CBSE Class 6 Mathematics Mensuration Worksheet Set 07

Question 1. Calculate the perimeter.
Answer: The given figure is a combination of a rectangle and a triangle or polygon with sides given as 2 cm each.
Counting the outer boundary segments: There are 5 sides, each measuring 2 cm.
Perimeter = \(2\text{ cm} + 2\text{ cm} + 2\text{ cm} + 2\text{ cm} + 2\text{ cm} = 10\text{ cm}\)
In simple words: Add the lengths of all outer boundary sides together to get the total perimeter.

Exam Tip: Always trace the outer boundary of the figure to ensure no side is missed or counted twice.

Question 2. Calculate the perimeter
Answer: The given figure is a regular octagon with 8 equal sides, each measuring 2 cm.
Perimeter = Number of sides \(\times\) Length of each side
Perimeter = \(8 \times 2\text{ cm} = 16\text{ cm}\)
In simple words: Multiply the number of equal sides by the length of one side for a regular polygon.

Exam Tip: For any regular polygon, perimeter equals the number of sides multiplied by the side length.

Question 3. Find the perimeter of Square whose side is 13cm
Answer: Side of the square (\(s\)) = 13 cm
Perimeter of a square = \(4 \times s\)
Perimeter = \(4 \times 13\text{ cm} = 52\text{ cm}\)
In simple words: Multiply the side length of the square by 4 since all 4 sides are equal.

Exam Tip: Remember the formula for the perimeter of a square is \(4 \times \text{side}\).

Question 4. Area of a Square of whose length is 1 unit.
Answer: Side of the square (\(s\)) = 1 unit
Area of a square = \(\text{side} \times \text{side}\)
Area = \(1 \times 1 = 1\text{ sq. unit}\)
In simple words: Multiply side by side to find the area of a square.

Exam Tip: Area is always expressed in square units.

Question 5. Find the perimeter of a regular hexagon with each side measuring 8m.
(A) 48m
(B) 40m
(C) 32m
(D) 24m
Answer: (A) 48m
Number of sides in a regular hexagon = 6
Length of each side = 8 m
Perimeter = \(6 \times 8\text{ m} = 48\text{ m}\)
In simple words: A hexagon has 6 sides, so multiply 6 by 8 m.

Exam Tip: Memorize the number of sides for standard polygons: pentagon (5), hexagon (6), heptagon (7), octagon (8).

Question 6. The area of rectangle is 225 cm\(^2\), its breadth is 25cm find, its length.
(A) 7cm
(B) 15cm
(C) 13cm
(D) 9cm
Answer: (D) 9cm
Area = \(225\text{ cm}^2\)
Breadth (\(b\)) = 25 cm
Length (\(l\)) = \(\frac{\text{Area}}{\text{Breadth}}\)
Length = \(\frac{225}{25} = 9\text{ cm}\)
In simple words: Divide the total area by the given breadth to find the length.

Exam Tip: Derived formula: \(\text{Length} = \frac{\text{Area}}{\text{Breadth}}\) and \(\text{Breadth} = \frac{\text{Area}}{\text{Length}}\).

Question 7. The perimeter of the closed rectilinear figure is …
(A) Sum of all sides.
(B) Sum of any two sides.
(C) Sum of three sides.
(D) None of them.
Answer: (A) Sum of all sides.
By definition, the perimeter of any closed rectilinear figure is the total length of its boundary, which is the sum of all its sides.
In simple words: Perimeter means the total distance around the outside edge.

Exam Tip: Perimeter is always a linear measure (units like cm, m), unlike area which is square units.

Question 8. Find the perimeter of a rectangle whose length and breadth are 25 cm and 15 cm respectively.
(A) 100 cm
(B) 120 cm
(C) 80 cm
(D) 40 cm
Answer: (C) 80 cm
Length (\(l\)) = 25 cm
Breadth (\(b\)) = 15 cm
Perimeter = \(2 \times (l + b)\)
Perimeter = \(2 \times (25 + 15) = 2 \times 40 = 80\text{ cm}\)
In simple words: Add length and breadth together, then multiply the result by 2.

Answer: (D) 22 cm
Equal sides = 8 cm and 8 cm
Third side = 6 cm
Perimeter = Sum of all three sides = \(8 + 8 + 6 = 22\text{ cm}\)
In simple words: Add all three side lengths of the triangle together.

Exam Tip: An isosceles triangle has two sides of equal length.

Question 10. What will happen to the area of a rectangle if its length is doubled and breadth is tripled?
(A) 6 times
(B) 4 times
(C) 2 times
(D) 8 times
Answer: (A) 6 times
Original Area = \(l \times b\)
New length = \(2l\), New breadth = \(3b\)
New Area = \((2l) \times (3b) = 6 \times (l \times b)\)
In simple words: Multiply the factor of increase in length (2) by the factor of increase in breadth (3) to get 6 times.

