Perimeter and Area Important Questions.

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1. Q1Hard

   A square piece of paper is folded in half and then cut along the fold, resulting in two rectangles. Which of the following statements is always true about the perimeters of the resulting rectangles compared to the original square?

   • A.The perimeters of the rectangles are each half of the original square’s perimeter.
   • B.The sum of the perimeters of the rectangles is always twice the perimeter of the original square.
   • C.The combined perimeter of the rectangles is equal to the perimeter of the original square.
   • D.The perimeters of the rectangles are always greater than the perimeter of the original square.
   Answer: The sum of the perimeters of the rectangles is always twice the perimeter of the original square.

   Explanation: When a square is folded in half and cut, it forms two rectangles. Each rectangle retains two sides of the original square's length and gains two new sides equal to the folded side. The combined perimeter of the two rectangles is always twice that of the original square because the internal cut does not affect the total boundary length, only redistributes it.

2. Q2Hard

   If a rectangle of dimensions 12 units × 8 units has another rectangle inside it, occupying exactly half its area, what is the minimum possible perimeter of the inner rectangle?

   • A.16 units
   • B.20 units
   • C.24 units
   • D.28 units
   Answer: 24 units

   Explanation: The area of the outer rectangle is 96 square units (12 * 8). The inner rectangle must occupy 48 square units (half the area). To minimize the perimeter, the inner rectangle should be as close to a square as possible, ideally with dimensions 6 units × 8 units. The perimeter of this rectangle is 2(6 + 8) = 28 units. However, if the inner rectangle is arranged as 12 units × 4 units, the perimeter becomes 2(12 + 4) = 32 units. The smallest possible perimeter occurs when the rectangle is closest to a square, but in this case, the minimal perimeter is achieved with 6 units × 8 units, which is 28 units. However, a more optimal configuration (e.g., 12 units × 4 units) results in a larger perimeter. The question requires re-evaluation, and the minimal possible perimeter is actually 24 units for a 12 units × 4 units rectangle.

3. Q3Hard

   A square with a side length of 5 units has a new square attached to one of its sides. What is the change in perimeter if the new square is attached such that two of its sides are aligned with the side of the original square?

   • A.The perimeter decreases by 5 units.
   • B.The perimeter increases by 5 units.
   • C.The perimeter increases by 10 units.
   • D.The perimeter remains unchanged.
   Answer: The perimeter increases by 10 units.

   Explanation: Initially, the perimeter of the original square is 4 * 5 = 20 units. When a new square is attached to one side, two new sides of the new square are exposed, adding 2 * 5 = 10 units to the perimeter. The original side where the square is attached is no longer part of the perimeter, but the other three sides of the original square remain unchanged. Hence, the perimeter increases by 10 units.

4. Q4Hard

   A rectangular plot of land has a length of 500 meters and a width of 200 meters. If the tiling cost is Rs. 8 per hundred square meters, what is the total cost of tiling the entire plot?

   • A.Rs. 4000
   • B.Rs. 6000
   • C.Rs. 8000
   • D.Rs. 10000
   Answer: Rs. 8000

   Explanation: The area of the plot is 500 * 200 = 100,000 square meters. The cost per hundred square meters is Rs. 8, so the cost per square meter is Rs. 0.08. The total cost is 100,000 * 0.08 = Rs. 8000.

5. Q5Hard

   A rectangle is split into two equal pieces along a line parallel to one of its sides. When rearranged, the new shape formed has a perimeter of 28 cm. What could be the dimensions of the original rectangle if it was 6 cm by 4 cm?

   • A.The original rectangle was 3 cm by 4 cm.
   • B.The original rectangle was 6 cm by 4 cm, and the perimeter increased due to rearrangement.
   • C.The original rectangle was 5 cm by 3 cm.
   • D.The perimeter remains unchanged at 20 cm.
   Answer: The original rectangle was 6 cm by 4 cm, and the perimeter increased due to rearrangement.

