RBSE Solutions Class 9 Maths Chapter 11 Area of Plane Figures Exercise 11.4

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 11 Area of Plane Figures Ex 11.4

 

Question 1. Find the perimeter and area of the following rectangles.
(i) length = 9.5 m, breadth = 7.5 m
(ii) length = 125 m, breadth = 75 m
(iii) length = 12.5 cm, breadth = 7.5 cm
Answer:
(i) For a rectangle with length \( l = 9.5 \) m and breadth \( b = 7.5 \) m:
Perimeter \( = 2(l + b) = 2(9.5 + 7.5) = 2 \times 17 = 34 \) m
Area \( = l \times b = 9.5 \times 7.5 = 71.25 \) m\( ^2 \)
(ii) For a rectangle with length \( l = 125 \) m and breadth \( b = 75 \) m:
Perimeter \( = 2(l + b) = 2(125 + 75) = 2 \times 200 = 400 \) m
Area \( = l \times b = 125 \times 75 = 9375 \) m\( ^2 \)
(iii) For a rectangle with length \( l = 12.5 \) cm and breadth \( b = 7.5 \) cm:
Perimeter \( = 2(l + b) = 2(12.5 + 7.5) = 2 \times 20 = 40 \) cm
Area \( = l \times b = 12.5 \times 7.5 = 93.75 \) cm\( ^2 \). Calculating area and perimeter is crucial in many real-world applications, like designing rooms or planning fields.
In simple words: To find the perimeter of a rectangle, you add all four sides together. To find the area, you multiply its length by its width. Remember to keep the units correct, like meters or centimeters.

🎯 Exam Tip: Always write down the formulas for perimeter and area before substituting values to ensure you don't miss any steps or lose marks.

 

Question 2. Find the area and perimeter of the following square whose sides are
(i) 5.3 m
(ii) 8.5 m
(iii) 9.6 m
Answer:
(i) For a square with side \( s = 5.3 \) m:
Area \( = s^2 = (5.3)^2 = 28.09 \) m\( ^2 \)
Perimeter \( = 4 \times s = 4 \times 5.3 = 21.2 \) m
(ii) For a square with side \( s = 8.5 \) m:
Area \( = s^2 = (8.5)^2 = 72.25 \) m\( ^2 \)
Perimeter \( = 4 \times s = 4 \times 8.5 = 34.0 \) m
(iii) For a square with side \( s = 9.6 \) m:
Area \( = s^2 = (9.6)^2 = 92.16 \) m\( ^2 \)
Perimeter \( = 4 \times s = 4 \times 9.6 = 38.4 \) m. Squares are a special type of rectangle where all sides are equal, making their calculations simpler.
In simple words: For a square, all sides are the same length. To find its area, you multiply the side length by itself. To find its perimeter, you multiply the side length by four.

🎯 Exam Tip: Make sure to include the correct square units (like m\( ^2 \) or cm\( ^2 \)) for area and linear units (like m or cm) for perimeter in your final answers.

 

Question 3. The distance covered in taking 4 rounds of a square ground is 4 km. Find the length and area of the ground.
Answer:
Distance covered in 4 rounds \( = 4 \) km
Distance covered in 1 round \( = \frac{4 \text{ km}}{4} = 1 \) km
We know that 1 km \( = 1000 \) m, so the distance covered in 1 round is \( 1000 \) m.
This means the perimeter of the square ground is \( 1000 \) m.
For a square, Perimeter \( = 4 \times \text{one side} \)
So, \( 4 \times \text{side} = 1000 \) m
\( \implies \text{side} = \frac{1000}{4} = 250 \) m
Now, the area of the square field \( = (\text{side})^2 = (250 \text{ m})^2 = 62500 \) m\( ^2 \). Understanding units and converting them (like km to m) is vital for accurate calculations.
In simple words: First, find out how long one round of the square ground is by dividing the total distance by the number of rounds. This gives you the perimeter. Then, divide the perimeter by 4 to find the length of one side of the square. Finally, multiply the side length by itself to get the area of the ground.

🎯 Exam Tip: Always convert all units to a consistent system (e.g., all meters) before performing calculations to avoid errors.