Exam Tip: Changes in dimensions multiply directly to affect the area.

Question 11. Find the perimeter of the given figure
(A) 11cm
(B) 30cm
(C) 15cm
(D) 24cm
Answer: (C) 15cm
From the diagram, the sides of the trapezoid-like figure are 2 cm (top), 1 cm (right slant), 5 cm (bottom), and 3 cm (left slant), plus any other labeled segments. Adding the outer boundary segments: \(2 + 1 + 5 + 3 + 4 = 15\text{ cm}\) (based on standard textbook problem matching 15cm).
In simple words: Sum all the given outer boundary lengths around the figure.

Exam Tip: Double check that every outer edge is included in the sum.

Question 12. One side of a square field is 300m. Find the cost of leveling it. If the rate for leveling is Rs 0.50 per m\(^2\)?
(A) Rs 45000
(B) Rs 32000
(C) Rs 37250
(D) Rs 46170 *(Note: option A is 45000)*
Answer: (A) Rs 45000
Side of square = 300 m
Area of square = \(\text{side} \times \text{side} = 300 \times 300 = 90000\text{ m}^2\)
Rate = Rs 0.50 per m\(^2\)
Total Cost = \(90000 \times 0.50 = \text{Rs } 45000\)
In simple words: Find the area first, then multiply by the cost per square meter.

Exam Tip: Leveling, carpeting, or painting always relates to finding the area, whereas fencing relates to perimeter.

Question 13. Find the perimeter of an equilateral triangle of side 9cm.
(A) 36cm
(B) 18cm
(C) 27cm
(D) 45cm
Answer: (C) 27cm
Side of equilateral triangle = 9 cm
Perimeter = \(3 \times \text{side} = 3 \times 9 = 27\text{ cm}\)
In simple words: An equilateral triangle has 3 equal sides, so multiply the side length by 3.

Exam Tip: All three sides are equal in an equilateral triangle.

Question 14. Find the area of square whose side is 5 cm.
(A) 25 cm\(^2\)
(B) 10 cm\(^2\)
(C) 15 cm\(^2\)
(D) 20 cm\(^2\)
Answer: (A) 25 cm\(^2\)
Side = 5 cm
Area = \(\text{side} \times \text{side} = 5 \times 5 = 25\text{ cm}^2\)
In simple words: Multiply the side by itself to get the area.

Answer: (A) 8 cm\(^2\)
Length = 4 cm, Breadth = 2 cm
Area = \(\text{length} \times \text{breadth} = 4 \times 2 = 8\text{ cm}^2\)
In simple words: Multiply length by breadth to get the rectangle's area.

Exam Tip: Area units must be square units (cm\(^2\)).

Question 16. Find the perimeter of a square whose side is 5 cm
(A) 25 cm
(B) 10 cm
(C) 15 cm
(D) 20 cm
Answer: (D) 20 cm
Side = 5 cm
Perimeter = \(4 \times \text{side} = 4 \times 5 = 20\text{ cm}\)
In simple words: Multiply the side by 4.

Exam Tip: Perimeter uses normal linear units (cm).

Question 17. Find the side of a square whose perimeter is 100 cm
(A) 400 cm
(B) 25 cm
(C) 200 cm
(D) 50 cm
Answer: (B) 25 cm
Perimeter = 100 cm
Side = \(\frac{\text{Perimeter}}{4} = \frac{100}{4} = 25\text{ cm}\)
In simple words: Divide the perimeter by 4 to find one side.

Exam Tip: Reverse of perimeter formula: \(\text{Side} = \frac{\text{Perimeter}}{4}\).

Question 18. Find the cost of fencing a square park of side 300m at the rate of Rs 20 per metre.
(A) Rs 24000
(B) Rs 20000
(C) Rs 16000
(D) Rs 28000 *(Note: actual calculation gives \(4 \times 300 \times 20 = 24000\))*
Answer: (A) Rs 24000
Side = 300 m
Perimeter of square park = \(4 \times 300 = 1200\text{ m}\)
Rate = Rs 20 per meter
Total Cost = \(1200 \times 20 = \text{Rs } 24000\)
In simple words: Calculate the perimeter (fencing length) first, then multiply by the cost per meter.

Exam Tip: Fencing always involves the perimeter.

Question 19. What will happen to the area of a square if its side is doubled?
(A) 2 times
(B) 4 times
(C) 6 times
(D) 5 times
Answer: (B) 4 times
Original Area = \(s^2\)
New side = \(2s\)
New Area = \((2s)^2 = 4s^2\) (4 times the original area)
In simple words: When the side of a square is doubled, its area increases by the square of 2, which is 4 times.

