OP Malhotra Class 9 Maths Solutions Chapter 16 Area of Plane Figures Exercise 16 (A)

Exercise 16(A)

Question 1. Find the perimeter and area of the rectangle whose dimensions are
(i) 8 cm by 5 cm
(ii) 12 m by 9 m
(iii) 3.5 cm by 2 m
Answer:
(i) For a rectangle with length \( l = 8 \) cm and breadth \( b = 5 \) cm:
Perimeter \( = 2(l + b) = 2(8 + 5) \) cm \( = 2 \times 13 = 26 \) cm.
Area \( = l \times b = 8 \times 5 = 40 \) cm\(^2\).
(ii) For a rectangle with length \( l = 12 \) m and breadth \( b = 9 \) m:
Perimeter \( = 2(l + b) = 2(12 + 9) \) m \( = 2 \times 21 = 42 \) m.
Area \( = l \times b = 12 \times 9 = 108 \) m\(^2\).
(iii) For a rectangle with length \( l = 3.5 \) m and breadth \( b = 2 \) m:
Perimeter \( = 2(l + b) = 2(3.5 + 2) \) m \( = 2 \times 5.5 = 11 \) m.
Area \( = l \times b = 3.5 \times 2 = 7 \) m\(^2\).
In simple words: To find the perimeter, you add up all the sides of the rectangle. To find the area, you multiply its length by its width. Remember to use the correct units.

🎯 Exam Tip: Always remember the formulas for perimeter \( (2(l+b)) \) and area \( (l \times b) \) of a rectangle, and ensure you include the correct units (\( \text{cm} \), \( \text{m} \) for perimeter and \( \text{cm}^2 \), \( \text{m}^2 \) for area) in your final answer.

Question 2. Find the perimeter and area of the square whose each side is
(i) 3 cm
(ii) 5 m
(iii) 1.3 m
(iv) 2.4 m
Answer:
(i) For a square with side \( a = 3 \) cm:
Perimeter \( = 4a = 4 \times 3 = 12 \) cm.
Area \( = a^2 = (3)^2 = 9 \) cm\(^2\).
(ii) For a square with side \( a = 5 \) m:
Perimeter \( = 4a = 4 \times 5 = 20 \) m.
Area \( = a^2 = (5)^2 = 25 \) m\(^2\).
(iii) For a square with side \( a = 1.3 \) m:
Perimeter \( = 4a = 4 \times 1.3 = 5.2 \) m.
Area \( = a^2 = (1.3)^2 = 1.69 \) m\(^2\).
(iv) For a square with side \( a = 2.4 \) m:
Perimeter \( = 4a = 4 \times 2.4 = 9.6 \) m.
Area \( = a^2 = (2.4)^2 = 5.76 \) m\(^2\).
In simple words: For a square, all sides are equal. So, the perimeter is 4 times one side, and the area is one side multiplied by itself. Units are important here too!

🎯 Exam Tip: Remember that a square is a special type of rectangle where length equals breadth. The formulas simplify to \( 4a \) for perimeter and \( a^2 \) for area, where 'a' is the side length.

Question 3. Find the length of one side and perimeter of a square whose area is :
(a) 25 m²
(b) 144 m²
(c) 256 cm²
(d) 961 dm²
Answer:
(i) Given Area \( = 25 \) m\(^2\):
Side \( a = \sqrt{\text{Area}} = \sqrt{25} = 5 \) m.
Perimeter \( = 4a = 4 \times 5 = 20 \) m.
(ii) Given Area \( = 144 \) m\(^2\):
Side \( a = \sqrt{\text{Area}} = \sqrt{144} = 12 \) m.
Perimeter \( = 4a = 4 \times 12 = 48 \) m.
(iii) Given Area \( = 256 \) cm\(^2\):
Side \( a = \sqrt{\text{Area}} = \sqrt{256} = 16 \) cm.
Perimeter \( = 4a = 4 \times 16 = 64 \) cm.
(iv) Given Area \( = 961 \) dm\(^2\):
Side \( a = \sqrt{\text{Area}} = \sqrt{961} = 31 \) dm.
Perimeter \( = 4a = 4 \times 31 = 124 \) dm.
In simple words: If you know the area of a square, you can find the length of one side by taking the square root. Once you have the side length, multiply it by 4 to get the perimeter.

🎯 Exam Tip: When given the area of a square, calculate the side length first by finding the square root. Always check the units and use them consistently in your calculations and final answers.

Question 4. The perimeter of a rectangle is 30 cm, and its breadth is 6 cm, find its length and its area.
Answer:
Given perimeter of rectangle \( = 30 \) cm and breadth \( b = 6 \) cm.
The formula for perimeter is \( 2(l + b) \). So, \( 2(l + 6) = 30 \).
Divide both sides by 2: \( l + 6 = \frac{30}{2} = 15 \) cm.
Now, find the length \( l = 15 - 6 = 9 \) cm.
Next, calculate the area \( = l \times b = 9 \times 6 = 54 \) cm\(^2\).
In simple words: If you know the perimeter and breadth of a rectangle, you can first find the length by working backwards with the perimeter formula. Then, use the length and breadth to calculate the area.

🎯 Exam Tip: Break down the problem into steps. First, use the perimeter formula to find the unknown side, then use both sides to find the area. Keep track of units at each step.

Question 5. The area of a rectangular courtyard, 30 m long, is 450 m². Find its breadth and perimeter.
Answer:
Given area of rectangular courtyard \( = 450 \) m\(^2\) and length \( l = 30 \) m.
The formula for area is \( l \times b \). So, \( 30 \times b = 450 \).
Divide both sides by 30 to find the breadth \( b = \frac{450}{30} = 15 \) m.
Now, calculate the perimeter \( = 2(l + b) = 2(30 + 15) \) m \( = 2 \times 45 = 90 \) m.
In simple words: If you know the area and length of a rectangle, you can find the breadth by dividing the area by the length. After that, use both the length and breadth to find the perimeter.

