RBSE Solutions Class 6 Maths Chapter 14 Perimeter and Area Exercise 14.3

Rajasthan Board RBSE Class 6 Maths Chapter 14 Perimeter and Area Ex 14.3

 

Question 1. Find the areas and perimeters of the following rectangular figures. Which of them have same perimeter but different area?
(i) 40 cm x 15 cm
(ii) 45 cm x 20 cm
(iii) 8 cm x 7 cm
(iv) 10 cm x 5 cm
Answer: We will calculate the perimeter and area for each rectangle. The perimeter is found by \( 2 \times (length + breadth) \) and the area by \( length \times breadth \). This helps us see which figures have the same perimeter but different areas.
(i) For the rectangle with length \( l = 40 \) cm and breadth \( b = 15 \) cm:
Perimeter \( = 2 (l + b) = 2 (40 + 15) = 2 (55) = 110 \) cm
Area \( = l \times b = 40 \times 15 = 600 \) sq. cm
(ii) For the rectangle with length \( l = 45 \) cm and breadth \( b = 20 \) cm:
Perimeter \( = 2 (l + b) = 2 (45 + 20) = 2 \times 65 = 130 \) cm
Area \( = l \times b = 45 \times 20 = 900 \) sq. cm
(iii) For the rectangle with length \( l = 8 \) cm and breadth \( b = 7 \) cm:
Perimeter \( = 2 (l + b) = 2 (8 + 7) = 2 (15) = 30 \) cm
Area \( = l \times b = 8 \times 7 = 56 \) sq. cm
(iv) For the rectangle with length \( l = 10 \) cm and breadth \( b = 5 \) cm:
Perimeter \( = 2 (l + b) = 2 (10 + 5) = 2 \times 15 = 30 \) cm
Area \( = l \times b = 10 \times 5 = 50 \) sq. cm
From the calculations, figures (iii) and (iv) both have a perimeter of 30 cm, but their areas are 56 sq. cm and 50 sq. cm, respectively. These two figures have the same perimeter but different areas.
In simple words: We calculated the distance around (perimeter) and the space inside (area) for each shape. We found that two shapes had the same perimeter (30 cm) but their areas were different (56 sq. cm and 50 sq. cm).

🎯 Exam Tip: Remember to use the correct formulas for perimeter \( (2(l+b)) \) and area \( (l \times b) \) of a rectangle, and always include units in your final answer.

 

Question 2. Gopi has a square field of side 75 m. Narayan has a rectangular field of length 85 m. If the perimeters of both the fields are same, whose field has greater area and by how much?
Answer: First, let's find the details for Gopi's square field.
Side of Gopi's square field \( = 75 \) m
Area of Gopi's field \( = \text{side} \times \text{side} = 75 \times 75 = 5625 \) sq. m
Perimeter of Gopi's field \( = 4 \times \text{side} = 4 \times 75 = 300 \) m
Now for Narayan's rectangular field:
Length of Narayan's rectangular field \( = 85 \) m
Let the breadth of Narayan's field be \( a \) m.
Perimeter of Narayan's field \( = 2 (\text{length} + \text{breadth}) = 2 (85 + a) \) m
The problem states that both perimeters are equal.
So, \( 2 (85 + a) = 300 \)

\( \implies 85 + a = \frac{300}{2} = 150 \)

\( \implies a = 150 - 85 = 65 \)
So, the breadth of Narayan's field is \( 65 \) m.
Area of Narayan's field \( = \text{length} \times \text{breadth} = 85 \times 65 = 5525 \) sq. m
To compare, Gopi's area is 5625 sq. m and Narayan's area is 5525 sq. m. A square field often maximizes area for a given perimeter.
Since \( 5625 > 5525 \), Gopi's field has a greater area.
The difference in area \( = 5625 - 5525 = 100 \) sq. m.
Therefore, Gopi's field has a greater area by 100 sq. m.
In simple words: Gopi's square field is 75 meters on each side. Narayan's field is a rectangle with a length of 85 meters. Both fields have the same perimeter. We found Gopi's field has a bigger area by 100 square meters.

🎯 Exam Tip: When comparing areas or perimeters, always calculate both for each figure before making a comparison. Pay attention to whether the question asks for "whose field" or "by how much."

