GSEB Class 7 Maths Solutions Chapter 11 Perimeter and Area Exercise 11.1

Question 1. The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find:
(i) its area
(ii) the cost of the land, if 1 m² of the land costs Rs. 10,000.
Answer:
(i) Length of the rectangle (l) = 500 m
Breadth of the rectangle (b) = 300 m
Therefore, Area = \( l \times b \)
= \( 500 \text{ m} \times 300 \text{ m} = 150000 \text{ m}^2 \)
(ii) Cost of \( 1 \text{ m}^2 \) of land = Rs. 10,000
Therefore, Cost of \( 150000 \text{ m}^2 \) of land
= Rs. \( 10000 \times 150000 \) = Rs. \( 1,50,00,00,000 \)
In simple words: We first calculate the land's area by multiplying its length and breadth. Then, we find the total cost by multiplying this area by the given cost per square meter.

Exam Tip: Remember to use the correct units for area (m²) and cost (Rs.). Always show your formulas clearly for full marks.

Question 2. Find the area of a square park whose perimeter is 320 m.
Answer:
Perimeter of the square park = 320 m
We know that the perimeter of a square is \( 4 \times \text{Side} \).
So, \( 4 \times \text{Side} = 320 \text{ m} \)
Therefore, Side = \( \frac{320}{4} \text{ m} = 80 \text{ m} \)
Now, the area of the square park = \( \text{Side} \times \text{Side} \)
= \( 80 \text{ m} \times 80 \text{ m} = 6400 \text{ m}^2 \)
In simple words: First, find the length of one side of the square by dividing its perimeter by four. Then, multiply the side length by itself to get the area.

Exam Tip: Be careful not to confuse perimeter with area. Perimeter is the distance around, while area is the space inside.

Question 3. Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.
Answer:
Length of the rectangle = 22 m
Area of the rectangle = 440 m²
Let the breadth of the rectangle be \( b \) m.
We know that Area of a rectangle = \( \text{Length} \times \text{Breadth} \)
So, \( 22 \times b = 440 \)
Therefore, \( b = \frac{440}{22} \text{ m} = 20 \text{ m} \)
Thus, the breadth of the rectangle is 20 m.
Now, the perimeter of a rectangle = \( 2(\text{Length} + \text{Breadth}) \)
= \( 2(22 + 20) \text{ m} \)
= \( 2(42) \text{ m} = 84 \text{ m} \)
In simple words: First, divide the area by the length to get the breadth. Then, use the formula for perimeter (two times length plus breadth) to calculate the perimeter.

Exam Tip: When given area and one dimension, divide to find the other dimension. Then use both dimensions to calculate the perimeter.

Question 4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Answer:
Perimeter of the rectangular sheet = 100 cm
Length (l) = 35 cm
Let the breadth be \( b \) cm.
We know that the perimeter of a rectangle = \( 2(l + b) \)
So, \( 2(35 + b) = 100 \text{ cm} \)
Dividing by 2 on both sides:
\( 35 + b = \frac{100}{2} \text{ cm} \)
\( 35 + b = 50 \text{ cm} \)
\( b = (50 - 35) \text{ cm} \)
\( b = 15 \text{ cm} \)
Therefore, the breadth of the rectangular sheet is 15 cm.
Now, the area of the rectangle = \( l \times b \)
= \( 35 \text{ cm} \times 15 \text{ cm}^2 = 525 \text{ cm}^2 \)
In simple words: First, use the perimeter formula to find the breadth. Then, multiply the length and breadth to find the area.

Exam Tip: Remember to solve for the unknown dimension using the perimeter formula first. Then, use both dimensions to compute the area.

Question 5. The area of a square park is the same as that of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Answer:
Side of the square park = 60 m
Area of the square park = \( 60 \text{ m} \times 60 \text{ m} = 3600 \text{ m}^2 \)
Given that the area of the rectangular park is the same as the area of the square park.
So, Area of the rectangular park = \( 3600 \text{ m}^2 \)
Length of the rectangular park = 90 m
We know that Area of a rectangle = \( \text{Length} \times \text{Breadth} \)
Therefore, Breadth of the rectangular park = \( \frac{\text{Area}}{\text{Length}} \)
= \( \frac{3600}{90} \text{ m} = 40 \text{ m} \)
In simple words: Calculate the area of the square park. Since the rectangular park has the same area, use this area and the given length to find the breadth of the rectangular park.