🎯 Exam Tip: Always remember that area is length times breadth, so if you have area and one side, you can find the other side. Then, use the standard perimeter formula.

Question 6. The perimeter of a rectangular field is \( \frac { 3 }{ 5 } \) km. If the length of the field is twice its width, find its area in sq. metres.
Answer:
Given perimeter of a rectangular field \( = \frac{3}{5} \) km.
First, convert the perimeter to metres: \( \frac{3}{5} \times 1000 = 600 \) m. (Since 1 km = 1000 m).
Let the width of the field be \( x \) metres. Then, the length is \( 2x \) metres.
The perimeter formula is \( 2(l + b) \). So, \( 2(2x + x) = 600 \).
\( 2(3x) = 600 \)
\( 6x = 600 \)
Divide by 6: \( x = \frac{600}{6} = 100 \) m.
So, the width \( b = 100 \) m and the length \( l = 2x = 2 \times 100 = 200 \) m.
Finally, find the area \( = l \times b = 200 \times 100 = 20000 \) m\(^2\).
In simple words: First, change the perimeter from kilometres to metres. Then, use the perimeter and the fact that the length is twice the width to find both measurements. Once you have the length and width in metres, multiply them to get the area in square metres.

🎯 Exam Tip: Pay close attention to unit conversions (km to m) at the beginning of the problem. Setting up algebraic expressions for length and width based on their relationship simplifies finding the dimensions.

Question 7. If the side of a square is doubled, find the ratio of the area of the resulting square to that of the given square.
Answer:
Let the original side of the square be \( a \) m.
The original area \( = (\text{side})^2 = a^2 \) m\(^2\).
If the side is doubled, the new side becomes \( 2a \) m.
The new area \( = (\text{new side})^2 = (2a)^2 = 4a^2 \) m\(^2\).
Now, find the ratio of the new area to the original area:
Ratio \( = \frac{\text{New Area}}{\text{Original Area}} = \frac{4a^2}{a^2} = \frac{4}{1} = 4:1 \).
In simple words: When you make the side of a square twice as long, its area becomes four times bigger. So, the new area is to the old area as 4 is to 1.

🎯 Exam Tip: When a dimension (like the side of a square) is multiplied by a factor, the area is multiplied by the square of that factor. For instance, doubling the side makes the area \( 2^2 = 4 \) times larger.

Question 8. The side of a square field is 89 m. By how many square metres does its area fall short of a hectare ?
Answer:
Given side of a square field \( a = 89 \) m.
Area of the square field \( = a^2 = (89)^2 = 7921 \) m\(^2\).
We know that 1 hectare \( = 10000 \) m\(^2\).
The difference between 1 hectare and the field's area \( = 10000 - 7921 = 2079 \) m\(^2\).
In simple words: First, calculate the area of the square field. Then, remember that one hectare is equal to 10,000 square metres. Subtract the field's area from 10,000 square metres to find out how much it lacks to be a full hectare.

🎯 Exam Tip: It is crucial to remember the conversion factor for hectares: \( 1 \text{ hectare} = 10000 \text{ m}^2 \). This is a standard unit conversion frequently tested in area problems.

Question 9. The side of a square is 25 cm. Find the length of its diagonal correct to 2 decimal places.
Answer:
Given side of a square \( a = 25 \) cm.
The length of its diagonal for a square is given by the formula \( \sqrt{2} \times a \).
Diagonal \( = \sqrt{2} \times 25 \) cm.
Using the approximate value of \( \sqrt{2} \approx 1.414 \):
Diagonal \( = 1.414 \times 25 = 35.350 \) cm.
Rounding to 2 decimal places, the diagonal is \( 35.35 \) cm.
In simple words: For a square, you can find the diagonal by multiplying the side length by the square root of 2. Then, round your answer to two decimal places as asked.

🎯 Exam Tip: Remember the formula for the diagonal of a square \( (\text{diagonal} = a\sqrt{2}) \). It's derived from the Pythagorean theorem. Also, be careful with decimal approximations and rounding to the specified number of places.

Question 10. If the diagonal of a rectangle is 10 cm and its width is 6 cm, find its area.
Answer:
Given diagonal of rectangle \( d = 10 \) cm and breadth \( b = 6 \) cm.
Using the Pythagorean theorem for a rectangle: \( (\text{diagonal})^2 = (\text{length})^2 + (\text{breadth})^2 \).
So, \( d^2 = l^2 + b^2 \).
\( (10)^2 = l^2 + (6)^2 \)
\( 100 = l^2 + 36 \)
Subtract 36 from both sides: \( l^2 = 100 - 36 = 64 \).
Take the square root: \( l = \sqrt{64} = 8 \) cm.
The length of the rectangle is 8 cm.
Now, calculate the area \( = l \times b = 8 \times 6 = 48 \) cm\(^2\).
In simple words: Use the diagonal and width of the rectangle with Pythagoras' theorem to find its length. Once you have the length, multiply it by the width to get the area.

🎯 Exam Tip: The diagonal, length, and width of a rectangle form a right-angled triangle. Applying the Pythagorean theorem \( (a^2 + b^2 = c^2) \) is key to finding missing dimensions before calculating the area.