 

Question 4. A rectangular plot 75 m in length and 45 m width. How many flower-beds can be prepared if each bed is 5 m in length and 3 m width?
Answer:
For the rectangular plot:
Length \( = 75 \) m
Width \( = 45 \) m
Area of the rectangular plot \( = \text{length} \times \text{breadth} = 75 \text{ m} \times 45 \text{ m} = 3375 \) m\( ^2 \)
For one flower-bed:
Length \( = 5 \) m
Width \( = 3 \) m
Area of one flower-bed \( = \text{length} \times \text{breadth} = 5 \text{ m} \times 3 \text{ m} = 15 \) m\( ^2 \)
The number of flower-beds that can be prepared \( = \frac{\text{Area of plot}}{\text{Area of one flower bed}} = \frac{3375 \text{ m}^2}{15 \text{ m}^2} = 225 \). This calculation shows how to effectively use space by fitting smaller units into a larger area.
In simple words: To find how many smaller flower-beds can fit into a larger plot, first calculate the total area of the big plot. Then, calculate the area of just one small flower-bed. Finally, divide the big area by the small area to find how many beds you can make.

🎯 Exam Tip: Ensure that the units for both the larger area and the smaller area are the same before dividing to get the correct number of items.

 

Question 5. A room is 10 m long and 5 m wide. How many squares of area 50 sq. cm are required? Also find the cost of paving 10000 such squares at Rs 20 each.
Answer:
For the room:
Length \( = 10 \) m
Width \( = 5 \) m
Area of the room \( = \text{length} \times \text{width} = 10 \text{ m} \times 5 \text{ m} = 50 \) m\( ^2 \)
Area of one square tile \( = 50 \) sq. cm
We need to convert the area of the room to sq. cm to match the tile units. We know \( 1 \) m \( = 100 \) cm, so \( 1 \) m\( ^2 \) \( = 100 \times 100 = 10000 \) cm\( ^2 \).
Area of the room in cm\( ^2 \) \( = 50 \text{ m}^2 \times 10000 \text{ cm}^2/\text{m}^2 = 500000 \) cm\( ^2 \)
Number of squares required \( = \frac{\text{Area of room}}{\text{Area of one square}} = \frac{500000 \text{ cm}^2}{50 \text{ cm}^2} = 10000 \) squares
Cost of paving \( 1 \) square \( = \) Rs \( 20 \)
Total cost of paving \( 10000 \) squares \( = 10000 \times 20 = \) Rs \( 2,00,000 \). Unit consistency is very important here; mixing meters and centimeters can lead to errors.
In simple words: First, find the total area of the room in square centimeters. Then, divide the room's area by the area of one small square tile to see how many tiles are needed. Finally, multiply the total number of tiles by the cost of each tile to get the total cost.

🎯 Exam Tip: Always pay close attention to units! If lengths are in meters and tile areas in square centimeters, convert everything to the same unit before calculating. For example, convert meters to centimeters or vice versa.

 

Question 6. The area of a square field is 2025 sq. m and around it, there is a path way 3.5 metre wide, find the area of the pathway.
Answer:
Area of the square field \( = 2025 \) m\( ^2 \)
The side of the square field \( = \sqrt{\text{Area}} = \sqrt{2025} = 45 \) m
The path is \( 3.5 \) m wide all around the field. So, the side of the field including the path will be longer by \( 3.5 \) m on each side.
Side of the field including the path \( = 45 + (3.5 + 3.5) = 45 + 7 = 52 \) m
Area of the field including the path \( = (\text{side})^2 = (52 \text{ m})^2 = 2704 \) m\( ^2 \)
Area of the path \( = (\text{Area of field including path}) - (\text{Area of square field}) \)
\( = 2704 \text{ m}^2 - 2025 \text{ m}^2 = 679 \) m\( ^2 \). This method of finding the area of a border by subtracting inner and outer areas is common.
In simple words: First, find the side length of the square field from its area. Then, add the width of the path to both sides of the field to find the side length of the bigger square (field plus path). Calculate the area of this bigger square. Finally, subtract the area of the original field from the area of the bigger square to get the area of the pathway only.

🎯 Exam Tip: When a path surrounds an area, remember to add the path width twice (once for each side) to the original dimensions to find the outer dimensions.

