{1}{2} \times 10 \times 10 \times \sin(45^\circ)
\]

We know that \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), so:

\[
\text{Area} = \frac{1}{2} \times 10 \times 10 \times \frac{\sqrt{2}}{2}
\]

\[
\text{Area} = 50 \times \frac{\sqrt{2}}{2}
\]

\[
\text{Area} = 25\sqrt{2} \, \text{sq. cm}
\]

Final Answer:
The area of the triangle is \( 25\sqrt{2} \, \text{sq. cm} \), or approximately \( 35.36 \, \text{sq. cm} \).

Answer: Option D
Solution:
We are given that a triangle and a parallelogram are constructed on the same base, and their areas are equal. The altitude (height) of the parallelogram is given as 100 meters, and we need to find the altitude of the triangle.

Step 1: Recall the Formulas for Area
– The area of a **parallelogram** is given by:

\[
\text{Area of parallelogram} = \text{Base} \times \text{Height}
\]

– The area of a triangle is given by:

\[
\text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]

Step 2: Set up the Equation
Let the base of both the parallelogram and the triangle be \( b \), and let the altitude of the triangle be \( h_{\text{triangle}} \). Since the areas of the parallelogram and the triangle are equal, we can equate the two areas:

\[
\text{Area of parallelogram} = \text{Area of triangle}
\]

\[
b \times 100 = \frac{1}{2} \times b \times h_{\text{triangle}}
\]

Step 3: Solve for the Altitude of the Triangle
Cancel out the base \( b \) from both sides (assuming \( b \neq 0 \)):

\[
100 = \frac{1}{2} \times h_{\text{triangle}}
\]

Now, multiply both sides by 2:

\[
200 = h_{\text{triangle}}
\]

Final Answer:
The altitude of the triangle is 200 meters.

Answer: Option C
Solution:
We are given the area of a circle as \( 24.64 \, \text{m}^2 \), and we need to find its circumference.

Step 1: Use the Formula for the Area of a Circle
The area of a circle is given by the formula:

\[
A = \pi r^2
\]

Where \( r \) is the radius of the circle.

We are given \( A = 24.64 \, \text{m}^2 \), so:

\[
24.64 = \pi r^2
\]

Step 2: Solve for the Radius
We know that \( \pi \approx 3.1416 \), so:

\[
24.64 = 3.1416 r^2
\]

Now, divide both sides by \( \pi \):

\[
r^2 = \frac{24.64}{3.1416} \approx 7.85
\]

Take the square root of both sides to find the radius:

\[
r = \sqrt{7.85} \approx 2.8 \, \text{meters}
\]

Step 3: Use the Formula for the Circumference of a Circle
The circumference of a circle is given by the formula:

\[
C = 2\pi r
\]

Substitute \( r \approx 2.8 \) meters into this formula:

\[
C = 2 \times 3.1416 \times 2.8 \approx 17.6 \, \text{meters}
\]

Final Answer:
The circumference of the circle is approximately 17.6 meters.

Answer: Option B
Solution:
The ratio of the radii of two circles is given as \( 3 : 2 \).

The circumference of a circle is given by the formula:

\[
C = 2\pi r
\]

Where:
– \( C \) is the circumference,
– \( r \) is the radius,
– \( \pi \) is a constant.

Step 1: Find the Ratio of the Circumferences
Since the circumference is directly proportional to the radius, the ratio of the circumferences will be the same as the ratio of the radii.

Thus, the ratio of the circumferences of the two circles is:

\[
\frac{C_1}{C_2} = \frac{r_1}{r_2}
\]

Given that the ratio of the radii is \( 3 : 2 \), the ratio of the circumferences is also:

\[
\frac{C_1}{C_2} = \frac{3}{2}
\]

Final Answer:
The ratio of the circumferences of the two circles is \( 3 : 2 \).

Answer: Option C
Solution:
We are given:
– The diameter of the circular grassy plot is 42 cm.
– The width of the path around the plot is 3.5 m.
– The cost of gravelling the path is Rs. 4 per square meter.