Question 11. The length of a rectangle is twice its width. If the length of its diagonal is \( 16\sqrt{5} \) cm, find its area.
Answer:
Given length of diagonal \( = 16\sqrt{5} \) cm.
Let the width (breadth) of the rectangle be \( x \) cm.
Since the length is twice its width, the length \( l = 2x \) cm.
Using the Pythagorean theorem: \( (\text{diagonal})^2 = (\text{length})^2 + (\text{breadth})^2 \).
\( (16\sqrt{5})^2 = (2x)^2 + x^2 \)
\( 16^2 \times (\sqrt{5})^2 = 4x^2 + x^2 \)
\( 256 \times 5 = 5x^2 \)
\( 1280 = 5x^2 \)
Divide by 5: \( x^2 = \frac{1280}{5} = 256 \).
Take the square root: \( x = \sqrt{256} = 16 \) cm.
So, the breadth \( b = x = 16 \) cm.
And the length \( l = 2x = 2 \times 16 = 32 \) cm.
Now, calculate the area \( = l \times b = 32 \times 16 = 512 \) cm\(^2\).
In simple words: First, set up the width as 'x' and length as '2x'. Use these with the given diagonal in Pythagoras' theorem to find 'x'. Once 'x' is known, you have both length and width, so multiply them to find the area.

🎯 Exam Tip: When the length and width have a proportional relationship, express both in terms of a single variable (e.g., \( x \) and \( 2x \)). This simplifies the Pythagorean equation and helps solve for the dimensions correctly.

Question 12. The sides of two squares are in the ratio 2 : 3 . Find the ratio of their perimeters and areas.
Answer:
Given the ratio of the sides of two squares \( = 2:3 \).
Let the side of the first square be \( 2x \).
Let the side of the second square be \( 3x \).

(i) **Ratio of their perimeters:**
Perimeter of the first square \( = 4 \times (2x) = 8x \).
Perimeter of the second square \( = 4 \times (3x) = 12x \).
Ratio of perimeters \( = 8x : 12x = 8:12 \).
Divide both sides by 4: \( 2:3 \).

(ii) **Ratio of their areas:**
Area of the first square \( = (2x)^2 = 4x^2 \).
Area of the second square \( = (3x)^2 = 9x^2 \).
Ratio of areas \( = 4x^2 : 9x^2 = 4:9 \).
In simple words: If the sides of two squares are in a certain ratio, their perimeters will be in the same ratio. However, their areas will be in the ratio of the squares of their sides.

🎯 Exam Tip: For similar figures, the ratio of perimeters is the same as the ratio of corresponding sides, while the ratio of areas is the square of the ratio of corresponding sides.

Question 13. How many carpets, 3 m by 2 m, are required to cover the floor of a hall 30 m by 12 m ?
Answer:
Given length of the hall floor \( L = 30 \) m and breadth \( B = 12 \) m.
Area of the hall floor \( = L \times B = 30 \times 12 = 360 \) m\(^2\).
Given length of one carpet \( l = 3 \) m and breadth \( b = 2 \) m.
Area of one carpet \( = l \times b = 3 \times 2 = 6 \) m\(^2\).
Number of carpets required \( = \frac{\text{Area of floor}}{\text{Area of one carpet}} = \frac{360}{6} = 60 \) carpets.
In simple words: First, find the total area of the hall floor. Then, find the area of one carpet. Divide the total floor area by the area of one carpet to know how many carpets are needed.

🎯 Exam Tip: Ensure all dimensions are in the same units before calculating areas. The number of items needed to cover a larger area is found by dividing the larger area by the area of one item.

Question 14. A room is 5 m by 4 m. Find the cost of cementing its floor at Rs. 7.50 per m².
Answer:
Given length of a room \( l = 5 \) m and breadth \( b = 4 \) m.
Area of the room's floor \( = l \times b = 5 \times 4 = 20 \) m\(^2\).
Cost of cementing per m\(^2\) \( = \text{Rs. } 7.50 \).
Total cost of cementing \( = \text{Area} \times \text{Cost per m}^2 = 20 \times 7.50 = \text{Rs. } 150.00 \).
In simple words: First, find the area of the room's floor. Then, multiply this area by the cost to cement one square metre to get the total cost.

🎯 Exam Tip: When calculating total cost based on area, ensure you correctly compute the area first and then multiply it by the rate per unit area. Check units to avoid errors.

Question 15. A room is 12 m by 10 m by 10 m. Find the cost of covering its floor with bricks 20 cm by 6 cm, if the cost of one thousand bricks is Rs. 300.
Answer:
Given length of room \( l = 12 \) m and breadth \( b = 10 \) m.
Area of the floor \( = l \times b = 12 \times 10 = 120 \) m\(^2\).
Convert floor area to cm\(^2\) for consistency with brick dimensions:
\( 120 \) m\(^2\) \( = 120 \times (100 \times 100) \) cm\(^2\) \( = 120 \times 10000 = 1200000 \) cm\(^2\).

Given length of one brick \( = 20 \) cm and breadth \( = 6 \) cm.
Area of one brick \( = 20 \times 6 = 120 \) cm\(^2\).

Number of bricks required \( = \frac{\text{Area of floor}}{\text{Area of one brick}} = \frac{1200000}{120} = 10000 \) bricks.
Cost of one thousand bricks \( = \text{Rs. } 300 \).
Cost of one brick \( = \frac{300}{1000} = \text{Rs. } 0.30 \).
Total cost for 10000 bricks \( = 10000 \times 0.30 = \text{Rs. } 3000 \).
In simple words: First, calculate the floor area in square centimetres. Then, find the area of one brick. Divide the floor area by the brick area to find out how many bricks are needed. Finally, use the cost per thousand bricks to calculate the total cost.

🎯 Exam Tip: Consistency in units is critical. Convert all dimensions to the same unit (e.g., cm) before calculating areas and the number of items. Also, correctly calculate the cost per single item if the price is given for a bulk quantity.