Exam Tip: When problems state that two shapes have the "same area," compute the area of the shape for which you have complete information first, then use that area for the second shape.

Question 6. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also find which shape encloses more area?
Answer:
For the rectangular shape:
Length (l) = 40 cm
Breadth (b) = 22 cm
Perimeter of the rectangle = \( 2(l + b) = 2(40 + 22) \text{ cm} \)
= \( 2(62) \text{ cm} = 124 \text{ cm} \)
Area of the rectangle = \( l \times b = 40 \text{ cm} \times 22 \text{ cm} = 880 \text{ cm}^2 \)
When the same wire is rebent to form a square, its perimeter remains the same.
So, Perimeter of the square = Perimeter of the rectangle = 124 cm
Let the side of the square be \( s \).
Perimeter of the square = \( 4 \times s \)
\( 4 \times s = 124 \text{ cm} \)
Therefore, \( s = \frac{124}{4} \text{ cm} = 31 \text{ cm} \)
The measure of each side of the square is 31 cm.
Area of the square = \( s \times s = 31 \text{ cm} \times 31 \text{ cm} = 961 \text{ cm}^2 \)
Comparing the areas:
\( 961 \text{ cm}^2 > 880 \text{ cm}^2 \)
Therefore, the square shape encloses more area than the rectangle.
In simple words: First, calculate the perimeter and area of the rectangle. Since the wire is the same, the square will have the same perimeter. Use this to find the side length of the square and then its area. Finally, compare the areas to see which shape is larger.

Exam Tip: Remember that bending a wire changes its shape but not its total length, which means its perimeter stays the same. The area, however, can change significantly.

Question 7. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length and the area of the rectangle.
Answer:
Perimeter of the rectangle = 130 cm
Breadth of the rectangle = 30 cm
Let the length of the rectangle be \( l \).
We know that the perimeter of a rectangle = \( 2(l + \text{breadth}) \)
So, \( 2(l + 30) = 130 \text{ cm} \)
Dividing by 2 on both sides:
\( l + 30 = \frac{130}{2} \text{ cm} \)
\( l + 30 = 65 \text{ cm} \)
\( l = 65 - 30 \text{ cm} \)
\( l = 35 \text{ cm} \)
Therefore, the length of the rectangle is 35 cm.
Now, the area of the rectangle = \( l \times \text{breadth} \)
= \( 35 \text{ cm} \times 30 \text{ cm} = 1050 \text{ cm}^2 \)
In simple words: Use the perimeter formula with the given breadth to first calculate the length. Once you have both length and breadth, multiply them together to find the area.

Exam Tip: Always write down the formulas for perimeter and area. This helps to systematically solve for unknown dimensions and then compute the required values.

Question 8. A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is Rs. 20 per m².
Answer:
For the wall:
Length of the wall (l) = 4.5 m
Breadth of the wall (b) = 3.6 m
Area of the wall = \( l \times b = 4.5 \text{ m} \times 3.6 \text{ m} = 16.2 \text{ m}^2 \)
For the door:
Length of the door = 2 m
Breadth of the door = 1 m
Area of the door = \( 2 \text{ m} \times 1 \text{ m} = 2 \text{ m}^2 \)
The area to be white washed is the area of the wall minus the area of the door.
Area to be white washed = [Area of the wall] – [Area of the door]
= \( 16.2 \text{ m}^2 - 2 \text{ m}^2 = 14.2 \text{ m}^2 \)
The rate of white washing = Rs. 20 per m²
Cost of white washing = Rate \( \times \) Area to be white washed
= Rs. \( 20 \times 14.2 \) = Rs. 284
In simple words: First, find the total area of the wall. Then, find the area of the door. Subtract the door's area from the wall's area to get the actual area to be painted. Finally, multiply this paintable area by the cost per square meter.

Exam Tip: Remember to subtract any unpaintable areas (like doors or windows) from the total wall area before calculating the cost of painting or whitewashing.