We need to find the total cost of gravelling the path.

Step 1: Convert the Diameter to Radius
The radius of the plot is half the diameter:

\[
\text{Radius of the plot} = \frac{42 \, \text{cm}}{2} = 21 \, \text{cm}
\]

Convert the radius from centimeters to meters (since the path’s width is in meters):

\[
\text{Radius of the plot} = \frac{21}{100} = 0.21 \, \text{m}
\]

Step 2: Calculate the Outer Radius (Including the Path)
The path around the plot is 3.5 m wide. Therefore, the outer radius (including the plot and the path) is:

\[
\text{Outer radius} = \text{Radius of the plot} + \text{Width of the path} = 0.21 \, \text{m} + 3.5 \, \text{m} = 3.71 \, \text{m}
\]

Step 3: Calculate the Areas of the Two Circles
We can now calculate the area of the larger circle (with the path) and the area of the smaller circle (just the grassy plot).

– The area of the larger circle (including the path) is:

\[
A_{\text{outer}} = \pi r_{\text{outer}}^2 = \pi (3.71)^2 \approx 3.1416 \times 13.7641 \approx 43.2 \, \text{m}^2
\]

– The area of the smaller circle (just the grassy plot) is:

\[
A_{\text{plot}} = \pi r_{\text{plot}}^2 = \pi (0.21)^2 \approx 3.1416 \times 0.0441 \approx 0.1385 \, \text{m}^2
\]

Step 4: Calculate the Area of the Path
The area of the path is the difference between the area of the larger circle and the area of the smaller circle:

\[
A_{\text{path}} = A_{\text{outer}} – A_{\text{plot}} = 43.2 \, \text{m}^2 – 0.1385 \, \text{m}^2 \approx 43.0615 \, \text{m}^2
\]

Step 5: Calculate the Cost of Gravelling the Path
The cost of gravelling the path is Rs. 4 per square meter. Therefore, the total cost is:

\[
\text{Cost} = A_{\text{path}} \times 4 = 43.0615 \times 4 \approx 172.246 \, \text{Rs.}
\]

Final Answer:
The cost of gravelling the path is approximately Rs. 172.25.

Answer: Option B
Solution:
We are given a rectangular field with:
– Length = 20 m,
– Width = 16 m,
and a horse is tied at the corner of the field with a rope of length 14 m. We need to find the area that the horse can graze.

Step 1: Understand the Grazing Area
The horse is tied at the corner of the field, so it can only graze in a quarter of a circle, as the rope extends in a circular manner. The radius of this circle is the length of the rope, which is 14 meters.

Step 2: Area of the Quarter Circle
The area of a full circle is given by the formula:

\[
A = \pi r^2
\]

Where:
– \( r \) is the radius of the circle (14 m),
– \( \pi \approx 3.1416 \).

The area of the full circle would be:

\[
A_{\text{full}} = \pi \times (14)^2 = 3.1416 \times 196 \approx 615.44 \, \text{m}^2
\]

Since the horse can graze only a quarter of the circle, the area the horse can graze is:

\[
A_{\text{grazed}} = \frac{1}{4} \times A_{\text{full}} = \frac{1}{4} \times 615.44 \approx 153.86 \, \text{m}^2
\]

Final Answer:
The area that the horse can graze is approximately 153.86 square meters.

Answer: Option D
Solution:
To find the radius of the circle in which a triangle is inscribed, we can use the following formula for the radius of the circumscribed circle (circumradius) of a triangle:

\[
R = \frac{abc}{4A}
\]

Where:
– \( a \), \( b \), and \( c \) are the lengths of the sides of the triangle,
– \( A \) is the area of the triangle.

Step 1: Use Heron’s Formula to Find the Area of the Triangle
First, we need to calculate the area \( A \) of the triangle with sides 13 cm, 14 cm, and 15 cm. To do this, we will use Heron’s formula.

