CBSE Class 8 Mathematics Mensuration Assignment Set 01

 Mensuration

Mensuration Assignment

Question 1. Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm.
Answer: Length (\(l\)) = 8 cm
Breadth (\(b\)) = 6 cm
Height (\(h\)) = 3.5 cm
Volume of a cuboid = \(l \times b \times h\)
Volume = \(8 \times 6 \times 3.5 = 168 \text{ cm}^3\)
In simple words: The volume of a cuboid is found by multiplying its length, breadth, and height together.

Exam Tip: Always include proper cubic units (\(\text{cm}^3\)) for volume calculations.

Question 2. Find the altitude of a trapezium, the sum of the lengths of whose bases is 6.5 cm and whose area is 26 cm\(^2\).
Answer: Area of trapezium = 26 cm\(^2\)
Sum of bases (\(a + b\)) = 6.5 cm
Area = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(26 = \frac{1}{2} \times 6.5 \times h\)
\(26 = 3.25 \times h\)
\(h = \frac{26}{3.25} = 8 \text{ cm}\)
In simple words: Divide twice the area by the sum of the bases to get the altitude.

Exam Tip: Rearrange the formula \(\text{Area} = \frac{1}{2}(a+b)h\) to isolate \(h\) before substituting values to avoid calculation errors.

Question 3. Find the height of a cuboid whose volume is 275 cm\(^3\) and base area is 25 cm\(^2\).
Answer: Volume of cuboid = 275 cm\(^3\)
Base area = 25 cm\(^2\)
Volume = \(\text{Base Area} \times \text{Height}\)
\(275 = 25 \times h\)
\(h = \frac{275}{25} = 11 \text{ cm}\)
In simple words: The height of a cuboid is the total volume divided by the area of its base.

Exam Tip: Base area is equal to length multiplied by breadth (\(l \times b\)).

Question 4. Find the area of a rhombus whose diagonals are of measurements 6 cm and 8 cm.
Answer: Diagonal 1 (\(d_1\)) = 6 cm
Diagonal 2 (\(d_2\)) = 8 cm
Area of rhombus = \(\frac{1}{2} \times d_1 \times d_2\)
Area = \(\frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2\)
In simple words: Multiply both diagonals together and divide by 2 to find the area of a rhombus.

Exam Tip: This formula only works when the lengths of both diagonals are known.

Question 5. Find the volume of the cylinder whose base diameter is 14 cm and height is 10 cm.
Answer: Diameter (\(d\)) = 14 cm, so Radius (\(r\)) = \(\frac{14}{2} = 7\) cm
Height (\(h\)) = 10 cm
Volume of cylinder = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times (7)^2 \times 10\)
Volume = \(\frac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 \text{ cm}^3\)
In simple words: First find the radius by halving the diameter, then apply the cylinder volume formula.

Exam Tip: Always check whether diameter or radius is given in the question.

Question 6. Find the area of a triangle whose base is 4 cm and altitude is 6 cm.
Answer: Base (\(b\)) = 4 cm
Altitude (\(h\)) = 6 cm
Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{altitude}\)
Area = \(\frac{1}{2} \times 4 \times 6 = 12 \text{ cm}^2\)
In simple words: Multiply base and altitude, then divide by 2.

Exam Tip: Altitude means the perpendicular height from the base to the opposite vertex.

Question 7. Find the total surface area of a cube whose volume is 343 cm\(^3\).
Answer: Volume of cube = 343 cm\(^3\)
Let the side of the cube be \(a\).
\(a^3 = 343\)
\(a = \sqrt[3]{343} = 7 \text{ cm}\)
Total surface area of cube = \(6a^2\)
Total surface area = \(6 \times (7)^2 = 6 \times 49 = 294 \text{ cm}^2\)
In simple words: Find the side length by taking the cube root of the volume, then use the surface area formula.

Exam Tip: Memorize cubes of numbers from 1 to 10 for quick calculation.

Question 8. Find the side of a cube whose surface area is 2400 cm\(^2\).
Answer: Total surface area of cube = 2400 cm\(^2\)
\(6a^2 = 2400\)
\(a^2 = \frac{2400}{6} = 400\)
\(a = \sqrt{400} = 20 \text{ cm}\)
In simple words: Divide the surface area by 6, then take the square root to find the side length.

Exam Tip: Surface area of a cube consists of 6 identical square faces.

