ML Aggarwal Class 6 Maths Solutions Chapter 14 Mensuration

Question 1(i). Find the perimeter of the following figure:
(i) A quadrilateral with sides 5 cm, 3 cm, 2 cm and 7 cm.
Answer: The perimeter represents the total distance around the outer edge of a closed shape. For this quadrilateral, you add up all four sides: 5 + 3 + 2 + 7 = 17 cm. The perimeter of the figure is 17 cm.
In simple words: Perimeter means the total length all the way around a shape. Add all the side lengths together to find it.

Exam Tip: Always identify all sides of the figure and add them carefully. Check your arithmetic twice to avoid mistakes in the final answer.

Question 1(ii). Find the perimeter of the following figure:
(ii) A quadrilateral with sides 31 cm, 38 cm, 48 cm and 38 cm.
Answer: The perimeter of a closed plane figure equals the length around its boundary. This quadrilateral has four sides that sum to: 31 + 38 + 48 + 38 = 155 cm. Therefore, the perimeter of the figure is 155 cm.
In simple words: Add up the lengths of all four sides. The sum gives you how far it is all the way around the shape.

Exam Tip: For quadrilaterals with multiple sides, organise your addition carefully. Use estimation to check if your final answer seems reasonable.

Question 1(iii). Find the perimeter of the following figure:
(iii) A rhombus with each side measuring 19 cm.
Answer: The perimeter of a closed plane figure is the length around its outer edge. Since a rhombus is a four-sided shape where all sides have equal length, the perimeter equals 4 multiplied by the side length. Perimeter = 4 \( \times \) 19 = 76 cm. The perimeter of the figure is 76 cm.
In simple words: A rhombus has four sides of the same length. Multiply one side by 4 to get the distance around it.

Exam Tip: For regular shapes where all sides are equal, remember the shortcut formulas: square = 4s, rhombus = 4s, regular polygon = (number of sides) × (side length).

Question 1(iv). Find the perimeter of the following figure:
(iv) A regular pentagon with each side measuring 7 cm.
Answer: The perimeter of a closed plane figure is the length of its boundary. A regular pentagon is a five-sided shape where every side is the same length. To find the perimeter, multiply the side length by 5: Perimeter = 5 \( \times \) 7 = 35 cm. The perimeter of the figure is 35 cm.
In simple words: A pentagon has five equal sides. Multiply one side by 5 to find the total distance around it.

Exam Tip: For any regular polygon, the perimeter formula is (number of sides) \( \times \) (length of one side). Identify the number of sides first, then apply the formula.

Question 2. Find the perimeter of each of the following shapes:
(i) A triangle of sides 3 cm, 4 cm and 6 cm.
(ii) A triangle of sides 7 cm, 5.4 cm and 10.2 cm.
(iii) An equilateral triangle of side 11 cm.
(iv) An isosceles triangle with equal sides 10 cm each and third side 7 cm.
Answer:
(i) The perimeter of a triangle equals the sum of all three sides. For this triangle: 3 + 4 + 6 = 13 cm. The perimeter of the triangle is 13 cm.

(ii) Adding the three side lengths: 7 + 5.4 + 10.2 = 22.6 cm. The perimeter of the triangle is 22.6 cm.

(iii) An equilateral triangle has all three sides of equal length. To find the perimeter, multiply the side by 3: Perimeter = 3 \( \times \) 11 = 33 cm. The perimeter of the equilateral triangle is 33 cm.

(iv) An isosceles triangle has two sides of equal length plus one different side. Adding all three: 10 + 10 + 7 = 27 cm. The perimeter of the isosceles triangle is 27 cm.
In simple words: To find any triangle's perimeter, add all three sides. For special triangles like equilateral (all sides equal) or isosceles (two sides equal), you can use multiplication as a shortcut.

Exam Tip: Always add carefully when sides include decimals. Recognise triangle types - equilateral means all sides equal (use 3s), isosceles means two sides equal (add carefully to avoid confusion).

Question 3. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Answer: Given: Length of the lid = 40 cm, Breadth of the lid = 10 cm. The tape seals the entire outer edge, so its length matches the perimeter. Using the rectangle perimeter formula: Perimeter = 2(length + breadth) = 2(40 + 10) = 2 \( \times \) 50 = 100 cm, which equals 1 m. The length of tape required is 1 m.
In simple words: The tape goes around the whole edge of the rectangular lid. Find the perimeter to know how much tape you need.

Exam Tip: In real-world problems about fencing, taping, or bordering, the perimeter is what you need. Always convert the final answer to the requested unit (here, metres).

Question 4. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Answer: Given: Length of the table-top = 2 m 25 cm = 2.25 m, Breadth of the table-top = 1 m 50 cm = 1.50 m. Using the rectangle perimeter formula: Perimeter = 2(length + breadth) = 2(2.25 + 1.50) = 2 \( \times \) 3.75 = 7.5 m. The perimeter of the tabletop is 7.5 m.
In simple words: Convert mixed measurements to a single unit first. Then apply the rectangle formula by adding length and breadth, then multiplying by 2.

Exam Tip: Convert mixed units (metres and centimetres) to decimals before calculating. Always double-check your conversion: 25 cm = 0.25 m, 50 cm = 0.50 m.

Question 5. The perimeter of a rectangle is 58 m. If its length is 17 m, find its breadth.
Answer: Given: Perimeter of rectangle = 58 m, Length = 17 m. Let b be the breadth. Using the perimeter formula:

Perimeter = 2(length + breadth)

\[ \implies 58 = 2(17 + b) \]

\[ \implies \frac{58}{2} = 17 + b \]

\[ \implies 29 = 17 + b \]

\[ \implies b = 29 - 17 \]

\[ \implies b = 12 \text{ m} \]

The breadth of the rectangle is 12 m.
In simple words: Work backwards from perimeter to find a missing side. Divide the perimeter by 2, then subtract the known length to get the breadth.

Exam Tip: When a side is missing, rearrange the perimeter formula. Always show each step clearly - examiners look for working, not just the final answer.

Question 6. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Answer: Given: Length of the land = 0.7 km, Breadth of the land = 0.5 km. First, find the perimeter: Perimeter = 2(length + breadth) = 2(0.7 + 0.5) = 2 \( \times \) 1.2 = 2.4 km. Since each side requires 4 rows of wires, the total wire length is: Length of wire = 4 \( \times \) Perimeter = 4 \( \times \) 2.4 = 9.6 km. The length of wire needed is 9.6 km.
In simple words: Find the perimeter first. Then multiply by the number of rows to get the total length of wire required.

Exam Tip: Read carefully - "4 rows of wires" means multiply the perimeter by 4. Distinguish between single-layer fencing and multi-layer fencing problems.