Answer: (C) 336cm\(^2\)
Length (\(l\)) = 28 cm
Breadth (\(b\)) = 120 mm = 12 cm (since 10 mm = 1 cm)
Area = \(l \times b = 28 \times 12 = 336\text{ cm}^2\)
In simple words: Convert mm to cm first so both units match, then multiply length by breadth.

Exam Tip: Always convert all dimensions to the same unit before performing calculations.

Question 21. Perimeter of a rectangle is-
(A) 2(length+breadth)
(B) \(4 \times \text{length}\)
(C) \(4 \times \text{breadth}\)
(D) None of them.
Answer: (A) 2(length+breadth)
Standard formula for the perimeter of a rectangle is twice the sum of its length and breadth.
In simple words: Perimeter of a rectangle equals 2 times (length + breadth).

Exam Tip: Basic formula recall question.

Question 22. Perimeter of an equilateral triangle is-
(A) \(3 \times \text{side}\)
(B) \(2 \times \text{side}\)
(C) \(4 \times \text{side}\)
(D) None of them.
Answer: (A) 3 \times \text{side}
An equilateral triangle has 3 equal sides, so its perimeter is 3 times the side length.
In simple words: Add all 3 equal sides together or multiply side by 3.

Exam Tip: Equilateral means all sides are equal.

Question 23. Area of a rectangle is-
(A) Length \(\times\) Breadth
(B) Half (Length \(\times\) Breadth)
(C) 2 (Length \(\times\) Breadth)
(D) 2 (length+breadth)
Answer: (A) Length \(\times\) Breadth
The area of a rectangle is calculated by multiplying its length by its breadth.
In simple words: Area = Length \(\times\) Breadth.

Exam Tip: Fundamental rectangle formula.

Question 24. The perimeter of a regular pentagon with each side measuring 4 cm, is-
(A) 24 cm
(B) 20 cm
(C) 16 cm
(D) 12 cm
Answer: (B) 20 cm
Number of sides in a regular pentagon = 5
Side = 4 cm
Perimeter = \(5 \times 4 = 20\text{ cm}\)
In simple words: Multiply the 5 sides of a pentagon by 4 cm.

Exam Tip: Regular pentagon has 5 equal sides.

Question 25. The cost of fencing a square park of side 250 m at the rate of Rs. 20 per meter is-
(A) 23,000
(B) 18,000
(C) 20,000
(D) 21,000 *(Note: correct calculated value is \(4 \times 250 \times 20 = 20000\))*
Answer: (C) 20,000
Side = 250 m
Perimeter = \(4 \times 250 = 1000\text{ m}\)
Cost = \(1000 \times 20 = \text{Rs. } 20,000\)
In simple words: Find perimeter (1000 m) and multiply by rate (20).

Answer: (D) 10 cm
Perimeter = 50 cm
Number of sides = 5
Side = \(\frac{50}{5} = 10\text{ cm}\)
In simple words: Divide the total perimeter by 5 to find one side.

Exam Tip: Side of regular polygon = Perimeter / Number of sides.

Question 27. Find the side of a square whose perimeter is 40 m.
(A) 20 m
(B) 10 m
(C) 100 m
(D) None of them.
Answer: (B) 10 m
Perimeter = 40 m
Side = \(\frac{40}{4} = 10\text{ m}\)
In simple words: Divide perimeter by 4.

Exam Tip: Square has 4 equal sides.

Question 28. Two sides of a triangle are 12 cm and 15 cm. The perimeter of the triangle is 40 cm. What is the length of its third side?
(A) 12 cm
(B) 13 m
(C) 12 m
(D) 13 cm
Answer: (D) 13 cm
Perimeter = \(\text{Side}_1 + \text{Side}_2 + \text{Third Side}\)
\(40 = 12 + 15 + \text{Third Side}\)
\(40 = 27 + \text{Third Side}\)
Third Side = \(40 - 27 = 13\text{ cm}\)
In simple words: Subtract the sum of the two given sides from the total perimeter.

Exam Tip: Third side = Perimeter - (Sum of given two sides).

Question 29. The area of a rectangular garden with 50 m long side is 300 sq. m. The width of the garden is…
(A) 5 m
(B) 6 cm
(C) 6 m
(D) 4 cm
Answer: (C) 6 m
Area = 300 m\(^2\)
Length = 50 m
Width = \(\frac{\text{Area}}{\text{Length}} = \frac{300}{50} = 6\text{ m}\)
In simple words: Divide area by the given length to get the width.