The semi-perimeter \( s \) of the triangle is given by:

\[
s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21 \, \text{cm}
\]

Now, the area \( A \) of the triangle is:

\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]

Substitute the values:

\[
A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6}
\]

\[
A = \sqrt{7056} = 84 \, \text{cm}^2
\]

Step 2: Calculate the Circumradius
Now that we have the area \( A = 84 \, \text{cm}^2 \), we can use the formula for the circumradius:

\[
R = \frac{abc}{4A}
\]

Substitute the values:

\[
R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336} \approx 8.125 \, \text{cm}
\]

Final Answer:
The radius of the circle in which the triangle is inscribed is approximately 8.125 cm.

Answer: Option C
Solution:
We are given:
– The radius of the circle is \( 28 \, \text{m} \),
– The skating champion moves along the circumference of the circle in 44 seconds.

We need to find how many seconds it will take her to move along the perimeter of a hexagon with a side length of \( 48 \, \text{m} \).

Step 1: Calculate the Speed of the Skating Champion
The distance she travels in 44 seconds is the circumference of the circle, which is given by the formula:

\[
C = 2 \pi r
\]

Substitute the given radius \( r = 28 \, \text{m} \):

\[
C = 2 \pi \times 28 \approx 2 \times 3.1416 \times 28 \approx 175.84 \, \text{m}
\]

The speed of the skating champion is the distance traveled per second, which is:

\[
\text{Speed} = \frac{175.84 \, \text{m}}{44 \, \text{sec}} \approx 4 \, \text{m/sec}
\]

Step 2: Calculate the Perimeter of the Hexagon
The perimeter \( P \) of a regular hexagon is the sum of the lengths of all six sides:

\[
P = 6 \times \text{side length} = 6 \times 48 = 288 \, \text{m}
\]

Step 3: Calculate the Time to Move Along the Perimeter of the Hexagon
Now that we know the skating champion’s speed is \( 4 \, \text{m/sec} \), we can calculate the time it will take her to move along the perimeter of the hexagon:

\[
\text{Time} = \frac{\text{Perimeter of hexagon}}{\text{Speed}} = \frac{288 \, \text{m}}{4 \, \text{m/sec}} = 72 \, \text{sec}
\]

Final Answer:
It will take her 72 seconds to move along the perimeter of the hexagon.

Answer: Option A
Solution:
We are given:
– The length of the floor is 20 feet,
– The breadth of the floor is 10 feet,
– Square tiles of 2 feet length are to be laid on the floor,
– Black tiles are laid along the first row on all sides,
– White tiles are laid in one-third of the remaining area, and
– Blue tiles are laid in the rest of the remaining area.

We need to determine how many blue tiles will be there.

Step 1: Calculate the Total Number of Tiles on the Floor
The area of the floor is:

\[
\text{Area of floor} = 20 \times 10 = 200 \, \text{sq. feet}
\]

Each tile has a side length of 2 feet, so the area of each tile is:

\[
\text{Area of each tile} = 2 \times 2 = 4 \, \text{sq. feet}
\]

Now, the total number of tiles on the floor is:

\[
\text{Total number of tiles} = \frac{\text{Area of floor}}{\text{Area of each tile}} = \frac{200}{4} = 50 \, \text{tiles}
\]

Step 2: Lay the Black Tiles
The black tiles are laid along the first row on all sides. Let’s first determine the number of tiles on the perimeter.

The length of the perimeter is:

\[
\text{Perimeter of the room} = 2 \times (20 + 10) = 60 \, \text{feet}
\]

Since each tile is 2 feet in length, the number of black tiles along the perimeter is:

\[
\text{Number of black tiles} = \frac{\text{Perimeter}}{\text{Length of one tile}} = \frac{60}{2} = 30 \, \text{tiles}
\]

Step 3: Lay the White Tiles
After laying the black tiles along the perimeter, the remaining area is the inner portion of the room. The remaining length and breadth are:

\[
\text{Remaining length} = 20 – 2 = 18 \, \text{feet}
\] \[
\text{Remaining breadth} = 10 – 2 = 8 \, \text{feet}
\]

The area of the remaining space is:

\[
\text{Remaining area} = 18 \times 8 = 144 \, \text{sq. feet}
\]