Question 7. Find the perimeter of a regular hexagon with each side measuring 7.5 m.
Answer: Given: Each side of the regular hexagon = 7.5 m. A regular hexagon has six equal sides. To find the perimeter, multiply the side length by 6: Perimeter = 6 \( \times \) 7.5 = 45 m. The perimeter of the regular hexagon is 45 m.
In simple words: A hexagon has six sides. If all sides are the same length, just multiply one side by 6.

Exam Tip: Memorise the number of sides for common shapes: triangle = 3, square = 4, pentagon = 5, hexagon = 6. For regular polygons, the formula is always (number of sides) \( \times \) (side length).

Question 8. The lengths of two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the length of its third side?
Answer: Given: Two sides of the triangle = 12 cm and 14 cm, Perimeter = 36 cm. Let x be the third side. Since perimeter equals the sum of all sides:

\[ \implies 36 = 12 + 14 + x \]

\[ \implies 36 = 26 + x \]

\[ \implies x = 36 - 26 \]

\[ \implies x = 10 \text{ cm} \]

The length of the third side is 10 cm.
In simple words: If you know two sides and the total perimeter, subtract both known sides from the perimeter to get the missing side.

Exam Tip: Use the basic perimeter equation: Perimeter = Side 1 + Side 2 + Side 3. Rearrange to find the unknown side algebraically.

Question 9. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Answer: Given: Perimeter of regular pentagon = 100 cm. A regular pentagon has five equal sides. Using the perimeter formula:

Perimeter = 5 \( \times \) side

\[ \implies 100 = 5 \times \text{side} \]

\[ \implies \text{side} = \frac{100}{5} \]

\[ \implies \text{side} = 20 \text{ cm} \]

Each side of the regular pentagon is 20 cm.
In simple words: For a regular pentagon, divide the perimeter by 5 to find the length of one side.

Exam Tip: When finding a side length from the perimeter, use division. The formula becomes: Side = Perimeter ÷ (number of sides).

Question 10. A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(i) a square?
(ii) an equilateral triangle?
(iii) a regular hexagon?
Answer:
(i) When the string forms a square, the perimeter = 30 cm. Since a square has four equal sides:

Perimeter = 4 \( \times \) side

\[ \implies 30 = 4 \times \text{side} \]

\[ \implies \text{side} = \frac{30}{4} \]

\[ \implies \text{side} = 7.5 \text{ cm} \]

Each side of the square is 7.5 cm.

(ii) When the string forms an equilateral triangle, the perimeter = 30 cm. Since an equilateral triangle has three equal sides:

Perimeter = 3 \( \times \) side

\[ \implies 30 = 3 \times \text{side} \]

\[ \implies \text{side} = \frac{30}{3} \]

\[ \implies \text{side} = 10 \text{ cm} \]

Each side of the equilateral triangle is 10 cm.

(iii) When the string forms a regular hexagon, the perimeter = 30 cm. Since a regular hexagon has six equal sides:

Perimeter = 6 \( \times \) side

\[ \implies 30 = 6 \times \text{side} \]

\[ \implies \text{side} = \frac{30}{6} \]

\[ \implies \text{side} = 5 \text{ cm} \]

Each side of the regular hexagon is 5 cm.
In simple words: The string length becomes the perimeter. Divide it by the number of sides to find each side's length for any regular polygon.

Exam Tip: Compare the three shapes - the same perimeter produces different side lengths depending on how many sides the shape has. More sides = shorter sides for a fixed perimeter.

Question 11. Find the cost of fencing a rectangular park of length 225 m and breadth 115 m at the rate of Rs. 13 per metre.
Answer: Given: Length of the park = 225 m, Breadth of the park = 115 m, Rate = Rs. 13 per metre. Find the perimeter first: Perimeter = 2(length + breadth) = 2(225 + 115) = 2 \( \times \) 340 = 680 m. Now multiply by the rate to find total cost: Cost = Perimeter \( \times \) Rate = 680 \( \times \) 13 = Rs. 8840. The cost of fencing the rectangular park is Rs. 8840.
In simple words: Calculate the perimeter. Then multiply it by the cost per unit length to find the total fencing cost.

Exam Tip: Real-world problems often combine perimeter with rates or costs. Always find the perimeter first, then apply the rate. Double-check your multiplication.

Question 12. Meera went to a rectangular park 140 m long and 90 m wide. She took 5 complete rounds on its boundary. What is the distance covered by her?
Answer: Given: Length of the park = 140 m, Breadth of the park = 90 m, Number of rounds = 5. Find the perimeter: Perimeter = 2(length + breadth) = 2(140 + 90) = 2 \( \times \) 230 = 460 m. For 5 complete rounds, multiply the perimeter by 5: Distance = 5 \( \times \) Perimeter = 5 \( \times \) 460 = 2300 m. The distance covered by Meera is 2300 m.
In simple words: Find one complete lap around the boundary. Then multiply by the number of laps to get the total distance.

Exam Tip: In running/walking problems, perimeter = one lap. Multiply perimeter by the number of laps. Convert the final answer to the requested unit if needed.

Question 13. Pinky runs 8 times around a rectangular park with length 80 m and breadth 55 m while Pankaj runs 7 times around a square park. Compare their distances.
Answer: First, find Pinky's distance. The rectangular park has: Perimeter = 2(length + breadth) = 2(80 + 55) = 2 \( \times \) 135 = 270 m. For 8 rounds: Distance covered by Pinky = 8 \( \times \) 270 = 2160 m. Without the side length of the square park provided in the source, the comparison cannot be fully completed based on available information. However, the method is: calculate the square's perimeter, multiply by 7 for Pankaj's distance, then compare the two distances.
In simple words: For each person, find the perimeter of their park, then multiply by the number of rounds. Compare the total distances.

Exam Tip: When comparing distances in such problems, ensure you have all dimensions. Calculate each person's distance separately before making comparisons.

Question 13. Pinky walks around a rectangular park with dimensions of 80 m by 55 m. Pankaj walks around a square park of side 75 m. Who covers more distance and by how much?
Answer: For Pinky, the boundary length of the rectangular park is calculated as 2(80 + 55) = 2 × 135 = 270 m. In 8 rounds, Pinky travels 8 × 270 = 2160 m. For Pankaj, the boundary length of the square park is 4 × 75 = 300 m. In 7 rounds, Pankaj travels 7 × 300 = 2100 m. Comparing the distances: 2160 - 2100 = 60 m. So, Pinky covers more ground by 60 m.
In simple words: Pinky walks a longer distance than Pankaj. She goes 60 m more than him in total.

Exam Tip: Always calculate perimeter correctly using the given formula for each shape, multiply by the number of rounds, then compare the total distances covered.