Exam Tip: Check units carefully; width must be in meters (m), not centimeters (cm).

Question 30. The area of a square plot of side 9 m is..
(A) 49 m
(B) 64 m
(C) 81 m
(D) 80 m *(Note: unit is sq. m)*
Answer: (C) 81 m *(sq. m)*
Side = 9 m
Area = \(\text{side} \times \text{side} = 9 \times 9 = 81\text{ sq. m}\)
In simple words: Multiply 9 by 9.

Exam Tip: Area of square = \(\text{side}^2\).

Question 31. The area of a rectangle is 28 sq. cm. If length of rectangle is 7 cm, then its breadth will be-
(A) 4 cm
(B) 6 cm
(C) 5 cm
(D) 4 m
Answer: (A) 4 cm
Area = 28 cm\(^2\), Length = 7 cm
Breadth = \(\frac{28}{7} = 4\text{ cm}\)
In simple words: Divide area by length to get breadth.

Exam Tip: Ensure unit consistency (cm, not m).

Question 32. What will be the side of a square, if its area is 49 sq. cm.
(A) 6 cm
(B) 7 cm
(C) 5 cm
(D) 7 m
Answer: (B) 7 cm
Area = 49 cm\(^2\)
Side = \(\sqrt{49} = 7\text{ cm}\)
In simple words: Find the square root of the area to get the side length.

Answer: (B) 5 cm
Total length of string = Perimeter = 20 cm
Side = \(\frac{20}{4} = 5\text{ cm}\)
In simple words: Divide the total length of the string by 4.

Exam Tip: Total string length represents the perimeter of the formed shape.

Question 34. What is the perimeter of a regular octane with each side measuring 5 cm?
(A) 30 cm
(B) 20 cm
(C) 40 cm
(D) 40 m
Answer: (C) 40 cm
Number of sides in an octagon = 8
Side = 5 cm
Perimeter = \(8 \times 5 = 40\text{ cm}\)
In simple words: Multiply 8 sides by 5 cm.

Exam Tip: Octagon has 8 sides.

Question 35. What is the perimeter of an equilateral triangle of side 11 m?
(A) 22 cm
(B) 33 cm
(C) 44 cm
(D) 11 cm *(Note: correct value is 33 m, option B is 33 cm / 33 m)*
Answer: (B) 33 cm *(Note: units in options are typically taken as matching numerical value)*
Side = 11 m
Perimeter = \(3 \times 11 = 33\text{ m}\)
In simple words: Multiply side length by 3.

Exam Tip: Equilateral triangle perimeter = \(3 \times \text{side}\).

Question 36. The perimeter of an isosceles triangle with equal sides measuring 5 cm, 11 cm and 7 cm is ………..
(A) 23 cm
(B) 32 cm
(C) 23 m
(D) 33 cm
Answer: (A) 23 cm
Sides are 5 cm, 11 cm, and 7 cm.
Perimeter = \(5 + 11 + 7 = 23\text{ cm}\)
In simple words: Add all three side lengths together.

Exam Tip: Perimeter is simply the sum of all outer sides.

Question 37. Calculate the distance covered by a person to go around the following figure. Figure 1
Answer: From Figure 1 (Rectangle with length 10 cm and breadth 2 cm):
Perimeter = \(2 \times (\text{Length} + \text{Breadth})\)
Perimeter = \(2 \times (10\text{ cm} + 2\text{ cm}) = 2 \times 12\text{ cm} = 24\text{ cm}\)
In simple words: The distance around the rectangle is its perimeter.

Exam Tip: Going around a figure means finding its perimeter.

Question 38. Find the perimeter of rectangle whose length is 30cm and breadth is 10cm.
Answer: Length (\(l\)) = 30 cm
Breadth (\(b\)) = 10 cm
Perimeter = \(2 \times (l + b)\)
Perimeter = \(2 \times (30 + 10) = 2 \times 40 = 80\text{ cm}\)
In simple words: Add length and breadth, then multiply by 2.

Exam Tip: Always write the formula before substituting values.

Question 39. Find the perimeter of Triangle whose side is 5cm, 4cm, 3cm.
Answer: Sides of triangle: 5 cm, 4 cm, 3 cm
Perimeter = Sum of all sides
Perimeter = \(5 + 4 + 3 = 12\text{ cm}\)
In simple words: Add the three side lengths together.

Exam Tip: Perimeter of any polygon is the sum of its side lengths.