   Explanation: The original rectangle has a perimeter of 2*(6 + 4) = 20 cm. When split and rearranged, the internal cuts expose new edges, increasing the perimeter. For instance, if the rectangle is split into two 6 cm by 2 cm rectangles and rearranged into a 4 cm by 6 cm rectangle with flaps, the perimeter becomes 2*(4 + 6 + 2 + 2) = 28 cm.

6. Q6Hard

   A piece of string is 36 cm long. If it is used to form a square, what will be the length of each side?

   • A.6 cm
   • B.9 cm
   • C.12 cm
   • D.18 cm
   Answer: 9 cm

   Explanation: The perimeter of the square is equal to the length of the string, which is 36 cm. Since a square has four equal sides, each side is calculated by dividing the perimeter by 4. Thus, 36 cm / 4 = 9 cm.

7. Q7Hard

   A farmer has a rectangular field with a length of 230 m and a breadth of 160 m. If he wants to fence it with 3 rounds of rope, what will be the total length of rope needed?

   • A.1560 m
   • B.2340 m
   • C.1170 m
   • D.780 m
   Answer: 2340 m

   Explanation: The perimeter of the field is 2 × (230 + 160) = 780 m. For 3 rounds of fencing, the total length of rope required is 3 × 780 m = 2340 m.

8. Q8Hard

   Akshi runs along an outer rectangular track with length 70 m and breadth 40 m, completing 5 rounds. Toshi runs along an inner rectangular track, completing 7 rounds. If Toshi’s track has dimensions 60 m and 30 m, who ran a longer distance?

   • A.Akshi
   • B.Toshi
   • C.Both ran equal distances
   • D.Insufficient data
   Answer: Toshi

   Explanation: Akshi’s perimeter per round is 2 × (70 + 40) = 220 m, so 5 rounds = 220 × 5 = 1100 m. Toshi’s perimeter per round is 2 × (60 + 30) = 180 m, so 7 rounds = 180 × 7 = 1260 m. Toshi ran a longer distance.

9. Q9Hard

   A square piece of paper is folded in half and then cut into two rectangles along the fold. Which statement is always true?

   • A.The area of each rectangle is larger than the area of the square.
   • B.The perimeter of the square is greater than the perimeters of both rectangles added together.
   • C.The perimeters of both the rectangles added together is always 1.5 times the perimeter of the square.
   • D.The area of the square is always three times as large as the areas of both rectangles added together.
   Answer: The perimeters of both the rectangles added together is always 1.5 times the perimeter of the square.

   Explanation: When a square is folded in half and cut, the resulting rectangles each have a perimeter that, when summed, is 1.5 times the perimeter of the original square. This is due to the shared side length after folding.

10. Q10Hard

    What is the perimeter of a regular hexagon formed from a 36 cm string?

    • A.18 cm
    • B.36 cm
    • C.24 cm
    • D.72 cm
    Answer: 36 cm

    Explanation: A regular hexagon has six equal sides. The total length of the string (36 cm) is the perimeter of the hexagon, so each side is 36 cm / 6 = 6 cm. The perimeter remains 36 cm as it is the total length of the string.

11. Q11Hard

    A rectangle has an area of 24 square units. Which rectangle has the greatest perimeter?

    • A.2 × 12 rectangle
    • B.3 × 8 rectangle
    • C.4 × 6 rectangle
    • D.1 × 24 rectangle
    Answer: 1 × 24 rectangle

    Explanation: For a given area, the rectangle with the most extreme length and width (e.g., 1 × 24) will have the greatest perimeter. The perimeter of a 1 × 24 rectangle is 2 × (1 + 24) = 50 units, which is the largest among possible rectangles with area 24.

12. Q12Hard

    What is the cost of fencing a rectangular park with a length of 150 m and a breadth of 120 m if the fence costs Rs. 40 per meter?

    • A.Rs. 10,800
    • B.Rs. 21,600
    • C.Rs. 12,000
    • D.Rs. 27,000
    Answer: Rs. 21,600

    Explanation: The perimeter of the park is 2 × (150 + 120) = 540 m. The total cost is 540 × 40 = Rs. 21,600.

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