Question 14. A rectangular park is 104 m long and 56 m wide. Anjali walks around it at the rate of 4 km/h. What time will she take in making 5 rounds of the park?
Answer: The boundary of the rectangular park measures 2(104 + 56) = 2 × 160 = 320 m. Over 5 rounds, the total distance walked is 5 × 320 = 1600 m. Anjali's walking speed is 4 km/h, which converts to 4000 m per 60 minutes. To find how many minutes Anjali needs to cover 1600 m: time = (1600 × 60) ÷ 4000 = 96000 ÷ 4000 = 24 minutes. Therefore, Anjali takes 24 minutes to complete 5 rounds of the park.
In simple words: First, find how far Anjali walks (1600 m). Then use her speed to work out the time - she needs 24 minutes.

Exam Tip: When converting speed units (km/h to m/h or m/min), be careful with the conversion factors. Always ensure units match before performing calculations.

Question 15. A wire is in the shape of a regular pentagon of side 12 cm. It is rebent into the shape of a rectangle whose length is \( \frac{3}{2} \) times its breadth. Find the length and the breadth of the rectangle.
Answer: The wire's length matches the pentagon's perimeter: 5 × 12 = 60 cm. Let the rectangle's width be b cm. Then its length is \( \frac{3}{2}b \) cm. Using the perimeter formula for the rectangle: \( 2 \left( \frac{3}{2}b + b \right) = 60 \)

\( 2 \times \frac{5b}{2} = 60 \)

\( 5b = 60 \)

\( b = 12 \) cm. The rectangle's length is \( \frac{3}{2} \times 12 = 18 \) cm. So the rectangle has a length of 18 cm and a breadth of 12 cm.
In simple words: The wire's length is 60 cm. Use this to find what the width is, then multiply by \( \frac{3}{2} \) to get the length.

Exam Tip: When a question involves reshaping wire or string, remember the total length stays the same - equate it to the new shape's perimeter.

Exercise 14.2

Question 1. Find the area of the region enclosed by the following figures by counting squares:
Answer: To estimate area using square paper, these rules are applied:

1. One full square = 1 sq. unit
2. A square more than half-filled = count as 1 sq. unit
3. A square less than half-filled = ignore it
4. A square exactly half-filled = count as \( \frac{1}{2} \) sq. unit

(i) This figure has 9 complete squares enclosed. Area = 9 × 1 = 9 sq. units

(ii) This figure has 5 complete squares enclosed. Area = 5 sq. units

(iii) This figure has 10 complete squares enclosed. Area = 10 sq. units

(iv) This figure has 6 complete squares enclosed. Area = 6 sq. units

(v) This figure has 4 complete squares enclosed. Area = 4 sq. units

(vi) This figure has 6 complete squares enclosed. Area = 6 sq. units
In simple words: Count the full squares inside. For half-squares and partial squares, apply the rules above.

Exam Tip: Be careful to distinguish between complete squares, more-than-half squares, and less-than-half squares - miscounting any category will affect your final answer.

Question 2. Find the areas of the rectangles whose lengths and breadths are:
(i) 9 m and 6 m
(ii) 17 m and 3 m
(iii) 14 m and 4 m
Which one has the largest area and which one has the smallest area?
Answer: Using the formula: Area = length × breadth

(i) Area = 9 × 6 = 54 sq. m

(ii) Area = 17 × 3 = 51 sq. m

(iii) Area = 14 × 4 = 56 sq. m

Comparing: 56 > 54 > 51. Rectangle (iii) has the largest area at 56 sq. m. Rectangle (ii) has the smallest area at 51 sq. m.
In simple words: Multiply length by width for each rectangle. Then arrange the answers from biggest to smallest.

Exam Tip: Always state both the largest and smallest areas explicitly with their values - partial answers cost marks.

Question 3. Find the areas of the rectangles whose two adjacent sides are:
(i) 14 cm and 23 cm
(ii) 3 km and 4 km
(iii) 2 m and 90 cm
Answer: Using Area = length × breadth:

(i) Area = 23 × 14 = 322 sq. cm

(ii) Area = 4 × 3 = 12 sq. km

(iii) First convert 2 m to 200 cm. Area = 200 × 90 = 18000 sq. cm. Converting to square metres: 18000 ÷ 10000 = 1.8 sq. m
In simple words: Multiply the two sides together. Watch out for different units - convert them to match before multiplying.

Exam Tip: When units differ, convert to a single unit first. Remember: 1 sq. m = 10000 sq. cm.

Question 4. Find the areas of the squares whose sides are:
(i) 8 cm
(ii) 14 m
(iii) 2 m 50 cm
Answer: Using Area = side × side:

(i) Area = 8 × 8 = 64 sq. cm

(ii) Area = 14 × 14 = 196 sq. m

(iii) Convert 2 m 50 cm to 2.5 m. Area = 2.5 × 2.5 = 6.25 sq. m
In simple words: Multiply the side by itself. If the side includes both metres and centimetres, convert it to one unit first.

Exam Tip: Remember that area of a square uses the same side length twice - it's side squared, not side doubled.

Question 5. A room is 4 m long and 3 m 25 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Answer: Convert 3 m 25 cm to 3.25 m. The carpet area needed equals the floor area. Area = 4 × 3.25 = 13 sq. m. So, 13 square metres of carpet is needed to cover the floor.
In simple words: The amount of carpet needed matches the size of the room's floor. Just multiply the length and width.

Exam Tip: When a practical application asks for carpet, paint, or tile needed, it's asking for the area of that surface.

Question 6. What is the cost of tiling a rectangular field 50 m long and 30 m wide at the rate of Rs. 20 per square metre?
Answer: The field's area measures 50 × 30 = 1500 sq. m. At a rate of Rs. 20 per square metre, the total cost is 1500 × 20 = Rs. 30000. Therefore, the tiling cost for the rectangular field is Rs. 30000.
In simple words: Find the field's size first. Then multiply by the cost per square metre to get the total cost.

Exam Tip: For cost problems, always calculate area first, then multiply by the given rate per unit area.

Question 7. A floor is 5 m long and 4 m wide. A square carpet of side 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Answer: The floor's total area is 5 × 4 = 20 sq. m. The carpet covers 3 × 3 = 9 sq. m. The uncarpeted area is 20 - 9 = 11 sq. m. So, 11 square metres of the floor remains uncovered by the carpet.
In simple words: Work out the floor's total size. Take away the carpet's size. What's left is the uncarpeted part.

Exam Tip: These "remaining" or "uncovered" problems require subtraction - find the total first, then subtract the covered/used portion.