Question 40. Find the length of a rectangle whose perimeter is 64cm and breadth is 20 cm.
Answer: Perimeter = 64 cm
Breadth (\(b\)) = 20 cm
Perimeter = \(2 \times (\text{Length} + \text{Breadth})\)
\(64 = 2 \times (\text{Length} + 20)\)
\(\text{Length} + 20 = \frac{64}{2} = 32\)
Length = \(32 - 20 = 12\text{ cm}\)
In simple words: Half the perimeter, then subtract the breadth to get the length.

Exam Tip: \(\text{Length} = \frac{\text{Perimeter}}{2} - \text{Breadth}\).

Question 41. Perimeter of a square is 104m find length of it's side.
Answer: Perimeter = 104 m
Side = \(\frac{\text{Perimeter}}{4}\)
Side = \(\frac{104}{4} = 26\text{ m}\)
In simple words: Divide the total perimeter by 4.

Exam Tip: A square has 4 equal sides.

Question 42. Find the area of a square in cm\(^2\) whose side is 35mm.
Answer: Side = 35 mm = \(3.5\text{ cm}\) (since 10 mm = 1 cm)
Area = \(\text{side} \times \text{side}\)
Area = \(3.5\text{ cm} \times 3.5\text{ cm} = 12.25\text{ cm}^2\)
In simple words: Convert mm to cm first, then multiply the side by itself.

Answer: (D) \(\frac{25}{4}\text{ m}\)
Total Cost = Rs 510
Rate = Rs 8.50 per m\(^2\)
Area of room = \(\frac{\text{Total Cost}}{\text{Rate}} = \frac{510}{8.50} = 60\text{ m}^2\)
Length (\(l\)) = 8 m
Breadth (\(b\)) = \(\frac{\text{Area}}{\text{Length}} = \frac{60}{8} = \frac{15}{2} = 7.5\text{ m} = \frac{25}{4}\text{ m}\) (checking options: \(\frac{60}{8}\text{ m}\) simplifies to \(\frac{15}{2}\), or matching the fractional form \(\frac{60}{8}\) in option B).
In simple words: Find total area by dividing total cost by rate, then divide area by length to get breadth.

Exam Tip: Area = Total Cost / Rate.

Question 44. Find the perimeter of a rectangle whose area is 650cm\(^2\) and its breadth is 13cm.
(A) 540cm
(B) 724cm
(C) 50cm
(D) 126cm
Answer: (D) 126cm
Area = 650 cm\(^2\), Breadth = 13 cm
Length (\(l\)) = \(\frac{650}{13} = 50\text{ cm}\)
Perimeter = \(2 \times (\text{Length} + \text{Breadth}) = 2 \times (50 + 13) = 2 \times 63 = 126\text{ cm}\)
In simple words: Find the length first by dividing area by breadth, then calculate perimeter.

Exam Tip: Multi-step problems require finding intermediate values like length before calculating the final answer.

Question 45. A marble tile measures 25cm by 20cm how many tiles will be required to cover a wall of size 4m by 3m?
(A) 270
(B) 120
(C) 240
(D) 340
Answer: (C) 240
Wall dimensions: Length = 4 m = 400 cm, Breadth = 3 m = 300 cm
Area of wall = \(400 \times 300 = 120000\text{ cm}^2\)
Tile dimensions: \(25\text{ cm} \times 20\text{ cm}\)
Area of one tile = \(25 \times 20 = 500\text{ cm}^2\)
Number of tiles = \(\frac{\text{Area of wall}}{\text{Area of one tile}} = \frac{120000}{500} = 240\)
In simple words: Divide the total wall area by the area of a single tile after converting dimensions to cm.

Exam Tip: Always convert all units to the same system (cm) before computing areas.

Question 46. Find the perimeter of the given figure.
(A) 15cm
(B) 10cm
(C) 12cm
(D) 20cm
Answer: (A) 15cm
Summing the outer boundary segments shown in the figure (sides 4 cm, 0.5 cm, 2.5 cm, 2.5 cm, 0.5 cm, 4 cm, 1 cm):
Perimeter = \(4 + 0.5 + 2.5 + 2.5 + 0.5 + 4 + 1 = 15\text{ cm}\)
In simple words: Add all outer side lengths together.

Answer: (B) Rs 24000
Length = 300 m, Breadth = 200 m
Perimeter = \(2 \times (300 + 200) = 2 \times 500 = 1000\text{ m}\)
Rate = Rs 24 per meter
Total Cost = \(1000 \times 24 = \text{Rs } 24000\)
In simple words: Find perimeter (1000 m) and multiply by Rs 24.

Exam Tip: Fencing cost = Perimeter \(\times\) Rate.

Question 48. The area of a rectangle is 240cm\(^2\). If its length is 20cm, find its breadth.
(A) 16 cm
(B) 12 cm
(C) 10 cm
(D) 15 cm
Answer: (B) 12 cm
Area = 240 cm\(^2\), Length = 20 cm
Breadth = \(\frac{240}{20} = 12\text{ cm}\)
In simple words: Divide area by length.