Question 16. In exchange for a square plot of land, one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as the square plot. Determine the width of the rectangular plot.
Answer:
Given side of the square field \( a = 84 \) m.
Area of the square plot \( = a^2 = (84)^2 = 7056 \) m\(^2\).
The rectangular plot has the same area as the square plot, so its area is also \( 7056 \) m\(^2\).
Given length of the rectangular plot \( l = 144 \) m.
Area of rectangle \( = l \times b \). So, \( 144 \times b = 7056 \).
Width (breadth) \( b = \frac{7056}{144} = 49 \) m.
In simple words: First, find the area of the square plot. Since the rectangular plot has the same area and its length is known, divide the area by the length to find the width of the rectangular plot.

🎯 Exam Tip: The key here is to recognize that the areas are equal. Calculate the area of the square first, then use that area with the given length of the rectangle to find its width.

Question 17. A plot 110 metres long and 80 metres broad is to be covered with grass leaving 5 metres area to be laid with grass.
Answer:
Given length of plot \( l = 110 \) m and breadth \( b = 80 \) m.
Area of the entire plot \( = l \times b = 110 \times 80 = 8800 \) m\(^2\).
Grass is to be laid, leaving a 5 m margin all around.
This means the inner rectangular area (where grass is laid) will have smaller dimensions.
New inner length \( = 110 - (2 \times 5) = 110 - 10 = 100 \) m.
New inner breadth \( = 80 - (2 \times 5) = 80 - 10 = 70 \) m.
Area of the inner part (where grass is laid) \( = 100 \times 70 = 7000 \) m\(^2\).
The image below shows the plot with the path:

In simple words: First, find the total area of the plot. Since there's a 5-metre border left around the grass area, subtract 10 metres (5 metres from each side) from both the length and breadth to get the dimensions of the inner grass area. Then, multiply these new dimensions to find the area where grass will be laid.

🎯 Exam Tip: For problems involving paths or margins around a rectangular area, remember that the dimensions of the inner or outer rectangle change by *twice* the path width (once for each side). Visualizing with a simple sketch can prevent errors.

Question 18. The dimensions of a rectangular field are in the ratio 6 : 5. Find the cost of constructing a fence around the field at the rate of Rs. 1.25 per metre, given that the area of the field is 27000 m².
Answer:
Given the ratio of the dimensions (length : breadth) of the field \( = 6:5 \).
Let the length \( l = 6x \) m and the breadth \( b = 5x \) m.
Area of the field \( = l \times b = (6x) \times (5x) = 30x^2 \) m\(^2\).
We are given that the area of the field is \( 27000 \) m\(^2\).
So, \( 30x^2 = 27000 \).
Divide by 30: \( x^2 = \frac{27000}{30} = 900 \).
Take the square root: \( x = \sqrt{900} = 30 \).
Now, find the actual dimensions:
Length \( l = 6x = 6 \times 30 = 180 \) m.
Breadth \( b = 5x = 5 \times 30 = 150 \) m.
To fence the field, we need to find its perimeter.
Perimeter \( = 2(l + b) = 2(180 + 150) \) m \( = 2(330) = 660 \) m.
Cost of fence per metre \( = \text{Rs. } 1.25 \).
Total cost of fencing \( = 660 \times 1.25 = \text{Rs. } 825 \).
In simple words: First, use the ratio and the given area to find the actual length and breadth of the field. Once you have these, calculate the perimeter of the field. Finally, multiply the perimeter by the cost per metre to find the total fencing cost.

🎯 Exam Tip: When given a ratio of dimensions, introduce a common multiple (like \( x \)) to represent the actual dimensions. Use the area to solve for \( x \), then calculate perimeter and cost. Accuracy in calculation is key.

Question 19. The cost of enclosing a rectangular garden with a fence all round at the rate of 75 paise per metre is Rs. 300. If the length of the garden is 120 metres, find the area of the garden in square metres.
Answer:
Given total cost of fencing \( = \text{Rs. } 300 \).
Rate of fencing \( = 75 \) paise per metre \( = \text{Rs. } 0.75 \) per metre.
The total cost is (Perimeter \( \times \) Rate per metre). So, Perimeter \( = \frac{\text{Total Cost}}{\text{Rate per metre}} \).
Perimeter of the garden \( = \frac{300}{0.75} = \frac{300}{\frac{3}{4}} = 300 \times \frac{4}{3} = 400 \) m.
Given length of the garden \( l = 120 \) m.
Perimeter of a rectangle \( = 2(l + b) \).
So, \( 2(120 + b) = 400 \).
Divide by 2: \( 120 + b = \frac{400}{2} = 200 \) m.
Breadth \( b = 200 - 120 = 80 \) m.
Now, calculate the area of the garden \( = l \times b = 120 \times 80 = 9600 \) m\(^2\).
In simple words: First, use the total fencing cost and the cost per metre to find the total perimeter of the garden. Then, use this perimeter and the given length to find the garden's breadth. Finally, multiply the length and breadth to get the area.

🎯 Exam Tip: Pay attention to unit consistency (rupees and paise). Work backward from total cost and rate to find the perimeter, then use the perimeter formula to find the missing dimension, and finally calculate the area.

Question 20. Find the area and perimeter of a square plot of land, the length of whose diagonal is 14 metres. Give your answer correct to 2 places of decimals.
Answer:
Given diagonal of a square plot \( d = 14 \) m.
For a square, the relationship between side (\( a \)) and diagonal (\( d \)) is \( d = a\sqrt{2} \).
So, \( a = \frac{d}{\sqrt{2}} = \frac{14}{\sqrt{2}} \) m.
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( a = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \) m.

Area of the square \( = a^2 = (7\sqrt{2})^2 = 7^2 \times (\sqrt{2})^2 = 49 \times 2 = 98 \) m\(^2\).
Alternatively, Area \( = \frac{(\text{diagonal})^2}{2} = \frac{14^2}{2} = \frac{196}{2} = 98 \) m\(^2\).