Question 8. In the adjoining figure, find the area of the path (shown shaded) which is 2 m wide all around.
Answer: The outer rectangle measures 100 × 60 = 6000 sq. m. Since the path is 2 m wide on all sides, the inner rectangle's length is 100 - (2 + 2) = 96 m, and breadth is 60 - (2 + 2) = 56 m. The inner rectangle's area is 96 × 56 = 5376 sq. m. The path's area equals the outer area minus the inner area: 6000 - 5376 = 624 sq. m. Therefore, the path covers 624 square metres.
In simple words: The path is the gap between the outer and inner rectangles. Subtract the inner area from the outer area.

Exam Tip: When a path goes around all sides, subtract twice the path width from both the length and breadth of the outer shape to get the inner dimensions.

Question 9. Four square flower beds of side 1 m 50 cm are dug on a rectangular piece of land 8 m long and 6 m 50 cm wide. What is the area of the remaining part of land?
Answer: The land's total area is 8 × 6.5 = 52 sq. m. Each flower bed has a side of 1.5 m, so each bed's area is 1.5 × 1.5 = 2.25 sq. m. Four flower beds cover 4 × 2.25 = 9 sq. m. The remaining land area is 52 - 9 = 43 sq. m. Therefore, 43 square metres of land remain uncovered.
In simple words: Find the whole land area. Find the total area of all flower beds. Subtract the beds from the total.

Exam Tip: When multiple objects are placed on a larger area, find the individual area first, multiply by the number of objects, then subtract from the total.

Question 10. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to cover a rectangular region whose length and breadth are respectively:
(i) 70 cm and 36 cm
(ii) 144 cm and 1 m.
Answer: To find how many tiles fit in a region, divide the region's area by one tile's area.

Area of one tile = 12 × 5 = 60 sq. cm

(i) Area of region = 70 × 36 = 2520 sq. cm

Number of tiles = 2520 ÷ 60 = 42 tiles

Therefore, 42 tiles are needed to cover this rectangular region.

(ii) First, convert 1 m to 100 cm. Area of region = 144 × 100 = 14400 sq. cm

Number of tiles = 14400 ÷ 60 = 240 tiles

Therefore, 240 tiles are needed to cover this rectangular region.
In simple words: Divide the total area you want to cover by the area of one tile to find how many tiles you need.

Exam Tip: Always make sure all measurements are in the same units before calculating. Convert metres to centimetres if needed.

Question 11. The area of a rectangular plot is 340 sq. m. If its breadth is 17 m, find its length and the perimeter.
Answer: Given: Area = 340 sq. m, Breadth = 17 m

Since Area = length × breadth, we have:

Length = Area ÷ Breadth = 340 ÷ 17 = 20 m

Perimeter = 2(length + breadth) = 2(20 + 17) = 2 × 37 = 74 m

The length of the plot is 20 m and the perimeter is 74 m.
In simple words: To find length, divide the total area by the width. Then add length and width, multiply by 2 to get the distance around the shape.

Exam Tip: Always show the formula first, then substitute the values and calculate step-by-step to earn full marks.

Question 12. If the area of a rectangular plot is 144 sq. m and its length is 16 m. Find the breadth of the plot and the cost of fencing it at the rate of Rs 60 per metre.
Answer: Given: Area = 144 sq. m, Length = 16 m, Rate of fencing = Rs 60 per metre

Breadth = Area ÷ Length = 144 ÷ 16 = 9 m

Perimeter = 2(length + breadth) = 2(16 + 9) = 2 × 25 = 50 m

Cost of fencing = Perimeter × Rate = 50 × 60 = Rs 3000

The breadth of the plot is 9 m and the cost of fencing is Rs 3000.
In simple words: Find the width by dividing area by length. Then calculate the distance around it. Multiply that distance by the cost per metre to get the total cost.

Exam Tip: Watch out - the question asks for both breadth AND cost. Don't forget the second part or you'll lose marks.

Question 13. The perimeter of a square is equal to that of a rectangle of length 17 m and breadth 11 m. Find the area of the square.
Answer: Given: The square's perimeter equals the rectangle's perimeter. Rectangle length = 17 m, breadth = 11 m

Perimeter of rectangle = 2(17 + 11) = 2 × 28 = 56 m

Since the square's perimeter is also 56 m:

Side of square = 56 ÷ 4 = 14 m

Area of square = side × side = 14 × 14 = 196 sq. m
In simple words: First find how far around the rectangle is. A square with the same distance around has equal sides - divide by 4 to get each side. Then multiply the side by itself to get the area.

Exam Tip: This type of question links two shapes - always find the common feature (here, the equal perimeter) and use it to connect them.

Question 14. A rectangle has same area as that of a square of side 10 m. If the breadth of the rectangle is 8 m, find the perimeter of the rectangle.
Answer: Given: Rectangle area = Square area, Square side = 10 m, Rectangle breadth = 8 m

Area of square = 10 × 10 = 100 sq. m

Area of rectangle = 100 sq. m

Length of rectangle = 100 ÷ 8 = 12.5 m

Perimeter of rectangle = 2(12.5 + 8) = 2 × 20.5 = 41 m

The perimeter of the rectangle is 41 m.
In simple words: The rectangle covers the same space as the square. Use this equal area to find the rectangle's length, then calculate how far around it is.

Exam Tip: When areas are equal, set them equal and solve for the unknown dimension - this is a common linking technique in geometry problems.

Question 15. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres):
Answer:

(i) L-shaped figure - divide into two rectangles:

Rectangle 1 (vertical part): Length = 12 cm, Breadth = 2 cm
Area of Rectangle 1 = 12 × 2 = 24 sq. cm

Rectangle 2 (horizontal bottom part): Length = 8 cm, Breadth = 2 cm
Area of Rectangle 2 = 8 × 2 = 16 sq. cm

Total area = 24 + 16 = 40 sq. cm

(ii) Plus-shaped figure - divide into three rectangles:

Central horizontal rectangle: Length = 7 + 7 + 7 = 21 cm, Breadth = 7 cm
Area = 21 × 7 = 147 sq. cm

Top square: Side = 7 cm
Area = 7 × 7 = 49 sq. cm

Bottom square: Side = 7 cm
Area = 7 × 7 = 49 sq. cm

Total area = 147 + 49 + 49 = 245 sq. cm

(iii) T-shaped figure - divide into two rectangles:

Top bar: Length = 5 cm, Breadth = 1 cm
Area = 5 × 1 = 5 sq. cm

Stem (vertical part): Length = 4 cm, Breadth = 1 cm
Area = 4 × 1 = 4 sq. cm

Total area = 5 + 4 = 9 sq. cm
In simple words: Irregular shapes are easier if you split them into simple rectangles. Find each rectangle's area, then add them all together to get the total.

Exam Tip: Always show how you split the shape and label each rectangle clearly. This helps the examiner follow your thinking and award full marks.