Now, the number of tiles that can be laid in this remaining area is:

\[
\text{Number of tiles in remaining area} = \frac{144}{4} = 36 \, \text{tiles}
\]

One-third of these 36 tiles will be white tiles, so the number of white tiles is:

\[
\text{Number of white tiles} = \frac{1}{3} \times 36 = 12 \, \text{tiles}
\]

Step 4: Lay the Blue Tiles
The remaining tiles will be blue. Since there are 36 tiles in the remaining area, and 12 of them are white, the number of blue tiles is:

\[
\text{Number of blue tiles} = 36 – 12 = 24 \, \text{tiles}
\]

Final Answer:
The number of blue tiles is 24.

Answer: Option B
Solution:
We are given:
– The area of a square is \( 3136 \, \text{cm}^2 \),
– The diameter of a circle is equal to the perimeter of the square.

We need to find the circumference of the circle.

Step 1: Find the Side Length of the Square
The area of the square is:

\[
\text{Area of square} = \text{side}^2 = 3136 \, \text{cm}^2
\]

Taking the square root to find the side length:

\[
\text{side} = \sqrt{3136} = 56 \, \text{cm}
\]

Step 2: Find the Perimeter of the Square
The perimeter of the square is:

\[
\text{Perimeter of square} = 4 \times \text{side} = 4 \times 56 = 224 \, \text{cm}
\]

Step 3: Use the Perimeter of the Square as the Diameter of the Circle
We are told that the diameter of the circle is equal to the perimeter of the square. Therefore, the diameter of the circle is:

\[
\text{Diameter of circle} = 224 \, \text{cm}
\]

Step 4: Find the Circumference of the Circle
The formula for the circumference of the circle is:

\[
C = \pi \times \text{diameter}
\]

Substitute the value of the diameter:

\[
C = 3.1416 \times 224 \approx 703.72 \, \text{cm}
\]

Final Answer:
The circumference of the circle is approximately 703.72 cm.

Answer: Option E
Solution:
We are given:
– The area of the rectangular field is 2100 sq. meters,
– The length of the field is 60 meters.

We need to find the perimeter of the rectangle.

Step 1: Calculate the Breadth of the Field
The area of a rectangle is given by the formula:

\[
\text{Area} = \text{length} \times \text{breadth}
\]

Substitute the known values:

\[
2100 = 60 \times \text{breadth}
\]

Solving for breadth:

\[
\text{breadth} = \frac{2100}{60} = 35 \, \text{meters}
\]

Step 2: Calculate the Perimeter of the Field
The perimeter \( P \) of a rectangle is given by the formula:

\[
P = 2 \times (\text{length} + \text{breadth})
\]

Substitute the values of length and breadth:

\[
P = 2 \times (60 + 35) = 2 \times 95 = 190 \, \text{meters}
\]

Final Answer:
The perimeter of the field is 190 meters.

Answer: Option B
Solution:
We are given:
– The width of the carpet is 63 cm (which we need to convert to meters),
– The dimensions of the floor are 14 meters by 9 meters.

We need to determine how many meters of carpet will be required to cover the floor.

Step 1: Convert the Carpet Width to Meters
The width of the carpet is given as 63 cm, which is equivalent to:

\[
\text{Width in meters} = \frac{63}{100} = 0.63 \, \text{meters}
\]

Step 2: Calculate the Area of the Floor
The area of the floor is:

\[
\text{Area of floor} = 14 \times 9 = 126 \, \text{sq. meters}
\]

Step 3: Calculate the Length of Carpet Required
The area that one meter of carpet will cover (with a width of 0.63 meters) is:

\[
\text{Area covered by 1 meter of carpet} = 0.63 \times 1 = 0.63 \, \text{sq. meters}
\]

To find the length of carpet required to cover the entire floor, we divide the total area of the floor by the area covered by one meter of carpet:

\[
\text{Length of carpet required} = \frac{\text{Area of floor}}{\text{Area covered by 1 meter of carpet}} = \frac{126}{0.63} = 200 \, \text{meters}
\]

Final Answer:
The length of carpet required to cover the floor is 200 meters.