 

Question 3. Area of a square is 65 sq. cm. Perimeter of a rectangle is equal to the perimeter of this square. Find the length of the rectangle if its breadth is 6.5 cm. Which figure has greater area?
Answer: There seems to be a minor error in the question, as the area of a square cannot be 65 sq. cm if its side is a whole number (since \( \sqrt{65} \) is not a whole number). Assuming the area of the square is 64 sq. cm, as used in the solution:
Area of the square \( = 64 \) sq. cm
We know that Area \( = \text{side} \times \text{side} = (\text{side})^2 \).
So, \( \text{side} = \sqrt{\text{Area}} = \sqrt{64} = \sqrt{8^2} = 8 \) cm.
The side of the square is \( 8 \) cm.
Perimeter of the square \( = 4 \times \text{side} = 4 \times 8 = 32 \) cm.
Now for the rectangle:
Breadth of the rectangle \( = 6.5 \) cm
Let the length of the rectangle be \( a \) cm.
Perimeter of the rectangle \( = 2 (\text{length} + \text{breadth}) = 2 (a + 6.5) \) cm.
According to the question, the perimeter of the rectangle is equal to the perimeter of the square.
So, \( 2 (a + 6.5) = 32 \)

\( \implies a + 6.5 = \frac{32}{2} = 16 \)

\( \implies a = 16 - 6.5 = 9.5 \) cm.
The length of the rectangle is \( 9.5 \) cm.
Now, let's find the area of the rectangle:
Area of rectangle \( = \text{length} \times \text{breadth} = 9.5 \times 6.5 = 61.75 \) sq. cm.
Comparing areas: Area of square \( = 64 \) sq. cm, Area of rectangle \( = 61.75 \) sq. cm.
Since \( 64 > 61.75 \), the square has the greater area.
In simple words: We started with a square whose area was 64 sq. cm, so its side was 8 cm and its perimeter was 32 cm. We then found a rectangle that had the same perimeter (32 cm) and a breadth of 6.5 cm. We calculated the length of the rectangle to be 9.5 cm. Comparing the areas, the square (64 sq. cm) had a larger area than the rectangle (61.75 sq. cm).

🎯 Exam Tip: Always double-check calculations involving decimals. When finding the side from the area of a square, ensure you use the square root. Be careful when the question asks to compare areas after matching perimeters.

 

Question 4. A rectangular piece of 20 cm x 15 cm is cut off from a bigger rectangle as shown in the figures below. In each case, find the difference in perimeter before and after the piece is cut off.
(i) (A rectangular piece cut from the corner of a larger rectangle, where the outer dimensions are 80 cm length and 30 cm width. The cut piece is 20 cm by 15 cm, forming an 'L' shape.)
(ii) (A rectangular piece cut from the middle of one side of a larger rectangle, where the outer dimensions are 60 cm length and 40 cm width. The cut piece is 30 cm by 20 cm.)
Answer: We need to calculate the perimeter before and after cutting the piece. The perimeter is the total length of the boundary of the figure. Understanding how the cut affects the visible edges is key. In most cases, removing a corner piece will increase the perimeter if the cut edges replace the original boundary in a way that adds to the total length.
(i) For the first figure:
Original rectangle dimensions: Length \( = 80 \) cm, Width \( = 30 \) cm.
Perimeter of the complete square (before cutting) \( = 2 (\text{length} + \text{width}) = 2 (80 + 30) = 2 (110) = 220 \) cm. (Assuming 'square' in the source solution text is a typo for 'rectangle', as dimensions are different).
After cutting a 20 cm \( \times \) 15 cm piece from the corner, the new perimeter is formed by adding the lengths of all outer sides of the 'L' shape. The outer sides now are: \( 80 + (30 - 15) + 20 + 15 + (80 - 20) + 30 \).
Perimeter after cutting \( = 80 + 15 + 20 + 15 + 60 + 30 = 220 \) cm.
Let's re-evaluate based on the common understanding for corner cuts: When a rectangle is cut from a corner, the perimeter usually stays the same. The two new cut lines replace the two original outer lines. For example, if a 20x15 piece is cut from an 80x30 rectangle, the original perimeter is \( 2(80+30) = 220 \). The new perimeter will be \( 80 + 30 + (80-20) + (30-15) + 20 + 15 = 80 + 30 + 60 + 15 + 20 + 15 = 220 \) cm. So the perimeter remains unchanged. The source solution's calculation seems to be: \( 80 + 30 + 15 + 20 + 15 + 30 + 80 = 270 \) cm. This calculation adds the length and breadth of the main rectangle and then the dimensions of the cut-out portion twice, which is incorrect. Let's follow the standard rule: when a rectangle is cut from a corner of a larger rectangle, the perimeter remains the same because the two new internal edges created are exactly equal in length to the two external edges removed. So, the perimeter before cutting \( = 2 (80 + 30) = 220 \) cm. The perimeter after cutting also remains \( 220 \) cm.
Difference in perimeter \( = 220 - 220 = 0 \) cm. There is no change. (ii) For the second figure:
Original rectangle dimensions: Length \( = 60 \) cm, Width \( = 40 \) cm.
Perimeter of the complete rectangle (before cutting) \( = 2 (60 + 40) = 2 (100) = 200 \) cm.
A piece of 30 cm \( \times \) 20 cm is cut from the middle of one side. This means two new vertical edges of 20 cm each are added, and the horizontal segment of 30 cm is replaced by two horizontal segments (of unknown length, if cut from the middle of the 60cm side). Let's assume the cut is made by replacing a 30 cm segment of the 60 cm side with a 20 cm + 30 cm + 20 cm "U" shape (where the 30 cm is the base of the U, and 20 cm are the two vertical sides). The perimeter after cutting would be: (Original Perimeter) - (length of cut out part on the edge) + (new boundary lengths). Here, a length of 30 cm from one of the 60 cm sides is replaced by a path of \( 20 + 30 + 20 = 70 \) cm. So, the perimeter after cutting \( = 200 - 30 + (20 + 30 + 20) = 200 - 30 + 70 = 240 \) cm. Difference in perimeter \( = 240 - 200 = 40 \) cm. Thus, the perimeter will increase by 40 cm.
In simple words: For the first shape where a corner was cut, the perimeter stayed the same, meaning no difference. For the second shape where a piece was cut from the middle of a side, the perimeter increased by 40 cm because new edges were added inside the shape.