Objective Type Questions - Mental Maths

Question 1. Fill in the blanks:
(i) The perimeter of a closed plane figure is the length of its .....
(ii) The unit of measurement of perimeter is same as that of .....
(iii) If the side of a rhombus is 7 cm then its perimeter is .....
(iv) The area of a closed plane figure is measured in ......
Answer:
(i) boundary
(ii) length
(iii) A rhombus has 4 equal sides, so perimeter = 4 × side = 4 × 7 = 28 cm
(iv) square units (sq. units)
In simple words: Perimeter means the distance around a shape. Length is measured in metres or centimetres. Area is measured in square units like sq. cm.

Exam Tip: Memorise that all four sides of a rhombus are equal - this is the key to solving rhombus perimeter problems quickly.

Question 2. State whether the following statements are true (T) or false (F):
(i) Centimetre is the unit of area.
(ii) The sum of lengths of a polygon is called its area.
(iii) If the side of a square is doubled, then its perimeter is also doubled.
(iv) If the side of a square is doubled, then its area is also doubled.
(v) To find the cost of constructing a road, we find its area.
(vi) To find the cost of fencing a field, we find its perimeter.
Answer:
(i) False. Centimetre is the unit of length. Area is measured in square centimetres (sq. cm).

(ii) False. The sum of all side lengths of a polygon is called its perimeter, not its area.

(iii) True. If side = s, then original perimeter = 4s. When side becomes 2s, new perimeter = 4(2s) = 8s = 2 × 4s. The perimeter doubles too.

(iv) False. If side = s, then original area = s². When side becomes 2s, new area = (2s)² = 4s². The area becomes four times larger, not double.

(v) True. Building a road covers the ground surface, so we calculate its area to know how much material is needed.

(vi) True. Fencing runs along the outer boundary of the field, so we find its perimeter to know how much fencing material is required.
In simple words: Remember: perimeter is the distance around, area is the space inside. Doubling a side doubles the perimeter but makes the area four times bigger.

Exam Tip: Statements (v) and (vi) are practical applications - always ask yourself: "Does this measure distance around something (perimeter) or space inside it (area)?"

Question 3. If the perimeter of a square is 50 cm, then its side is
(1) 200 cm
(2) 150 cm
(3) 25 cm
(4) 12.5 cm
Answer: (4) 12.5 cm

Given: Perimeter = 50 cm

Perimeter of square = 4 × side

50 = 4 × side

side = 50 ÷ 4 = 12.5 cm
In simple words: A square has 4 equal sides. If the distance all the way around is 50 cm, each side must be 50 divided by 4, which is 12.5 cm.

Exam Tip: For a square, always divide the perimeter by 4 to get one side - this is faster and less error-prone than guessing.

Question 4. The area of the rectangle with length 25 cm and breadth 12 cm is
(1) 300 sq. m
(2) 74 cm
(3) 300 sq. cm
(4) 74 sq. cm
Answer: (3) 300 sq. cm

Given: Length = 25 cm, Breadth = 12 cm

Area = length × breadth = 25 × 12 = 300 sq. cm
In simple words: Multiply the length by the breadth to find the area. The answer must be in square centimetres, not just centimetres.

Exam Tip: Watch the units carefully - option (1) says sq. m (way too big), and option (4) confuses area with perimeter. Always check your units before choosing.

Question 5. If the perimeter of a square is 36 cm, then its area is
(1) 6 sq. cm
(2) 9 sq. cm
(3) 18 sq. cm
(4) 81 sq. cm
Answer: (4) 81 sq. cm

Given: Perimeter = 36 cm

Side = Perimeter ÷ 4 = 36 ÷ 4 = 9 cm

Area = side × side = 9 × 9 = 81 sq. cm
In simple words: First find how long each side is by dividing the perimeter by 4. Then multiply that side by itself to get the area.

Exam Tip: This is a two-step problem: find the side first, then find the area. Don't skip the first step or you'll get it wrong.

Question 5. If the perimeter of a square is 36 cm, then its area is
(a) 81 sq. cm
(b) 9 sq. cm
(c) 36 sq. cm
(d) 18 sq. cm
Answer: (a) 81 sq. cm
In simple words: The perimeter tells you the total distance around the square. Divide 36 by 4 to get one side (9 cm). Then multiply 9 by 9 to get the area.

Exam Tip: Always use the perimeter formula (4 × side) to find the side first, then use the area formula (side × side). Common error: confusing perimeter with area.

Question 6. If the area of a rectangular plot is 180 sq. m and its length is 15 m, then its breadth is
(a) 12 m
(b) 12 cm
(c) 60 m
(d) 9 m
Answer: (a) 12 m
In simple words: Area equals length times breadth. If you know the area and the length, divide the area by the length to get the breadth.

Exam Tip: Watch the units - the answer should match the units of the given measurements (metres, not centimetres in this case).

Question 7. If the length and the breadth of a rectangle are doubled, then its perimeter
(a) remains the same
(b) doubles
(c) becomes four times
(d) becomes half
Answer: (b) doubles
In simple words: When you double both the length and width, the perimeter doubles as well. The new perimeter is 2 times the old perimeter.

Exam Tip: Use variables (l and b) instead of actual numbers to prove this. This makes it clear for any rectangle size.

Question 8. If the length and the breadth of a rectangle are doubled then its area
(a) remains same
(b) becomes half
(c) doubles
(d) becomes four times
Answer: (d) becomes four times
In simple words: When you double both length and width, the area becomes 4 times larger. This is because you multiply the new length by the new width, which gives you 2 × 2 = 4 times the original area.

Exam Tip: Remember that doubling two dimensions multiplies the area by 4, not 2. This is a common error among students.

Question 9. If the sides of a square are halved, then its area
(a) remains same
(b) becomes half
(c) becomes one-fourth
(d) doubles
Answer: (c) becomes one-fourth
In simple words: When you make the side half as long, the area shrinks to one-quarter of what it was. This happens because the new area is (s/2) × (s/2) = s²/4.

Exam Tip: When any linear dimension is halved, the area reduces to one-quarter. When halved, area always becomes 1/4 of the original - this is a key relationship to remember.

Question 10. In the adjoining figure, a square of side 1 cm is joined to a square of side 3 cm. The perimeter of the new figure is
(a) 13 cm
(b) 14 cm
(c) 15 cm
(d) 16 cm
Answer: (b) 14 cm
In simple words: When the smaller square touches the larger one, part of each square's edge is no longer on the outer boundary. You must subtract these hidden edges from the sum of both perimeters.

Exam Tip: Always identify which edges are hidden when shapes are joined. The new perimeter is not simply the sum of the two original perimeters - subtract the coinciding edges twice (once from each shape).