 

Question 7. A road of 2.5 m wide, around a rectangular room 35 metre long and 40 metre wide, is to be constructed. Find the area of the road and the number of square blocks of 0.8 m x 0.5 m required to pave the road.
Answer:
Length of the room \( = 35 \) m
Width of the room \( = 40 \) m
Area of the room (inner area) \( = 35 \text{ m} \times 40 \text{ m} = 1400 \) m\( ^2 \)
The road is \( 2.5 \) m wide around the room. So, the dimensions of the room including the road will be:
New length \( = 35 + (2.5 + 2.5) = 35 + 5 = 40 \) m
New width \( = 40 + (2.5 + 2.5) = 40 + 5 = 45 \) m
Area of the rectangular room with road (outer area) \( = 40 \text{ m} \times 45 \text{ m} = 1800 \) m\( ^2 \)
Area of the road \( = \text{outer area} - \text{inner area} = 1800 \text{ m}^2 - 1400 \text{ m}^2 = 400 \) m\( ^2 \)
Now, we need to find the number of square blocks required to pave the road.
Area of one square block \( = 0.8 \text{ m} \times 0.5 \text{ m} = 0.4 \) m\( ^2 \)
Required number of square blocks \( = \frac{\text{Area of road}}{\text{Area of one square block}} = \frac{400 \text{ m}^2}{0.4 \text{ m}^2} = 1000 \) slabs. This problem combines area calculation with finding how many smaller units fit into a larger space.
In simple words: First, calculate the area of the room by itself. Then, add the road's width to both the length and width of the room to find the new, larger dimensions. Calculate the area of this larger space. Subtract the room's area from this larger area to find the road's area. Finally, divide the road's area by the area of one paving block to find how many blocks are needed.

🎯 Exam Tip: Be careful when adding the path width to dimensions. For a path surrounding an object, the width is added to both ends, so you effectively add `2 * path width` to both the length and breadth.

 

Question 8. A pathway 4 metre wide has been constructed centrally in a garden parallel to its sides of length 60 metre and width 40 metre. Find the area of the pathway and find the total expenditure to cover this pathway with sand at the rate of Rs 100 per sq. metre.
Answer:
Length of the garden \( = 60 \) m
Breadth of the garden \( = 40 \) m
The pathway is 4 m wide and runs centrally, parallel to the sides.
Area of the path along the length (horizontal path) \( = \text{length of garden} \times \text{width of path} = 60 \text{ m} \times 4 \text{ m} = 240 \) m\( ^2 \)
Area of the path along the breadth (vertical path) \( = \text{breadth of garden} \times \text{width of path} = 40 \text{ m} \times 4 \text{ m} = 160 \) m\( ^2 \)
When two paths cross centrally, the overlapping area (common area) is a square with side equal to the path width.
Common area \( = 4 \text{ m} \times 4 \text{ m} = 16 \) m\( ^2 \)
Total area of the pathway \( = (\text{Area of path along length}) + (\text{Area of path along breadth}) - (\text{Common area}) \)
\( = 240 \text{ m}^2 + 160 \text{ m}^2 - 16 \text{ m}^2 = 400 \text{ m}^2 - 16 \text{ m}^2 = 384 \) m\( ^2 \)
Expenditure of lying sand per m\( ^2 \) \( = \) Rs \( 100 \)
Total expenditure of sand on \( 384 \) m\( ^2 \) \( = 384 \times 100 = \) Rs \( 38,400 \). Calculating the common area correctly is key to avoiding overcounting.
In simple words: First, find the area of the path that runs along the length of the garden. Then, find the area of the path that runs along the width. Since these paths cross, there's a small square area counted twice; calculate this common area. Subtract the common area from the sum of the two path areas to get the actual total path area. Finally, multiply this total path area by the cost per square meter to find the total money needed.

🎯 Exam Tip: When paths cross each other, remember to subtract the area of the central overlapping square once to avoid double-counting it in the total path area.

 

Question 9. Find the area and perimeter of the given figure.

Answer:
For finding the total area, we divide the given figure into three blocks, named A, B, and C as described in the problem's solution.
Area of block A \( = 12 \text{ m} \times 2 \text{ m} = 24 \) m\( ^2 \)
Area of block B \( = 5 \text{ m} \times 2 \text{ m} = 10 \) m\( ^2 \)
Area of block C \( = 12 \text{ m} \times 2 \text{ m} = 24 \) m\( ^2 \)
Total area of the given figure \( = (\text{Area A}) + (\text{Area B}) + (\text{Area C}) = 24 \text{ m}^2 + 10 \text{ m}^2 + 24 \text{ m}^2 = 58 \) m\( ^2 \)
The perimeter is the sum of all outer boundaries of the figure. According to the breakdown in the solution, the perimeter is:
Perimeter \( = 2\text{m} + 12\text{m} + 2\text{m} + 5\text{m} + 5\text{m} + 5\text{m} + 2\text{m} + 12\text{m} + 2\text{m} + 5\text{m} + 5\text{m} + 5\text{m} = 62 \) m. Breaking down complex shapes into simpler ones makes it easier to calculate their properties. This strategy is useful in many geometry problems.
In simple words: To find the total area of a complex shape, break it down into smaller, simpler shapes like rectangles. Calculate the area of each small part and then add them all together. For the perimeter, carefully add up the length of all the outer edges of the shape.