🎯 Exam Tip: When a piece is cut from a figure, always visualize how the boundary changes. If new internal edges are formed, they contribute to the perimeter. A corner cut often keeps the perimeter the same, while a cut from the middle of a side generally increases it.

 

Question 5. On a centimeter squared paper, make as many rectangles as you can, such that the area of the rectangle is 64 sq.cm (consider only natural number lengths).
(i) Which rectangle has the greatest perimeter?
(ii) Which rectangle has the least perimeter?
(iii) Find the change in the width of rectangle with decreasing perimeter of the rectangle.
Answer: To find all possible rectangles with an area of 64 sq. cm using natural number lengths, we need to find pairs of factors for 64. A natural number length means the sides must be positive whole numbers. The perimeter of a rectangle is \( 2(l+b) \). Generally, for a fixed area, the perimeter is greatest when the rectangle is long and thin, and least when it is closer to a square.
Here are the possible lengths (l) and breadths (b) for rectangles with an area of 64 sq. cm, and their corresponding perimeters:

(i) The rectangle with the greatest perimeter is when length \( = 64 \) cm and breadth \( = 1 \) cm. Its perimeter is \( 130 \) cm.
(ii) The rectangle with the least perimeter is the square, when length \( = 8 \) cm and breadth \( = 8 \) cm. Its perimeter is \( 32 \) cm.
(iii) As the perimeter of the rectangle decreases, the width (breadth) of the rectangle increases. For example, when perimeter goes from 130 cm to 32 cm, the breadth goes from 1 cm to 8 cm. This shows that shapes closer to a square have smaller perimeters for the same area.
In simple words: We found different ways to make a rectangle with an area of 64 sq. cm using whole number sides. The longest, thinnest rectangle (64 cm by 1 cm) had the biggest perimeter. The square-shaped rectangle (8 cm by 8 cm) had the smallest perimeter. As the perimeter got smaller, the width of the rectangle got bigger, making it more square-like.

🎯 Exam Tip: For a given area, a rectangle with sides that are very different (long and thin) will have a larger perimeter, while a square (or a rectangle with sides close to equal) will have the smallest perimeter.

 

Question 6. On a centimeter squared paper, make as many rectangles as you can, such that the perimeter of the rectangle is 16 cm (consider only natural number lengths).
(i) Which rectangle has the greatest area?
(ii) Which rectangle has the least area?
(iii) Find the change in length of rectangle with increasing area of the rectangle.
Answer: We need to find all possible rectangles with a perimeter of 16 cm using natural number lengths. The perimeter of a rectangle is \( 2(l+b) \). For a fixed perimeter, the area is greatest when the rectangle is closer to a square, and least when it is long and thin. This is the opposite pattern to fixed area problems.
Given perimeter \( = 16 \) cm.
So, \( 2(l+b) = 16 \implies l+b = 8 \). We need pairs of natural numbers \( l \) and \( b \) that add up to 8. The length (l) should generally be greater than or equal to the breadth (b).
Here are the possible lengths (l) and breadths (b) for rectangles with a perimeter of 16 cm, and their corresponding areas:

(i) The rectangle with the greatest area is the square, when length \( = 4 \) cm and breadth \( = 4 \) cm. Its area is \( 16 \) sq. cm.
(ii) The rectangle with the least area is when length \( = 7 \) cm and breadth \( = 1 \) cm. Its area is \( 7 \) sq. cm.
(iii) As the area of the rectangle increases, the length of the rectangle decreases (moving towards a square shape). For example, as the area goes from 7 sq. cm to 16 sq. cm, the length decreases from 7 cm to 4 cm.
In simple words: We made different rectangles that all had a perimeter of 16 cm, using whole numbers for sides. The square shape (4 cm by 4 cm) had the largest area (16 sq. cm). The long, thin rectangle (7 cm by 1 cm) had the smallest area (7 sq. cm). We saw that as the area got bigger, the length of the rectangle got shorter, making it more like a square.

🎯 Exam Tip: For a given perimeter, a square will always enclose the maximum area. A rectangle that is very long and thin will enclose the minimum area for that perimeter.