Question 11. Two regular hexagons of perimeter 30 cm each are joined as shown in the adjoining figure. The perimeter of the new figure is
(a) 65 cm
(b) 60 cm
(c) 55 cm
(d) 50 cm
Answer: (d) 50 cm
In simple words: Each hexagon has 6 equal sides. With a perimeter of 30 cm, each side is 5 cm. When they join along one side, that 5 cm edge vanishes from both shapes. So you subtract 10 cm total from 60 cm.

Exam Tip: For regular polygons, always find the side length first by dividing perimeter by the number of sides. When shapes merge, two edges (one from each shape) are hidden - account for both.

Question 12. If the area of a square is numerically equal to its perimeter, then the length of each side is
(a) 1 unit
(b) 2 units
(c) 3 units
(d) 4 units
Answer: (d) 4 units
In simple words: Set up an equation where area (s²) equals perimeter (4s). Solve by factoring: s² = 4s becomes s² - 4s = 0, which gives s(s - 4) = 0. Since s cannot be zero, s = 4.

Exam Tip: This is an equation-solving problem disguised as a geometry question. Always set up the equality and solve algebraically - do not guess and check.

Question 13. Four square tables with side 1.3 m are placed end to end to form one big rectangular table. The perimeter of the rectangular table is
(a) 5.2 m
(b) 10.4 m
(c) 13 m
(d) 20.8 m
Answer: (c) 13 m
In simple words: Four squares in a line create a long rectangle. The length becomes 4 × 1.3 = 5.2 m. The width stays 1.3 m. The perimeter is 2(5.2 + 1.3) = 2 × 6.5 = 13 m.

Exam Tip: Visualize how the squares are arranged. "End to end" usually means they form a long rectangle, not a 2 × 2 grid. Always confirm the final shape before calculating.

Question 14. The area of a rectangle is 180 cm². If its length is 15 cm, then the ratio of its breadth to its length is
(a) 1 : 12
(b) 12 : 1
(c) 5 : 4
(d) 4 : 5
Answer: (d) 4 : 5
In simple words: First find the breadth by dividing area by length: 180 ÷ 15 = 12 cm. Then form the ratio breadth : length = 12 : 15. Reduce by dividing both by 3 to get 4 : 5.

Exam Tip: Always simplify ratios to lowest terms by finding the GCD. Check that both parts of the ratio are correct before finalizing your answer.

Question 15. A picture is 60 cm wide and 1.8 m long. The ratio of its width to its perimeter in lowest form is
(a) 1 : 2
(b) 1 : 3
(c) 1 : 6
(d) 1 : 8
Answer: (d) 1 : 8
In simple words: Convert length to the same units: 1.8 m = 180 cm. Calculate perimeter: 2(180 + 60) = 480 cm. The ratio width : perimeter = 60 : 480. Simplify by dividing both by 60 to get 1 : 8.

Exam Tip: Always ensure both quantities are in the same units before calculating. Unit mismatch is a common source of errors in ratio problems.

Question 16. Statement I: I have two wires of 27 m each. I bend one into a circle and the other into an equilateral triangle. Both shapes have the same perimeter. Statement II: The perimeter of a closed shape is the length of its boundary.
(a) Statement I is true but statement II is false.
(b) Statement I is false but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
Answer: (c) Both Statement I and Statement II are true.
In simple words: Statement I is true because both wires are 27 m long, so both shapes have a perimeter (or circumference) of 27 m. Statement II is true because it correctly defines perimeter as the total distance around a closed shape.

Exam Tip: For Statement I-II questions, evaluate each statement independently first, then determine their combined truth value. Both statements must be checked carefully.

Question 17. Statement I: Two rectangular sheets, of length 20 m and width 10 m each, are placed edge to edge along their longer sides. The perimeter of the resulting figure is 80 m. Statement II: Perimeter of a pentagon is five times the length of any side.
(a) Statement I is true but statement II is false.
(b) Statement I is false but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
Answer: (a) Statement I is true but statement II is false.
In simple words: Statement I is correct: when two 20 m × 10 m rectangles join along their 20 m sides, they create a 20 m × 20 m square with perimeter 80 m. Statement II is only true for regular pentagons where all sides are equal - irregular pentagons have different side lengths, so this rule does not apply to all pentagons.

Exam Tip: Statement II contains a common trap. A property that works for regular polygons does not automatically work for all polygons of that type. Always check the condition carefully.

Question 18. Statement I: A mobile phone is 15 cm long and 8 cm wide. The sum of the areas of the front and the back of the mobile phone is 240 cm². Statement II: Area of a square with side a is a².
(a) Statement I is true but statement II is false.
(b) Statement I is false but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
Answer: (c) Both Statement I and Statement II are true.
In simple words: Statement I is true: the area of one side is 15 × 8 = 120 cm², so the front and back together make 240 cm². Statement II is true: this is the standard formula for the area of a square with side length a.

Exam Tip: When a question mentions "front and back" or "both sides," remember to double the area of one face. This is a realistic scenario that appears often in practical geometry problems.

Question 19. Statement I: A room has two doors of equal dimensions. The perimeter of each door is 7 m. The sum of their widths is 3 m. Therefore, the height of each door is 2 m. Statement II: Length of the rectangle = perimeter + width
(1) Statement I is true but statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (1) Statement I is true but statement II is false
In simple words: When you check Statement I by dividing 3 m by 2 doors, each door is 1.5 m wide. Using the perimeter formula, the height works out to be 2 m, so Statement I is correct. For Statement II, the correct formula subtracts the width from the perimeter (after dividing by 2), not adds it, so Statement II is wrong.

Exam Tip: Always verify both statements separately using the formulas given — don't assume one is correct based on the other. Test with actual numbers to confirm.

Question 20. Statement I: Area and perimeter of a given shape always have the same units. Statement II: Area of a closed plane figure measures the region enclosed by its boundary.
(1) Statement I is true but statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (2) Statement I is false but Statement II is true
In simple words: Perimeter is measured in single units like metres or centimetres, but area is measured in square units like square metres or square centimetres. They have different units. Area does mean the space inside a shape's boundary, so Statement II is right.

Exam Tip: Remember the key difference: perimeter uses linear units (m, cm) while area uses square units (sq. m, sq. cm). This is a fact that appears in many exam questions.

Check Your Progress

Question 1. The perimeter of a square ABCD is twice the perimeter of △PQR. Find the area of the square ABCD.
Answer: Given that triangle PQR has sides measuring PQ = 6 cm, QR = 7 cm, and PR = 5 cm.