Answer: Option B
Solution:
We are given:
– The area of a rectangular field increases by 50%,
– The length of the field is increased by 20%,
– We need to find the percentage increase in the breadth.

Step 1: Let the Original Length and Breadth be \( L \) and \( B \)
Let the original length of the field be \( L \) and the original breadth be \( B \).

The area of the original field is:

\[
\text{Original Area} = L \times B
\]

Step 2: New Length and Breadth After the Increases
The length is increased by 20%, so the new length is:

\[
\text{New Length} = L \times \left(1 + \frac{20}{100}\right) = L \times 1.2
\]

Let the new breadth be \( B’ \). Since the area increases by 50%, the new area is 1.5 times the original area:

\[
\text{New Area} = 1.5 \times \text{Original Area} = 1.5 \times (L \times B)
\]

The new area is also given by:

\[
\text{New Area} = \text{New Length} \times \text{New Breadth} = (L \times 1.2) \times B’
\]

Equating the two expressions for the new area:

\[
1.5 \times L \times B = (L \times 1.2) \times B’
\]

Step 3: Solve for \( B’ \)
Canceling \( L \) from both sides:

\[
1.5 \times B = 1.2 \times B’
\]

Now, solving for \( B’ \):

\[
B’ = \frac{1.5 \times B}{1.2} = 1.25 \times B
\]

Step 4: Calculate the Percentage Increase in the Breadth
The increase in breadth is:

\[
\text{Increase in Breadth} = B’ – B = 1.25 \times B – B = 0.25 \times B
\]

The percentage increase in the breadth is:

\[
\text{Percentage increase in breadth} = \frac{0.25 \times B}{B} \times 100 = 25\%
\]

Final Answer:
The breadth was increased by 25%.

Answer: Option D
Solution:
We are given:
– The total area of 64 small squares on the chessboard is 400 sq. cm,
– There is a 3 cm wide border around the chessboard.

We need to determine the length of the side of the chessboard.

Step 1: Find the Area of One Small Square
The chessboard consists of 64 small squares, so the area of each small square is:

\[
\text{Area of each small square} = \frac{\text{Total area of small squares}}{64} = \frac{400}{64} = 6.25 \, \text{sq. cm}
\]

Step 2: Find the Side Length of One Small Square
The area of a square is the side length squared. So, the side length of one small square is:

\[
\text{Side length of one small square} = \sqrt{6.25} = 2.5 \, \text{cm}
\]

Step 3: Find the Side Length of the Chessboard Without the Border
Since the chessboard consists of 8 rows and 8 columns of small squares, the side length of the chessboard without the border is:

\[
\text{Side length without border} = 8 \times 2.5 = 20 \, \text{cm}
\]

Step 4: Find the Total Side Length of the Chessboard (Including the Border)
The border is 3 cm wide on all sides of the chessboard. Therefore, the total side length of the chessboard, including the border, is:

\[
\text{Total side length} = 20 + 2 \times 3 = 20 + 6 = 26 \, \text{cm}
\]

Final Answer:
The length of the side of the chessboard is 26 cm.

Answer: Option A
Solution:
We are given:
– The area of the square-shaped lake is 50 sq. km.
– We need to find the distance the driver would swim if crossing the lake diagonally.

Step 1: Calculate the Side Length of the Square Lake
The area of a square is given by the formula:

\[
\text{Area} = \text{side}^2
\]

Given the area is 50 sq. km, we can find the side length by taking the square root of the area:

\[
\text{Side length} = \sqrt{50} \approx 7.071 \, \text{km}
\]

Step 2: Calculate the Diagonal of the Square Lake
The diagonal \( d \) of a square can be calculated using the Pythagorean theorem:

\[
d = \sqrt{\text{side}^2 + \text{side}^2} = \sqrt{2 \times \text{side}^2} = \text{side} \times \sqrt{2}
\]

Substitute the side length into the formula:

\[
d = 7.071 \times \sqrt{2} \approx 7.071 \times 1.414 \approx 10 \, \text{km}
\]

Final Answer:
The driver will have to swim a distance of 10 km to cross the lake diagonally.