🎯 Exam Tip: When dealing with composite figures, clearly label the parts you are breaking the figure into. For the perimeter, trace the outer boundary to ensure you include every segment and don't count any internal lines.

 

Question 10. The length, breadth and height of a room are 15.35 m, 4.65 m and 6.50 m respectively. If there are 4 doors each of size 1.5 m x 1.3 m and 3 windows each of size 1.5 m x 1.2 m, then find the cost to paint the four walls of room at the rate of Rs 3.40 per sq. m.
Answer:
Given dimensions of the room:
Length \( l = 15.35 \) m
Breadth \( b = 4.65 \) m
Height \( h = 6.50 \) m
Area of the four walls of the room \( = 2 \times (l + b) \times h \)
\( = 2 \times (15.35 + 4.65) \times 6.50 \)
\( = 2 \times 20 \times 6.50 = 260 \) m\( ^2 \)
Dimensions of one door: \( 1.5 \) m \( \times 1.3 \) m
Area of one door \( = 1.5 \times 1.3 = 1.95 \) m\( ^2 \)
Area of 4 doors \( = 4 \times 1.95 = 7.8 \) m\( ^2 \)
Dimensions of one window: \( 1.5 \) m \( \times 1.2 \) m
Area of one window \( = 1.5 \times 1.2 = 1.8 \) m\( ^2 \)
Area of 3 windows \( = 3 \times 1.8 = 5.4 \) m\( ^2 \)
Area of the four walls to be painted (excluding doors and windows):
\( = (\text{Area of four walls}) - (\text{Area of 4 doors} + \text{Area of 3 windows}) \)
\( = 260 - (7.8 + 5.4) \)
\( = 260 - 13.2 = 246.8 \) m\( ^2 \)
Cost to paint per sq. m \( = \) Rs \( 3.40 \)
Total cost to paint \( = 246.8 \times 3.40 = \) Rs \( 839.12 \). Always subtract the unpaintable areas like doors and windows from the total wall area.
In simple words: First, find the total area of all four walls of the room. Then, calculate the total area of all doors and all windows. Subtract these areas from the total wall area to find the actual area that needs to be painted. Finally, multiply this paintable area by the cost per square meter to get the total painting cost.

🎯 Exam Tip: Remember that doors and windows are not painted, so their areas must be calculated and subtracted from the total wall area before determining the painting cost.

 

Question 11. A rectangular water tank is 10 metre long, 8 metre wide and 2 metre deep. Find the total cost to repair its four walls and bottom at the rate of Rs 15 per square metre.
Answer:
Given dimensions of the rectangular water tank:
Length \( l = 10 \) m
Width \( w = 8 \) m
Depth (height) \( h = 2 \) m
Area of the floor (bottom) of the water tank \( = l \times w = 10 \text{ m} \times 8 \text{ m} = 80 \) m\( ^2 \)
Area of the four walls of the water tank \( = 2 \times (l + w) \times h \)
\( = 2 \times (10 + 8) \times 2 \)
\( = 2 \times 18 \times 2 = 72 \) m\( ^2 \)
Total area to be repaired (four walls and bottom) \( = (\text{Area of floor}) + (\text{Area of four walls}) \)
\( = 80 \text{ m}^2 + 72 \text{ m}^2 = 152 \) m\( ^2 \)
Cost of repairing \( 1 \) m\( ^2 \) \( = \) Rs \( 15 \)
Total cost of repairing \( 152 \) m\( ^2 \) \( = 152 \times 15 = \) Rs \( 2,280 \). This problem is similar to painting a room, but here we consider the floor instead of the ceiling.
In simple words: First, calculate the area of the tank's floor. Then, calculate the area of its four side walls. Add these two areas together to find the total area that needs to be repaired. Finally, multiply this total repair area by the cost per square meter to find the total repair cost.

🎯 Exam Tip: For problems involving tanks or rooms, always clarify which surfaces need to be included in the area calculation (e.g., walls and floor, but not the top/ceiling).