Calculate the perimeter of △PQR by summing all sides:
\( \text{Perimeter of } \triangle PQR = 6 + 7 + 5 = 18 \text{ cm} \)

The perimeter of square ABCD is double the triangle's perimeter:
\( \text{Perimeter of square ABCD} = 2 \times 18 = 36 \text{ cm} \)

For a square, the perimeter equals 4 times the side length:
\( 36 = 4 \times \text{side} \)
\( \text{side} = 9 \text{ cm} \)

The area of the square is the side multiplied by itself:
\( \text{Area of square ABCD} = 9 \times 9 = 81 \text{ sq. cm} \)
In simple words: First find the triangle's perimeter by adding its three sides. Double that to get the square's perimeter. Divide by 4 to find one side of the square, then multiply the side by itself to get the area.

Exam Tip: Always work step-by-step through these multi-stage problems. Calculate perimeter first, then use it to find the side length, then find area. Show each step clearly.

Question 2. The perimeter of an equilateral triangle is 42 cm. Find the perimeter of a square, each side of which is double the side of the triangle.
Answer: An equilateral triangle has a perimeter of 42 cm. Since all three sides of an equilateral triangle are equal:
\( \text{Perimeter of equilateral triangle} = 3 \times \text{side} \)
\( 42 = 3 \times \text{side} \)
\( \text{side of triangle} = 14 \text{ cm} \)

The side of the square is twice the triangle's side:
\( \text{side of square} = 2 \times 14 = 28 \text{ cm} \)

The perimeter of the square is 4 times its side length:
\( \text{Perimeter of square} = 4 \times 28 = 112 \text{ cm} \)
In simple words: Divide the triangle's perimeter by 3 to get one side. Double it to get the square's side. Multiply that by 4 to get the square's perimeter.

Exam Tip: In equilateral triangles, all sides are the same, so dividing perimeter by 3 gives the side. For squares, multiply the side by 4 to get the perimeter.

Question 3. A wire of length 60 cm is cut into two pieces. One piece is used to form a rectangle of length 10 cm and width 8 cm. The other piece is bent into the shape of a regular hexagon. What is the length of each side of hexagon?
Answer: The total length of the wire is 60 cm.

For the rectangular piece, calculate its perimeter using length 10 cm and width 8 cm:
\( \text{Perimeter of rectangle} = 2(10 + 8) = 2 \times 18 = 36 \text{ cm} \)

The wire remaining for the hexagon is:
\( 60 - 36 = 24 \text{ cm} \)

A regular hexagon has 6 equal sides. If the perimeter is 24 cm:
\( \text{Perimeter of hexagon} = 6 \times \text{side} \)
\( 24 = 6 \times \text{side} \)
\( \text{side} = 4 \text{ cm} \)
In simple words: First figure out how much wire the rectangle uses by finding its perimeter. The leftover wire becomes the hexagon. Divide that leftover length by 6 to find one side of the hexagon.

Exam Tip: Always track how much material is used at each step. Subtract from the total to find what is left for the next part.

Question 4. A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which encloses more area, the square or the rectangle and by how much?
Answer: The square has a side of 10 cm, so its perimeter is:
\( \text{Perimeter of square} = 4 \times 10 = 40 \text{ cm} \)

When the same wire is formed into a rectangle with length 12 cm, the perimeter stays 40 cm. Using the rectangle perimeter formula:
\( \text{Perimeter of rectangle} = 2(\text{length} + \text{breadth}) \)
\( 40 = 2(12 + \text{breadth}) \)
\( 20 = 12 + \text{breadth} \)
\( \text{breadth} = 8 \text{ cm} \)

Now compare the areas. The square's area is:
\( \text{Area of square} = 10 \times 10 = 100 \text{ sq. cm} \)

The rectangle's area is:
\( \text{Area of rectangle} = 12 \times 8 = 96 \text{ sq. cm} \)

Since 100 sq. cm is greater than 96 sq. cm, the square encloses more area. The difference is:
\( 100 - 96 = 4 \text{ sq. cm} \)
In simple words: The same length of wire forms both shapes, so their perimeters are equal. Divide the perimeter by 2 and subtract the length to find the breadth. Calculate both areas and compare them.

Exam Tip: When the same wire is reshaped, the perimeter remains constant but the area changes. Always compare both measurements when asked.

Question 5. A rectangular room is 9 m long and 6 m wide. Find the cost of covering the floor with carpet 2 m wide at Rs. 350 per metre.
Answer: The floor's area is:
\( \text{Area of floor} = 9 \times 6 = 54 \text{ sq. m} \)

Since the carpet exactly covers the floor, it must also cover 54 sq. m. The carpet is 2 m wide, so its length is:
\( \text{Length of carpet} = \frac{\text{Area of carpet}}{\text{Width of carpet}} = \frac{54}{2} = 27 \text{ m} \)

The cost is calculated by multiplying length by the rate per metre:
\( \text{Cost} = 27 \times 350 = \text{Rs. } 9450 \)
In simple words: Work out the floor's area. Divide that area by the carpet's width to find how many metres of carpet you need. Multiply by the cost per metre.

Exam Tip: When dealing with area and length together, remember that area divided by one dimension gives you the other dimension. This is the key to finding the carpet length.

Question 6. If the cost of fencing a square plot at the rate of Rs. 30 per metre is Rs. 2400, then find the length of each side of the field.
Answer: The perimeter of the square plot can be found by dividing total cost by the rate per metre:
\( \text{Perimeter of square plot} = \frac{\text{Total cost}}{\text{Rate}} = \frac{2400}{30} = 80 \text{ m} \)

For a square, the perimeter equals 4 times the side:
\( \text{Perimeter of square} = 4 \times \text{side} \)
\( 80 = 4 \times \text{side} \)
\( \text{side} = 20 \text{ m} \)
In simple words: Divide the total cost by the cost per metre to get the perimeter. Then divide the perimeter by 4 to get one side of the square.

Exam Tip: Always work backwards from the cost. Divide cost by rate to find distance (perimeter), then use the shape's formula to find the side.

Question 7. If the cost of fencing a rectangular park at the rate of Rs. 30 per metre is Rs. 2400 and the length of the park is 24 m, find the breadth of the park.
Answer: First, calculate the perimeter using the total cost and rate:
\( \text{Perimeter of rectangular park} = \frac{2400}{30} = 80 \text{ m} \)

Use the rectangle perimeter formula with the known length of 24 m:
\( \text{Perimeter of rectangle} = 2(\text{length} + \text{breadth}) \)
\( 80 = 2(24 + \text{breadth}) \)
\( 40 = 24 + \text{breadth} \)
\( \text{breadth} = 16 \text{ m} \)
In simple words: Find the perimeter by dividing cost by rate. Then use the perimeter formula for rectangles to find the breadth when the length is known.

Exam Tip: When length is given, substitute it into the perimeter formula and solve for breadth. Always show the substitution step clearly.