Perimeter of the square \( = 4a = 4 \times 7\sqrt{2} = 28\sqrt{2} \) m.
Using \( \sqrt{2} \approx 1.4142 \):
Perimeter \( = 28 \times 1.4142 = 39.5976 \) m.
Rounding to 2 decimal places, the perimeter is \( 39.60 \) m.

The image below shows the square plot with its diagonal:

In simple words: For a square, the area can be found by squaring the diagonal and dividing by 2. To find the side length, divide the diagonal by the square root of 2. Then, multiply the side length by 4 to get the perimeter. Remember to round the perimeter to two decimal places.

🎯 Exam Tip: Remember the two key formulas for a square involving its diagonal: \( \text{side} = \frac{\text{diagonal}}{\sqrt{2}} \) and \( \text{Area} = \frac{(\text{diagonal})^2}{2} \). Use the approximation for \( \sqrt{2} \) to the required decimal places for accurate rounding.

Question 21. A lawn 10 m by 8 m is surrounded by a path 1 m wide. Find the area of the path.
Answer:
Given length of lawn \( l = 10 \) m and breadth of lawn \( b = 8 \) m.
Area of the lawn \( = l \times b = 10 \times 8 = 80 \) m\(^2\).
The path is 1 m wide around the lawn.
So, the outer length (including the path) \( L = l + (2 \times \text{path width}) = 10 + (2 \times 1) = 10 + 2 = 12 \) m.
The outer breadth (including the path) \( B = b + (2 \times \text{path width}) = 8 + (2 \times 1) = 8 + 2 = 10 \) m.
Area of the lawn with path (outer area) \( = L \times B = 12 \times 10 = 120 \) m\(^2\).
Area of the path \( = \text{Outer Area} - \text{Inner Area (Lawn)} = 120 - 80 = 40 \) m\(^2\).
The image below illustrates the lawn and the surrounding path:

In simple words: First, calculate the area of the lawn. Then, add the path's width to both sides of the lawn's length and breadth to get the dimensions of the larger rectangle (lawn plus path). Calculate the area of this larger rectangle, and finally, subtract the lawn's area from it to find the area of just the path.

🎯 Exam Tip: For problems with paths around a figure, remember to add *twice* the path width to both the length and the breadth of the inner figure to get the outer dimensions. The area of the path is always the difference between the outer and inner areas.

Question 22. The diagram given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. Find the area of the shaded portion.
Answer:
Given length of the rectangular field \( l = 50 \) m and breadth \( b = 35 \) m.
Width of each path \( = 5 \) m.
Area of the horizontal path \( = \text{length of field} \times \text{width of path} = 50 \times 5 = 250 \) m\(^2\).
Area of the vertical path \( = \text{breadth of field} \times \text{width of path} = 35 \times 5 = 175 \) m\(^2\).
The central square where the paths overlap has dimensions \( 5 \text{ m} \times 5 \text{ m} \).
Area of the overlapping square \( = 5 \times 5 = 25 \) m\(^2\).
Area of the two paths (shaded portion) \( = \text{Area of horizontal path} + \text{Area of vertical path} - \text{Area of overlapping square} \).
Area of shaded portion \( = 250 + 175 - 25 = 425 - 25 = 400 \) m\(^2\).
The image below shows the rectangular field with two intersecting paths:

In simple words: Calculate the area of the horizontal path and the vertical path separately. Since the middle part where they cross is counted twice, subtract its area once from the sum of the two path areas to get the actual shaded area.

🎯 Exam Tip: For intersecting paths problems, the crucial step is to subtract the area of the central overlapping region (which is a square if the path widths are equal) once, as it gets counted in both the horizontal and vertical path areas.

Question 23. Find the area of the four walls of a room, 7 m long, 5 m broad, and 3 m high. Find the cost of distempering the walls at the rate of Rs. 3.15 per m².
Answer:
Given length of room \( l = 7 \) m, breadth \( b = 5 \) m, and height \( h = 3 \) m.
Area of the four walls \( = 2(l + b)h \).
Area \( = 2(7 + 5) \times 3 \)
Area \( = 2(12) \times 3 \)
Area \( = 24 \times 3 = 72 \) m\(^2\).
Rate of distempering \( = \text{Rs. } 3.15 \) per m\(^2\).
Total cost of distempering \( = \text{Area of walls} \times \text{Rate per m}^2 = 72 \times 3.15 = \text{Rs. } 226.80 \).
In simple words: First, use the formula for the area of four walls (twice the sum of length and breadth, multiplied by height) to find the total wall area. Then, multiply this area by the cost per square metre to get the total cost for distempering.

🎯 Exam Tip: Remember the formula for the lateral surface area of a cuboid, which represents the area of four walls: \( 2(l+b)h \). Make sure to correctly substitute the dimensions and perform multiplication accurately.

Question 24. Find the cost of the paper for the walls of the room if each piece of paper is 40 cm wide and 60 cm long : the room is 6 m long, 4 m wide, 3 m high and the cost of paper is Rs. 1.40 per piece; allow 11 m² for doors, etc., and assume that a whole number of pieces has to be bought.
Answer:
Given length of room \( l = 6 \) m \( = 600 \) cm, breadth \( b = 4 \) m \( = 400 \) cm, and height \( h = 3 \) m \( = 300 \) cm.
Area of four walls \( = 2(l + b)h \).
Area \( = 2(600 + 400) \times 300 \) cm\(^2\).
Area \( = 2(1000) \times 300 = 2000 \times 300 = 600000 \) cm\(^2\).

Area allowed for doors, etc. \( = 11 \) m\(^2\).
Convert to cm\(^2\): \( 11 \times 100 \times 100 = 110000 \) cm\(^2\).

Remaining area to be papered \( = 600000 - 110000 = 490000 \) cm\(^2\).

Given length of one piece of paper \( = 60 \) cm and breadth \( = 40 \) cm.
Area of one piece of paper \( = 60 \times 40 = 2400 \) cm\(^2\).