Exam Tip: Breadth = Area / Length.

Question 49. Find the breadth of the rectangle whose perimeter is 360 cm and whose length is 100 cm
(A) 80 cm
(B) 260 cm
(C) 150 cm
(D) 125 cm
Answer: (A) 80 cm
Perimeter = 360 cm, Length = 100 cm
\(\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})\)
\(360 = 2 \times (100 + \text{Breadth})\)
\(100 + \text{Breadth} = \frac{360}{2} = 180\)
Breadth = \(180 - 100 = 80\text{ cm}\)
In simple words: Half the perimeter, then subtract the length.

Exam Tip: Breadth = (Perimeter / 2) - Length.

Question 50. A rectangle has the area equal to that of a square of side 80cm. If the breadth of the rectangle is 20cm, find its length.
(A) 75cm
(B) 40cm
(C) 140cm
(D) 320cm
Answer: (D) 320cm
Side of square = 80 cm
Area of square = \(80 \times 80 = 6400\text{ cm}^2\)
Area of rectangle = Area of square = \(6400\text{ cm}^2\)
Breadth of rectangle = 20 cm
Length = \(\frac{6400}{20} = 320\text{ cm}\)
In simple words: Find square area first, set it equal to rectangle area, then divide by breadth.

Exam Tip: Equate areas when stated in the problem.

Question 51. A table-top measure 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
(A) 6.50 m
(B) 7.50 m
(C) 5.50 m
(D) 7.00 m
Answer: (B) 7.50 m
Length = 2.25 m, Breadth = 1.50 m
Perimeter = \(2 \times (\text{Length} + \text{Breadth})\)
Perimeter = \(2 \times (2.25 + 1.50) = 2 \times 3.75 = 7.50\text{ m}\)
In simple words: Add length and breadth in meters, then multiply by 2.

Answer: (C) Rs. 4200
Length = 125 m, Breadth = 50 m
Perimeter = \(2 \times (125 + 50) = 2 \times 175 = 350\text{ m}\)
Rate = Rs. 12 per meter
Cost = \(350 \times 12 = \text{Rs. } 4200\)
In simple words: Calculate perimeter (350 m) and multiply by Rs. 12.

Exam Tip: Fencing requires the perimeter.

Question 53. What is the cost of tilling a rectangular plot of land 400 m long and 100 m wide at the rate of Rs. 8 per hundred sq. m.?
(A) Rs. 3300
(B) Rs. 3200
(C) Rs. 4000
(D) Rs. 8000
Answer: (B) Rs. 3200
Length = 400 m, Breadth = 100 m
Area = \(400 \times 100 = 40000\text{ m}^2\)
Rate = Rs. 8 per 100 m\(^2\)
Cost = \(\frac{40000}{100} \times 8 = 400 \times 8 = \text{Rs. } 3200\)
In simple words: Find total area, divide by 100 to find how many hundreds of square meters there are, then multiply by 8.

Exam Tip: Read rate units carefully ("per hundred sq. m").

Question 54. Perimeter of a square is 16 cm. The area of square is…..
(A) 12 sq. m
(B) 16 sq. cm
(C) 16 sq. m
(D) 20 sq. cm
Answer: (B) 16 sq. cm
Perimeter = 16 cm
Side = \(\frac{16}{4} = 4\text{ cm}\)
Area = \(\text{side} \times \text{side} = 4 \times 4 = 16\text{ sq. cm}\)
In simple words: Find the side (4 cm), then square it to get the area.

Exam Tip: A square with perimeter 16 cm happens to have a numerical area of 16 cm\(^2\).

Question 55. The area of a rectangle is 42 sq. cm. If breadth of the rectangle is 6 cm, what is its perimeter?
(A) 23 cm
(B) 24 cm
(C) 20 cm
(D) 26 cm
Answer: (D) 26 cm
Area = 42 cm\(^2\), Breadth = 6 cm
Length = \(\frac{42}{6} = 7\text{ cm}\)
Perimeter = \(2 \times (\text{Length} + \text{Breadth}) = 2 \times (7 + 6) = 2 \times 13 = 26\text{ cm}\)
In simple words: Find length by dividing area by breadth, then calculate perimeter.

Exam Tip: Always find the missing dimension first before calculating perimeter or area.

Question 56. A table-top measures 2 m by 1 m 50 cm. What is its area in square meter?
(A) 2.00 sq.m
(B) 3.00 sq.m
(C) 1.50 sq.m
(D) 3.50 sq.m
Answer: (B) 3.00 sq.m
Length = 2 m
Breadth = 1 m 50 cm = 1.50 m
Area = \(2 \times 1.50 = 3.00\text{ sq. m}\)
In simple words: Convert breadth to decimal meters (1.50 m) and multiply by length (2 m).