Answer: Option B
Solution:
We are given a triangle with side lengths of 3 cm, 4 cm, and 5 cm, which is a right-angled triangle (since \( 3^2 + 4^2 = 5^2 \)).

To calculate the area of the triangle, we can use the formula for the area of a right triangle:

\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]

Here, the base and height are the two shorter sides of the triangle, which are 3 cm and 4 cm.

\[
\text{Area} = \frac{1}{2} \times 3 \times 4 = \frac{1}{2} \times 12 = 6 \, \text{sq. cm}
\]

Final Answer:
The area of the triangle is 6 square centimeters.

Answer: Option C
Solution
We are given:
– The perimeter of the triangle is 30 cm,
– The area of the triangle is 30 cm²,
– The largest side of the triangle is 13 cm.

We need to find the length of the smallest side of the triangle.

Step 1: Use Heron’s Formula to Calculate the Area
Heron’s formula for the area of a triangle is given by:

\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]

where \(a\), \(b\), and \(c\) are the sides of the triangle, and \(s\) is the semi-perimeter, which is:

\[
s = \frac{a + b + c}{2}
\]

Here, we know the perimeter is 30 cm, so:

\[
s = \frac{30}{2} = 15 \, \text{cm}
\]

Let the three sides of the triangle be \(a\), \(b\), and \(c\), where \(c = 13\) cm (the largest side). Now, we have:

\[
A = 30 \, \text{cm}^2 = \sqrt{15(15-a)(15-b)(15-13)}
\]

Simplifying the equation:

\[
30 = \sqrt{15(15-a)(15-b)(2)}
\]

Squaring both sides:

\[
900 = 30(15-a)(15-b)
\]

Dividing both sides by 30:

\[
30 = (15-a)(15-b)
\]

Step 2: Solve for the Other Two Sides
Let the sides be \(a\) and \(b\). Since \(a + b + c = 30\), we can write:

\[
a + b + 13 = 30 \quad \Rightarrow \quad a + b = 17
\]

Now we have two equations:
1. \(a + b = 17\)
2. \((15-a)(15-b) = 30\)

We can solve this system of equations. First, from \(a + b = 17\), express \(b\) in terms of \(a\):

\[
b = 17 – a
\]

Substitute this into the second equation:

\[
(15-a)(15-(17-a)) = 30
\]

Simplifying the expression:

\[
(15-a)(a-2) = 30
\]

Expanding the equation:

\[
(15-a)(a-2) = 15a – 30 – a^2 + 2a = 30
\]

\[
-a^2 + 17a – 30 = 30
\]

\[
-a^2 + 17a – 60 = 0
\]

Multiplying through by -1:

\[
a^2 – 17a + 60 = 0
\]

Step 3: Solve the Quadratic Equation
Solve the quadratic equation \(a^2 – 17a + 60 = 0\) using the quadratic formula:

\[
a = \frac{-(-17) \pm \sqrt{(-17)^2 – 4(1)(60)}}{2(1)}
\]

\[
a = \frac{17 \pm \sqrt{289 – 240}}{2}
\]

\[
a = \frac{17 \pm \sqrt{49}}{2}
\]

\[
a = \frac{17 \pm 7}{2}
\]

Thus, the two possible values for \(a\) are:

\[
a = \frac{17 + 7}{2} = 12 \quad \text{or} \quad a = \frac{17 – 7}{2} = 5
\]

Step 4: Find the Smallest Side
The smallest side is \(a = 5\) cm.

Final Answer:
The length of the smallest side of the triangle is 5 cm.

Answer: Option B
Solution:
Let each have base = b and height = h
Then,
P = b × h, R = b × h, T=\( \frac{1}{2} \)× b × h
So, P = R, P = 2T and T=\( \frac{1}{2}R are all correct statements.
∴ Option B is false

Answer: Option A
Solution:
We are given:
– The dimensions of the rectangular field: length = 38 m and width = 32 m,
– The area of the path that runs around the inside of the field: 600 m².