Number of pieces of paper required \( = \frac{\text{Remaining area}}{\text{Area of one piece}} = \frac{490000}{2400} = 204.166 \).
Since a whole number of pieces must be bought, we round up to \( 205 \) sheets.

Cost of one sheet of paper \( = \text{Rs. } 1.40 \).
Total cost \( = 205 \times 1.40 = \text{Rs. } 287 \).
In simple words: First, find the total area of the four walls in square centimetres. Subtract the area for doors and windows (also in square centimetres) to find the actual area to be papered. Then, find the area of one piece of wallpaper. Divide the wall area by the paper area to get the number of pieces needed, rounding up to a whole number. Finally, multiply this number by the cost per piece.

🎯 Exam Tip: Pay very close attention to unit conversions (m to cm) and ensure consistency throughout the calculation. Remember to subtract the area of doors/windows from the total wall area before calculating the number of wallpaper pieces. Always round up the number of pieces if partial pieces are needed.

Question 25. A cistern 6 m long, 4 m wide, contains water to a depth of 1 m 25 cm : find the area of the wet surface.
Answer:
Given length of cistern \( l = 6 \) m, breadth \( b = 4 \) m.
Depth of water \( h = 1 \) m \( 25 \) cm \( = 1.25 \) m (since \( 25 \text{ cm} = 0.25 \text{ m} \)).
The wet surface of the cistern consists of the base area and the area of the four side walls that are in contact with water.
Area of the base \( = l \times b = 6 \times 4 = 24 \) m\(^2\).
Area of the four wet walls \( = 2(l + b)h \).
Area \( = 2(6 + 4) \times 1.25 \)
Area \( = 2(10) \times 1.25 = 20 \times 1.25 = 25 \) m\(^2\).
Total wet surface area \( = \text{Area of base} + \text{Area of four wet walls} = 24 + 25 = 49 \) m\(^2\).
In simple words: First, calculate the area of the bottom of the cistern. Then, find the area of the four walls that are covered by water (using the water depth as height). Add these two areas together to get the total wet surface area.

🎯 Exam Tip: When finding the wet surface area of a partially filled cuboid, remember to only include the base and the four side walls up to the water level. Ensure all units are consistent (e.g., convert cm to m).

Question 26. Find the area and perimeter of the shape illustrated in figure given below. The corners are all right angles.
Answer:
To find the area, we can imagine joining the upper portion to form a larger rectangle and then subtracting the missing inner rectangle.
From the figure, the outer length \( L = 6 + 2 + 2 = 10 \) cm.
The outer breadth \( B = 5 \) cm.
Area of the imagined outer rectangle \( = L \times B = 10 \times 5 = 50 \) cm\(^2\).
The missing inner rectangle has length \( = 6 \) cm and breadth \( = (5 - (2+2)) = 5 - 4 = 1 \) cm. (Alternatively, the inner missing rectangle has dimensions 6 cm by (5-4) cm). The figure shows the vertical side of the inner rectangle is \( 5 - (2+2) = 1 \) cm. This is not quite right given the shape. Let's re-evaluate based on the provided solution approach of "joining the upper portion". It seems the solution is considering the figure as a large rectangle with a cut-out. Length of outer rectangle \( l = 6+2+2 = 10 \) cm (assuming the top piece adds to the sides). The overall width is 5 cm. The inner missing rectangle is 6 cm long. Its height is the total height 5 cm minus the top height (2cm) and bottom height (2cm) is 1 cm. Area of given figure = Area of (10 x 5) rectangle - Area of (6 x 3) rectangle. This assumes the 3 cm is the missing vertical length. Given the figure's dimensions: Outer length (at the bottom) = 6 cm. The top horizontal part is \( 2 + 6 + 2 = 10 \) cm. Total height is 5 cm. The height of the 'middle' part is \( 5 - 2 - 2 = 1 \) cm. So, the figure can be split into three rectangles or calculated as a larger rectangle minus a smaller one. Let's use the interpretation from the solution: Area \( = L \times B - l \times h \). Outer rectangle (formed by extending the shape to fill the top-left corner): Max Length \( L = 6 + 2 + 2 = 10 \) cm. Max Breadth \( B = 5 \) cm. Area of this conceptual outer rectangle \( = 10 \times 5 = 50 \) cm\(^2\). The cut-out (the empty space in the top left) has length \( l = 6 \) cm. Its height \( h \) would be the total height minus the height of the projecting parts, so \( h = 5 - 2 = 3 \) cm (This is inconsistent with the other method of calculating inner rectangle height). Let's use the simpler method of summing up parts based on the image: The shape can be divided into three rectangles: 1. Bottom rectangle: \( 6 \text{ cm} \times 2 \text{ cm} \) (height is the lowest 2cm section) 2. Middle rectangle: \( 2 \text{ cm} \times (5-2-2) \text{ cm} = 2 \text{ cm} \times 1 \text{ cm} \) (this is the leftmost vertical strip of 2cm width and 1cm height, next to the 6cm wide base) 3. Top rectangle: \( 6 \text{ cm} \times 2 \text{ cm} \) This interpretation is also complicated. Let's stick to the interpretation of the solution provided where it states: Length of outer rectangle (l) = \( 6 + 2 + 2 = 10 \) cm. Breadth (B) = \( 5 \) cm. Length of inner rectangle (I) = \( 6 \) cm. Breadth (b) = \( 3 \) cm. (This implies the missing portion is \( 6 \times 3 \). If total is 5 and the 2cm is from top, and 2cm from bottom, then middle portion is 1cm. So a 6x3 cut-out seems like an assumption) Let's go by "Area of the given figure = L × B - l × h = 10 × 5 - 6 × 3 = 50 - 18 = 32 cm²" This implies a larger rectangle of 10x5 and a smaller "empty" rectangle of 6x3. From the diagram: Max length is \( 2+6+2 = 10 \) cm. Max height is \( 5 \) cm. So the "L x B" for the encompassing rectangle is \( 10 \times 5 = 50 \) cm\(^2\). The empty space is at the top center. Its width is \( 6 \) cm. Its height is \( 5 - 2 = 3 \) cm (as the top section has 2cm height, so remaining height of the cut out is 3cm). So the empty area is \( 6 \times 3 = 18 \) cm\(^2\). Area of the figure \( = 50 - 18 = 32 \) cm\(^2\). This matches the solution's area calculation. Now for perimeter: Perimeter \( = (5 + 10 + 5 + 2 + 3 + 6 + 3 + 2) \) cm. Starting from bottom-left corner and going clockwise: Bottom: 10 cm (total horizontal span) Right side: 5 cm (total vertical span) Top-right horizontal: 2 cm Top-right vertical (down): 3 cm (This 3cm is derived from 5cm total height - 2cm (topmost section height)) Top-middle horizontal (left): 6 cm Top-middle vertical (down): 3 cm Top-left horizontal: 2 cm Top-left vertical (down): 5 cm. Let's trace the outer boundary: 1. Bottom horizontal: \( (2+6+2) = 10 \) cm 2. Right vertical: \( 5 \) cm 3. Top horizontal (outer, right part): \( 2 \) cm 4. Inner vertical (going down from top-right corner of cut-out): This is \( 5 - 2 = 3 \) cm. 5. Inner horizontal (top of cut-out): \( 6 \) cm 6. Inner vertical (going down from top-left corner of cut-out): This is also \( 5 - 2 = 3 \) cm. 7. Top horizontal (outer, left part): \( 2 \) cm 8. Left vertical: \( 5 \) cm Summing these up: \( 10 + 5 + 2 + 3 + 6 + 3 + 2 + 5 = 36 \) cm. This matches the solution's perimeter calculation. The image for Question 26 is:

The SVG recreation needs to match the visual given, including the numbers. The number labels in the diagram are: Left vertical: 5 cm Bottom horizontal: 6 cm (this is the middle part) Top-left horizontal: 2 cm Top-right horizontal: 2 cm This implies the shape is a large 10x5 rectangle with a 6x3 rectangle cut from the top centre. The '6 cm' at the bottom of the diagram is clearly the width of the lower central part. Total width = \( 2 + 6 + 2 = 10 \) cm. Total height = \( 5 \) cm. Area = (Total Rectangle) - (Cutout) = \( (10 \times 5) - (6 \times (5-2)) \) where 2 is the height of the top bars. So the cut out is \( 6 \times 3 = 18 \). Area = \( 50 - 18 = 32 \) cm\(^2\). This is correct. Perimeter trace: 1. Left vertical edge: 5 cm 2. Bottom edge: \( 2 + 6 + 2 = 10 \) cm (assuming 6 is the inner base, and 2cm sections extend to left/right) 3. Right vertical edge: 5 cm 4. Top outer right edge: 2 cm 5. Top inner vertical edge: 3 cm (5cm - 2cm) 6. Top inner horizontal edge: 6 cm 7. Top inner vertical edge: 3 cm (5cm - 2cm) 8. Top outer left edge: 2 cm Total = \( 5 + 10 + 5 + 2 + 3 + 6 + 3 + 2 = 36 \) cm. This is correct. So the values in the solution are consistent with the visual diagram if interpreted this way. The figure for Question 26 should be: Length of outer rectangle (l) = 6 + 2 + 2 = 10 cm. Breadth (B) = 5 cm. Length of inner rectangle (I) = 6 cm. Breadth (b) = 3 cm. (This is \( 5-2 \) from the outer height) Area of the given figure \( = (10 \times 5) - (6 \times 3) = 50 - 18 = 32 \) cm\(^2\). Perimeter: \( = \text{sum of all outer edges} \) \( = (5 \text{ (left)} + 10 \text{ (bottom)} + 5 \text{ (right)} + 2 \text{ (top-right)} + 3 \text{ (inner-right-vertical)} + 6 \text{ (inner-top)} + 3 \text{ (inner-left-vertical)} + 2 \text{ (top-left)}) \) \( = 5+10+5+2+3+6+3+2 = 36 \) cm.

In simple words: To find the area, imagine it as a big rectangle (10 cm by 5 cm) with a smaller rectangle (6 cm by 3 cm) cut out from the top middle. Subtract the cut-out area from the big rectangle's area. For the perimeter, simply add up the lengths of all the outer edges of the shape.

🎯 Exam Tip: For irregular shapes made of straight lines, you can either divide the shape into simpler rectangles and sum their areas, or enclose the shape in a larger rectangle and subtract any empty regions. Always trace all outer edges for the perimeter.

Question 27. Find the area of the figure, in which all corners are right-angles and the dimensions are shown in centimetres.
Answer:
The figure can be divided into three rectangles (I, II, III). Let's find the dimensions of each rectangle from the diagram: Rectangle I: Length \( 4\frac{1}{4} \text{ cm} \) and Breadth \( 4 \text{ cm} \). Length \( = 4.25 \) cm. Area of Rectangle I \( = 4.25 \times 4 = 17 \) cm\(^2\).
Rectangle II: Length \( 7 \text{ cm} \) and Breadth \( 3\frac{1}{2} \text{ cm} \). Breadth \( = 3.5 \) cm. Area of Rectangle II \( = 7 \times 3.5 = 24.5 \) cm\(^2\).
Rectangle III: Length \( 3 \text{ cm} \) and Breadth \( 4\frac{1}{2} \text{ cm} \). Breadth \( = 4.5 \) cm. Area of Rectangle III \( = 3 \times 4.5 = 13.5 \) cm\(^2\).
Total area of the figure \( = \text{Area I} + \text{Area II} + \text{Area III} = 17 + 24.5 + 13.5 = 55 \) cm\(^2\).
The image below shows the figure divided into three rectangles:

In simple words: Break the complex shape into three simpler rectangles. Calculate the area of each small rectangle by multiplying its length and breadth. Then, add up the areas of all three rectangles to find the total area of the figure.