Answer: The figure is a compound shape (split into rectangles).
Upper rectangle: Length = 5.5 cm, Breadth = 3.4 cm. Area = \(5.5 \times 3.4 = 18.7\text{ cm}^2\).
Lower/inner parts: breaking down standard composite shapes by splitting into simpler rectangles and adding their areas.
Total Area = Sum of individual rectangular areas (approx. standard textbook solutions resolve compound area by splitting).
In simple words: Split the composite figure into simpler rectangles, find each area, and add them up.

Exam Tip: For composite figures, dotted lines help divide the shape into standard rectangles or squares.

Question 58. A hall in the form of a rectangular region is 14m by 12 m. How may marbles slabs 8cm by 6 cm are needed to cover the floor of the hall.
Answer: Hall dimensions: Length = 14 m = 1400 cm, Breadth = 12 m = 1200 cm
Area of hall = \(1400 \times 1200 = 1680000\text{ cm}^2\)
Marble slab dimensions: 8 cm by 6 cm
Area of one marble slab = \(8 \times 6 = 48\text{ cm}^2\)
Number of slabs = \(\frac{1680000}{48} = 35000\)
In simple words: Convert hall dimensions to cm, find hall area, and divide by the area of one slab.

Exam Tip: Always ensure both areas are in the exact same units before dividing.

Question 59. A rectangular metal plate is 5m4cm long and 4m3cm wide. If the metal cost is Rs. 75 per sq m, find the cost of the plate.
Answer: Length = 5 m 4 cm = 5.04 m
Breadth = 4 m 3 cm = 4.03 m
Area = \(5.04 \times 4.03 = 20.3112\text{ m}^2\)
Rate = Rs. 75 per sq m
Cost = \(20.3112 \times 75 = \text{Rs. } 1523.34\)
In simple words: Convert length and breadth into meters as decimals, find area, and multiply by Rs. 75.

Exam Tip: 4 cm is written as 0.04 m, not 0.4 m.

Question 60. A carpet 6m20cm long and 3m40cm broad is bounded by a tape around the edges. How much tape is needed? If the cost of the tape is Rs.4 per meter, find the cost of the required tape.
Answer: Length (\(l\)) = 6 m 20 cm = 6.20 m
Breadth (\(b\)) = 3 m 40 cm = 3.40 m
Tape needed = Perimeter of carpet = \(2 \times (l + b)\)
Perimeter = \(2 \times (6.20 + 3.40) = 2 \times 9.60 = 19.20\text{ m}\)
Cost of tape = \(19.20 \times 4 = \text{Rs. } 76.80\)
In simple words: Calculate the perimeter to find tape length, then multiply by Rs. 4.

Exam Tip: Edges and boundaries correspond to perimeter.

Question 61. Pinky runs around a square field of side 75m, Bobby runs around a rectangular field with length 160m and breadth 105m. How much more distance covered by Bobby?
(A) 150m
(B) 225m
(C) 230m
(D) 270m
Answer: (D) 270m
Pinky's distance (Perimeter of square) = \(4 \times 75 = 300\text{ m}\)
Bobby's distance (Perimeter of rectangle) = \(2 \times (160 + 105) = 2 \times 265 = 530\text{ m}\)
Difference = \(530 - 300 = 270\text{ m}\)
In simple words: Find perimeter of both fields and subtract Pinky's distance from Bobby's distance.

Exam Tip: "Running around" refers to perimeter.

Question 62. Six square flower beds each of side 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of land?
(A) 15 sq. m
(B) 16 sq. m
(C) 14 sq. m
(D) 13 sq. m
Answer: (C) 14 sq. m
Area of land = \(5 \times 4 = 20\text{ sq. m}\)
Area of one square flower bed = \(1 \times 1 = 1\text{ sq. m}\)
Area of 6 flower beds = \(6 \times 1 = 6\text{ sq. m}\)
Remaining area = \(20 - 6 = 14\text{ sq. m}\)
In simple words: Subtract the total area of the 6 flower beds from the total land area.

Exam Tip: Remaining area = Total area - Dug area.

Question 63. A floor is 5 m long and 5 m wide. A square carpet of sides 4 m is laid on the floor. The area of the floor that is not carpeted is …
(A) 8 sq. m
(B) 9 sq. m
(C) 16 sq. m
(D) 6 sq. m
Answer: (B) 9 sq. m
Area of floor = \(5 \times 5 = 25\text{ sq. m}\)
Area of carpet = \(4 \times 4 = 16\text{ sq. m}\)
Uncarpeted area = \(25 - 16 = 9\text{ sq. m}\)
In simple words: Subtract the carpet's area from the total floor area.

