Option C: Both 1 and 2
      Statement 1: It is possible to cut the sheet exactly into 4 square sheets.
         ○ The dimensions of the rectangular sheet are 20 cm by 8 cm.
         ○ To cut the sheet into square sheets, the side of each square must be a common divisor of both 20 cm and 8 cm.
         ○ The greatest common divisor (GCD) of 20 and 8 is 4.
         ○ Therefore, it is possible to cut the sheet into squares of side 4 cm.
         ○ The number of such squares that can be cut is:
           ○ Along the length: \( \frac{20}{4} = 5 \) squares
           ○ Along the breadth: \( \frac{8}{4} = 2 \) squares
         ○ Total number of squares = \( 5 \times 2 = 10 \).
         ○ However, the statement asks for exactly 4 square sheets, which can be achieved by cutting the sheet into squares of side 8 cm (2 squares along the length and 1 along the breadth), resulting in 2 squares, and then further cutting each into 2 smaller squares of side 4 cm.
         ○ Thus, it is possible to cut the sheet into exactly 4 square sheets of side 4 cm each.
     Statement 2: It is possible to cut the sheet into 10 triangular sheets of equal area.
         ○ The area of the rectangular sheet is \( 20 \times 8 = 160 \) square cm.
         ○ To cut the sheet into 10 triangular sheets of equal area, each triangle must have an area of \( \frac{160}{10} = 16 \) square cm.
         ○ A right-angled triangle with base and height such that \( \frac{1}{2} \times \text{base} \times \text{height} = 16 \) can be used.
         ○ For example, a triangle with base 8 cm and height 4 cm has an area of \( \frac{1}{2} \times 8 \times 4 = 16 \) square cm.
         ○ The sheet can be divided into such triangles by cutting along the diagonals of rectangles formed by dividing the sheet into smaller rectangles of dimensions 8 cm by 4 cm.
         ○ Therefore, it is possible to cut the sheet into 10 triangular sheets of equal area.
     Since both statements are correct, the correct answer is Option C: Both 1 and 2.

1. Correct Answer: Option C: 1 sq cm
 2. Explanation:
     To solve this problem, we need to determine the side length of each square and then calculate its area.
     ● Step 1: Understand the arrangement of squares.  
           ○ There are three rows, each containing four squares.
           ○ Therefore, there are a total of \(3 \times 4 = 12\) squares.
     ● Step 2: Determine the dimensions of the rectangle.  
           ○ Since there are four squares in each row, the length of the rectangle is \(4s\), where \(s\) is the side length of each square.
           ○ Since there are three rows, the width of the rectangle is \(3s\).
     ● Step 3: Use the diagonal to find the side length.  
           ○ The diagonal of the rectangle is given as 5 cm.
           ○ Using the Pythagorean theorem for the rectangle, we have:
         \[
         \text{Diagonal}^2 = \text{Length}^2 + \text{Width}^2
         \]
         \[
         5^2 = (4s)^2 + (3s)^2
         \]
         \[
         25 = 16s^2 + 9s^2
         \]
         \[
         25 = 25s^2
         \]
         \[
         s^2 = 1
         \]
     ● Step 4: Calculate the area of each square.  
           ○ The area of each square is \(s^2\).
           ○ Since \(s^2 = 1\), the area of each square is 1 sq cm.
     Therefore, the correct answer is Option C: 1 sq cm.

•   Area of the field will be maximum when both sides of rectangle are equal.

•   “will not exceed” in the question means “what is the maximum possible”.

•   Therefore x₁ = x₂ = 20 meters (because x₁ + x₂ = 40 is given). Max. Area = 20 × 20 = 400 sq. m.

Correct Answer: c: 0.16
To find the probability that an arrow falls in the red region, we need to calculate the area of the red region and divide it by the total area of the target.
     Step 1: Calculate the total area of the target.
         ○ The target is a circle with a diameter of 1 m, so its radius is 0.5 m.
         ○ The area of the circle (target) is given by the formula:
       \[ \text{Area}_{\text{total}} = \pi \times (\text{radius})^2 = \pi \times (0.5)^2 = 0.25\pi \, \text{m}^2 \]
     Step 2: Calculate the area of the red region.
         ○ The radius of the red region is 0.20 m.
         ○ The area of the red region is given by the formula:
       \[ \text{Area}_{\text{red}} = \pi \times (\text{radius of red})^2 = \pi \times (0.20)^2 = 0.04\pi \, \text{m}^2 \]
     Step 3: Calculate the probability.
         ○ The probability that an arrow falls in the red region is the ratio of the area of the red region to the total area of the target:
       \[ \text{Probability} = \frac{\text{Area}_{\text{red}}}{\text{Area}_{\text{total}}} = \frac{0.04\pi}{0.25\pi} = \frac{0.04}{0.25} = 0.16 \]
     Therefore, the probability that the arrows fall in the red region of the archery target is 0.16.

1. Correct Answer: Option B: has increased by 12%.
 2. Explanation:
     Let's denote the original length of the garden as \( L \) and the original width as \( W \). Therefore, the original area \( A_{\text{original}} \) of the garden is given by:
     \[
     A_{\text{original}} = L \times W
     \]
     Step 1: Calculate the new length.
     The length is increased by 40%, so the new length \( L_{\text{new}} \) is:
     \[
     L_{\text{new}} = L + 0.4L = 1.4L
     \]
     Step 2: Calculate the new width.
     The width is decreased by 20%, so the new width \( W_{\text{new}} \) is:
     \[
     W_{\text{new}} = W - 0.2W = 0.8W
     \]
     Step 3: Calculate the new area.
     The new area \( A_{\text{new}} \) of the garden is:
     \[
     A_{\text{new}} = L_{\text{new}} \times W_{\text{new}} = 1.4L \times 0.8W = 1.12LW
     \]
     Step 4: Determine the percentage increase in area.
     The percentage increase in area is calculated as follows:
     \[
     \text{Percentage Increase} = \left( \frac{A_{\text{new}} - A_{\text{original}}}{A_{\text{original}}} \right) \times 100\%
     \]
     Substituting the values, we get:
     \[
     \text{Percentage Increase} = \left( \frac{1.12LW - LW}{LW} \right) \times 100\% = \left( \frac{0.12LW}{LW} \right) \times 100\% = 12\%
     \]
     Therefore, the area of the new garden has increased by 12%.

Area 15 units can be 15 X 1 or 5 X 3. Area 48 units can be 48 X 1 or 24 X 2 or 16 X 3 or 12 X 4 or 8 X 6 Since the bigger figure is a square choose a combination with the length and breadth that forms a square: So if you take 5 units (from 15 units area) + 6 units (from 48 units area) or 8 units (from 48 units area) + 3 units (from 15 units area), it is equal to a side 11 square.