🎯 Exam Tip: For complex polygons with right angles, divide them into a few non-overlapping rectangles. Calculate the area of each smaller rectangle and sum them up. Ensure you use the correct dimensions for each part, carefully reading them from the diagram.

Question 28. How many tiles, each 12 cm by 6 cm are needed for a floor of a room 162 cm long, 144 cm wide
Answer:
First, calculate the area of the floor:
Length of the room \( = 162 \) cm
Breadth of the room \( = 144 \) cm
Area of the floor \( = \text{length} \times \text{breadth} = 162 \times 144 \) cm\( ^2 \)

Next, calculate the area of one tile:
Length of one tile \( = 12 \) cm
Breadth of one tile \( = 6 \) cm
Area of one tile \( = 12 \times 6 = 72 \) cm\( ^2 \)

Finally, find the number of tiles needed:
Number of tiles \( = \frac{\text{Area of floor}}{\text{Area of one tile}} = \frac{162 \times 144}{72} \)
\( = \frac{23328}{72} = 324 \) tiles
To cover the entire floor perfectly, you need 324 tiles. This calculation helps ensure no part of the floor is left uncovered.
In simple words: First, find the total space of the floor and the space of one tile. Then, divide the floor's space by the tile's space to find out how many tiles you need.

🎯 Exam Tip: Always ensure that all measurements are in the same units (e.g., all in cm or all in m) before performing calculations to avoid errors.

Question 29. How many dusters, each 16 cm square can be cut from material 84 cm long, 36 cm wide? What areas remains over?
Answer:
First, find the total area of the material:
Length of material \( = 84 \) cm
Breadth of material \( = 36 \) cm
Total area of material \( = \text{length} \times \text{breadth} = 84 \times 36 = 3024 \) cm\( ^2 \)

Next, find the area of one square duster:
Side of a square duster \( = 16 \) cm
Area of one duster \( = \text{side} \times \text{side} = 16 \times 16 = 256 \) cm\( ^2 \)

Now, calculate how many dusters can be cut:
Number of dusters \( = \frac{\text{Total area of material}}{\text{Area of one duster}} = \frac{3024}{256} = 11.8125 \)
Since only whole dusters can be cut, we can make 11 complete dusters.

Calculate the area used by 11 dusters:
Area used \( = 11 \times 256 = 2816 \) cm\( ^2 \)

Finally, find the area that remains over:
Remaining area \( = \text{Total area of material} - \text{Area used} = 3024 - 2816 = 208 \) cm\( ^2 \)
This shows that even when cutting items, there's often some material left over.
In simple words: We figure out the total space of the cloth and the space one duster takes. We divide to see how many dusters we can cut. Since we can only cut whole dusters, we use that number to calculate how much cloth is actually used. The cloth left over is the remaining area.

🎯 Exam Tip: When a problem asks to cut items from a larger piece, always round down the number of pieces to the nearest whole number. Then, calculate the used area based on this whole number, not the decimal.

Question 30. A floor which measure 15 m × 8 m is to be laid with tiles measuring 50 cm × 25 cm. Find the number of tiles required. Further, if a carpet is laid on the floor so that a space of 1 m exist between its edges and the edges of the floor, what fraction of the floor is uncovered?
Answer:
**Part 1: Number of tiles required**
First, convert all dimensions to the same unit (cm).
Floor length \( = 15 \) m \( = 15 \times 100 = 1500 \) cm
Floor breadth \( = 8 \) m \( = 8 \times 100 = 800 \) cm
Area of the floor \( = \text{length} \times \text{breadth} = 1500 \times 800 = 1200000 \) cm\( ^2 \)

Tile length \( = 50 \) cm
Tile breadth \( = 25 \) cm
Area of one tile \( = 50 \times 25 = 1250 \) cm\( ^2 \)

Number of tiles required \( = \frac{\text{Area of floor}}{\text{Area of one tile}} = \frac{1200000}{1250} = 960 \) tiles

**Part 2: Fraction of the floor uncovered by carpet**
Total area of the floor \( = 15 \times 8 = 120 \) m\( ^2 \)
The carpet leaves a 1 m space from all edges of the floor.
Inner length (carpet length) \( = 15 - (1 \text{ m from each side}) = 15 - (2 \times 1) = 15 - 2 = 13 \) m
Inner breadth (carpet breadth) \( = 8 - (1 \text{ m from each side}) = 8 - (2 \times 1) = 8 - 2 = 6 \) m
Area covered by the carpet \( = 13 \times 6 = 78 \) m\( ^2 \)

Area of the space left uncovered \( = \text{Total floor area} - \text{Area covered by carpet} = 120 - 78 = 42 \) m\( ^2 \)

Fraction of the floor which is uncovered \( = \frac{\text{Uncovered area}}{\text{Total floor area}} = \frac{42}{120} \)
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, 6.
\( = \frac{42 \div 6}{120 \div 6} = \frac{7}{20} \)
Such calculations are crucial for planning renovations and interior design.
In simple words: First, convert all measurements to the same unit (like centimeters) to find how many tiles are needed for the floor. Second, imagine a carpet laid on the floor with a 1-meter gap all around. Calculate the carpet's size, then the uncovered floor area, and finally, express this as a fraction of the total floor area.

🎯 Exam Tip: Be very careful with units; always convert all measurements to a single unit (e.g., all meters or all centimeters) before calculating areas or ratios. For problems with margins, remember to subtract the margin from both sides (twice) when finding the inner dimensions.