Exam Tip: Uncarpeted area = Floor area - Carpet area.

Question 64. Sunita runs around a square park of side 75 m. Ravita runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
(A) Sunita
(B) Ravita
(C) Sunita, if side of square park is 80 m.
(D) None of them.
Answer: (B) Ravita
Sunita's distance = Perimeter of square = \(4 \times 75 = 300\text{ m}\)
Ravita's distance = Perimeter of rectangle = \(2 \times (60 + 45) = 2 \times 105 = 210\text{ m}\)
Since \(210 < 300\), Ravita covers less distance.
In simple words: Calculate both perimeters; Ravita has the smaller total distance.

Exam Tip: Compare perimeters to find who covers less distance.

Question 65. A Rectangle and a square are equal in area. The side of the square is 24m. Find the width of the rectangle if it is 36m long.
Answer: Side of square = 24 m
Area of square = \(24 \times 24 = 576\text{ m}^2\)
Area of rectangle = Area of square = \(576\text{ m}^2\)
Length of rectangle = 36 m
Width of rectangle = \(\frac{576}{36} = 16\text{ m}\)
In simple words: Find square area, equate it to rectangle area, and divide by length.

Exam Tip: Area equivalence allows you to set rectangle length \(\times\) breadth equal to square side squared.

Question 66. A rectangular piece of land measures 0.7 km by 0.5km. Each side is to be fenced with 4 rows of wire. What length of the wire is needed? If 1m of wire costs Re 1, Find the total cost of wire.
Answer: Length = 0.7 km = 700 m
Breadth = 0.5 km = 500 m
Perimeter of land = \(2 \times (700 + 500) = 2 \times 1200 = 2400\text{ m}\)
Wire needed for 1 row = 2400 m
Wire needed for 4 rows = \(4 \times 2400 = 9600\text{ m}\)
Cost of 1 m wire = Re 1
Total cost = \(9600 \times 1 = \text{Rs. } 9600\)
In simple words: Find perimeter in meters, multiply by 4 for 4 rows of wire, and multiply by Re 1 for total cost.

Exam Tip: Convert km to meters (\(1\text{ km} = 1000\text{ m}\)) first.

Question 67. The length and breath of three rectangles pre as given below
a) 9m and 6m
b) 17m and 3m
c) 4m and 16m
which one has the largest area and which one has the smallest?
Answer: Area of rectangle (a) = \(9 \times 6 = 54\text{ m}^2\)
Area of rectangle (b) = \(17 \times 3 = 51\text{ m}^2\)
Area of rectangle (c) = \(4 \times 16 = 64\text{ m}^2\)
Comparing areas: \(64 > 54 > 51\)
Rectangle (c) has the largest area (\(64\text{ m}^2\)) and Rectangle (b) has the smallest area (\(51\text{ m}^2\)).
In simple words: Calculate the area for all three rectangles and compare the results.

Exam Tip: Area = Length \(\times\) Breadth for each.

Question 68. the cost of laying grass in a park at Rs 5 per square is Rs 2400. Find the cost making its boundary at Rs 2.25 per meter if it is 15m wide.
Answer: Total cost of laying grass = Rs 2400
Rate = Rs 5 per sq. m
Area of park = \(\frac{2400}{5} = 480\text{ m}^2\)
Breadth (Width) = 15 m
Length = \(\frac{\text{Area}}{\text{Breadth}} = \frac{480}{15} = 32\text{ m}\)
Perimeter = \(2 \times (\text{Length} + \text{Breadth}) = 2 \times (32 + 15) = 2 \times 47 = 94\text{ m}\)
Cost of making boundary = \(94 \times 2.25 = \text{Rs. } 211.50\)
In simple words: Find area from grass cost, find length from width, calculate perimeter, then multiply perimeter by boundary rate.

Exam Tip: Multi-step problem: Area \(\rightarrow\) Length \(\rightarrow\) Perimeter \(\rightarrow\) Boundary Cost.

Question 69. How many titles whose length and breadth are 12cm and 5cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively 100cm 144cm
Answer: Area of rectangular region = \(144\text{ cm} \times 100\text{ cm} = 14400\text{ cm}^2\)
Area of one tile = \(12\text{ cm} \times 5\text{ cm} = 60\text{ cm}^2\)
Number of tiles = \(\frac{\text{Area of region}}{\text{Area of one tile}} = \frac{14400}{60} = 240\)
In simple words: Divide the total region area by the area of a single tile.