We need to find the width of the path.

Step 1: Understand the problem
The path runs inside the field, so it reduces the length and width of the usable area (the area inside the path). Let’s assume the width of the path is \( x \) meters. After the path is constructed, the remaining area inside the field will have dimensions:
– Length of the remaining area = \( 38 – 2x \),
– Width of the remaining area = \( 32 – 2x \).

The area of the remaining rectangular field (inside the path) is:
\[
\text{Remaining Area} = (38 – 2x)(32 – 2x)
\]

Step 2: Calculate the total area of the field
The total area of the rectangular field is:
\[
\text{Total Area} = 38 \times 32 = 1216 \, \text{m}^2
\]

Step 3: Relate the areas
The area of the path is the difference between the total area and the remaining area:
\[
\text{Area of the path} = \text{Total Area} – \text{Remaining Area}
\] We are given that the area of the path is 600 m², so:
\[
600 = 1216 – (38 – 2x)(32 – 2x)
\]

Step 4: Simplify the equation
First, expand \( (38 – 2x)(32 – 2x) \):

\[
(38 – 2x)(32 – 2x) = 38 \times 32 – 38 \times 2x – 32 \times 2x + 4x^2
\] \[
= 1216 – 76x – 64x + 4x^2
\] \[
= 1216 – 140x + 4x^2
\]

Now substitute this into the equation:
\[
600 = 1216 – (1216 – 140x + 4x^2)
\] Simplify:
\[
600 = 1216 – 1216 + 140x – 4x^2
\] \[
600 = 140x – 4x^2
\] Rearrange the equation:
\[
4x^2 – 140x + 600 = 0
\]

Step 5: Solve the quadratic equation
Divide the entire equation by 4 to simplify:
\[
x^2 – 35x + 150 = 0
\]

Now, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-35) \pm \sqrt{(-35)^2 – 4(1)(150)}}{2(1)}
\] \[
x = \frac{35 \pm \sqrt{1225 – 600}}{2}
\] \[
x = \frac{35 \pm \sqrt{625}}{2}
\] \[
x = \frac{35 \pm 25}{2}
\]

Thus, the two possible values for \(x\) are:
\[
x = \frac{35 + 25}{2} = 30 \quad \text{or} \quad x = \frac{35 – 25}{2} = 5
\]

Step 6: Conclusion
Since the path cannot be wider than the field itself, the width of the path must be 5 meters.

Final Answer:
The width of the path is 5 meters.

Answer: Option C
Solution:
We are given:
– The circumference of the smaller circle is 132 meters.
– The circumference of the larger circle is 176 meters.

We need to find the difference in the areas of the two circles.

Step 1: Use the formula for the circumference of a circle
The formula for the circumference \( C \) of a circle is:
\[
C = 2\pi r
\] where \( r \) is the radius of the circle.

Step 2: Find the radii of both circles
For the smaller circle:
\[
132 = 2\pi r_1
\] Solving for \( r_1 \):
\[
r_1 = \frac{132}{2\pi} = \frac{132}{6.28} \approx 21 \, \text{meters}
\]

For the larger circle:
\[
176 = 2\pi r_2
\] Solving for \( r_2 \):
\[
r_2 = \frac{176}{2\pi} = \frac{176}{6.28} \approx 28 \, \text{meters}
\]

Step 3: Find the areas of the circles
The formula for the area \( A \) of a circle is:
\[
A = \pi r^2
\]

For the smaller circle:
\[
A_1 = \pi r_1^2 = \pi (21)^2 = \pi \times 441 \approx 1385 \, \text{square meters}
\]

For the larger circle:
\[
A_2 = \pi r_2^2 = \pi (28)^2 = \pi \times 784 \approx 2464 \, \text{square meters}
\]

Step 4: Find the difference in the areas
The difference in the areas is:
\[
\text{Difference} = A_2 – A_1 = 2464 – 1385 = 1079 \, \text{square meters}
\]

Final Answer:
The difference between the area of the larger circle and the smaller circle is 1079 square meters.