Question 8. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
Answer:

(i) The U-shaped figure can be divided into three separate rectangles:

Rectangle 1 (top section, DEFG): length = 5 cm, width = 1 cm
\( \text{Area} = 5 \times 1 = 5 \text{ sq. cm} \)

Rectangle 2 (bottom left, JIHG): length = 2 cm, width = 1 cm
\( \text{Area} = 2 \times 1 = 2 \text{ sq. cm} \)

Rectangle 3 (bottom right, ABCD): length = 2 cm, width = 1 cm
\( \text{Area} = 2 \times 1 = 2 \text{ sq. cm} \)

Total area of figure (i):
\( 5 + 2 + 2 = 9 \text{ sq. cm} \)

(ii) For the second irregular figure, first determine the missing measurements by examining how the sides relate.

Given: LC = 4 cm, EM = 3 cm, and EM = FD = LF + LD

Since IF + FL = JK = 3 cm and HG = IF = 2 cm:
\( 2 + FL = 3 \)
\( FL = 1 \text{ cm} \)

From LC = 4 cm and FL + LD = 3 cm:
\( LD = 2 \text{ cm} \)
\( DC = 4 - 2 = 2 \text{ cm} \)

Rectangle ABCD: length = 2 cm, width = 4 cm
\( \text{Area} = 2 \times 4 = 8 \text{ sq. cm} \)

Rectangle HIGF: length = 2 cm, width = 1 cm
\( \text{Area} = 2 \times 1 = 2 \text{ sq. cm} \)

Square MEFD: side = 3 cm
\( \text{Area} = 3 \times 3 = 9 \text{ sq. cm} \)

Rectangle IJKL: length = 3 cm, width = 3 cm
\( \text{Area} = 3 \times 3 = 9 \text{ sq. cm} \)

Total area of figure (ii):
\( 8 + 2 + 9 + 9 = 28 \text{ sq. cm} \)
In simple words: Break a complex shape into simpler rectangles. Find each rectangle's area separately. Add all the areas together to get the total. If dimensions are missing, use the known measurements to work them out logically.

Exam Tip: When dimensions are not clearly marked, use the property that opposite sides of rectangles are equal to calculate missing values. Draw lines to show how you split the figure.

Question 9. How many envelopes of size 25 cm × 15 cm can be made from a rectangular sheet of size 4 m × 1.2 m?
Answer: First, convert all measurements to the same unit (centimetres):
\( 4 \text{ m} = 400 \text{ cm} \)
\( 1.2 \text{ m} = 120 \text{ cm} \)

Calculate the area of the large sheet:
\( \text{Area of sheet} = 400 \times 120 = 48,000 \text{ sq. cm} \)

Calculate the area of one envelope:
\( \text{Area of one envelope} = 25 \times 15 = 375 \text{ sq. cm} \)

Find how many envelopes can be made by dividing total area by one envelope's area:
\( \text{Number of envelopes} = \frac{48,000}{375} = 128 \)
In simple words: Change metres to centimetres so all units match. Work out the area of the big sheet and the area of one envelope. Divide the sheet's area by the envelope's area.

Exam Tip: Always convert units before doing area calculations. Make sure both shapes use the same measurement unit, or your answer will be wrong.

Question 10. The perimeter of a rectangle is 36 cm. What will be length and breadth (in natural numbers) of that rectangle whose area is (i) maximum? (ii) minimum?
Answer: From the perimeter formula, we get length + breadth = 18 cm. When we list all possible pairs of natural numbers that add up to 18, we can calculate the area for each pair. Looking at the results, the largest area is 81 sq. cm, which occurs when length = 9 cm and breadth = 9 cm. The smallest area is 17 sq. cm, which occurs when length = 17 cm and breadth = 1 cm.
In simple words: For the biggest area, make the rectangle as close to a square as possible. For the smallest area, make it very long and thin.

Exam Tip: Always create a table listing all possible dimension pairs and their areas - this systematic approach ensures you don't miss the maximum or minimum values.

Question 11. In the adjoining figure, A, B and C are squares. If the area of the square A is 25 sq. cm and the perimeter of the square B is 12 cm, then find the perimeter and the area of the square C.
Answer: From the area of square A (25 sq. cm), the side of A is 5 cm. From the perimeter of square B (12 cm), the side of B is 3 cm. According to the figure layout, the side of square C equals the sum of the sides of squares A and B, which is 5 + 3 = 8 cm. Therefore, the perimeter of C = 4 × 8 = 32 cm, and the area of C = 8 × 8 = 64 sq. cm.
In simple words: Find each side length first, then add them to get C's side. Once you know C's side, multiply to find its perimeter and area.

Exam Tip: Always identify what you're given (area or perimeter) and convert it to the side length before attempting to find the required measurements for the combined figure.

Question 12. Study the following floor plan of a house prepared by an architect. Some of the measurements are given. (i) Find the missing measurements. (ii) Find out the total area of the house (including the Garden).
Answer:
(i) Using the known dimensions and the total width of 44 ft, we calculate each missing measurement:

For Utility: Width = Area ÷ Height = 77 ÷ 11 = 7 ft, so dimensions are 7 ft × 11 ft, area = 77 sq. ft.

For Toilet: Width = 44 - (13 + 18 + 7) = 6 ft, height = 11 ft (same as Kitchen), so dimensions are 6 ft × 11 ft, area = 66 sq. ft.

For Hall: Width = 24 ft, height = 16 ft (same as Master Bedroom), so dimensions are 24 ft × 16 ft, area = 384 sq. ft.

For Garden: Width = 44 - (13 + 24) = 7 ft, height = 16 ft (same as Hall), so dimensions are 7 ft × 16 ft, area = 112 sq. ft.

For Small Bedroom: Dimensions are 13 ft × 11 ft, area = 143 sq. ft.

The missing measurements are:
- Toilet: 6 ft × 11 ft, Area = 66 sq. ft
- Utility: 7 ft × 11 ft, Area = 77 sq. ft
- Hall: 24 ft × 16 ft, Area = 384 sq. ft
- Garden: 7 ft × 16 ft, Area = 112 sq. ft
- Small Bedroom: 13 ft × 11 ft, Area = 143 sq. ft

(ii) The total length of the house equals the Master Bedroom height plus the Small Bedroom height: 16 + 11 = 27 ft. The total width is 44 ft. Therefore, the total area = 44 × 27 = 1188 sq. ft.
In simple words: Find each room's missing side by using the area or by subtracting known widths from the total. Once you have all dimensions, add them correctly to find the total house area.

Exam Tip: Verify your answers by checking that all widths on each row add up to 44 ft and all heights on each column add up to 27 ft - this confirms your calculations are internally consistent.