Question 9. How many bricks will be required for a wall which is 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm \(\times\) 11.25 cm \(\times\) 6 cm?
Answer: Dimensions of the wall:
Length (\(L\)) = 8 m = 800 cm
Height (\(H\)) = 6 m = 600 cm
Thickness (\(B\)) = 22.5 cm
Volume of the wall = \(800 \times 600 \times 22.5 \text{ cm}^3\)
Dimensions of one brick:
Length = 25 cm, Breadth = 11.25 cm, Height = 6 cm
Volume of one brick = \(25 \times 11.25 \times 6 \text{ cm}^3\)
Number of bricks = \(\frac{\text{Volume of the wall}}{\text{Volume of one brick}}\)
Number of bricks = \(\frac{800 \times 600 \times 22.5}{25 \times 11.25 \times 6}\)
Number of bricks = \(\frac{10800000}{1687.5} = 6400\)
In simple words: Convert all wall dimensions to centimeters, find the volume of the wall and one brick, then divide them.

Exam Tip: Always convert all dimensions to the same unit (meters to centimeters) before calculating volumes.

Question 10. The diameter of garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions?
Answer: Diameter of roller (\(d\)) = 1.4 m, so Radius (\(r\)) = \(\frac{1.4}{2} = 0.7\) m
Length (Height of cylinder, \(h\)) = 2 m
Area covered in 1 revolution = Curved surface area of cylinder = \(2\pi rh\)
Area in 1 revolution = \(2 \times \frac{22}{7} \times 0.7 \times 2\)
Area in 1 revolution = \(2 \times \frac{22}{7} \times \frac{7}{10} \times 2 = 8.8 \text{ m}^2\)
Area covered in 5 revolutions = \(5 \times 8.8 = 44 \text{ m}^2\)
In simple words: One revolution covers an area equal to the curved surface area of the roller. Multiply by 5 for five revolutions.

Exam Tip: A garden roller flattens ground using its curved surface area.

Question 11. Find the volume of a cuboid whose length is 8 cm, width is 3 cm and height is 5 cm.
Answer: Length (\(l\)) = 8 cm
Width (\(b\)) = 3 cm
Height (\(h\)) = 5 cm
Volume = \(l \times b \times h\)
Volume = \(8 \times 3 \times 5 = 120 \text{ cm}^3\)
In simple words: Multiply length, width, and height to get the volume.

Exam Tip: Width and breadth mean the same thing in geometry problems.

Question 12. A cylindrical tank has a capacity of 5632 m\(^3\). If the diameter of its base is 16 m, find its depth.
Answer: Volume of cylinder = 5632 m\(^3\)
Diameter (\(d\)) = 16 m, so Radius (\(r\)) = \(\frac{16}{2} = 8\) m
Volume = \(\pi r^2 h\)
\(5632 = \frac{22}{7} \times (8)^2 \times h\)
\(5632 = \frac{22}{7} \times 64 \times h\)
\(h = \frac{5632 \times 7}{22 \times 64}\)
\(h = \frac{39424}{1408} = 28 \text{ m}\)
In simple words: Use the cylinder volume formula with the given volume and radius to solve for height (depth).

Exam Tip: Depth of a tank is treated mathematically as the height of the cylinder.

Question 13. Find the volume of 64 cubes whose one side is 4 cm.
Answer: Side of one cube (\(a\)) = 4 cm
Volume of one cube = \(a^3 = 4^3 = 64 \text{ cm}^3\)
Volume of 64 cubes = \(64 \times 64 = 4096 \text{ cm}^3\)
In simple words: Find the volume of one cube and multiply it by 64.

Exam Tip: The volume formula for a cube is side cubed (\(a^3\)).

Question 14. Find the volume of a cylinder whose base radius is 14 cm and height is 35 cm.
Answer: Radius (\(r\)) = 14 cm
Height (\(h\)) = 35 cm
Volume = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times (14)^2 \times 35\)
Volume = \(\frac{22}{7} \times 196 \times 35 = 22 \times 28 \times 35 = 21560 \text{ cm}^3\)
In simple words: Substitute radius and height directly into the cylinder volume formula.

Exam Tip: Simplify fractions early by canceling out terms with 7 in the denominator.

Question 15. Find the area of a parallelogram whose measurements are given in the following figure.
Answer: Base = 5 cm
Corresponding height (altitude) = 3 cm
Area of parallelogram = \(\text{Base} \times \text{Height}\)
Area = \(5 \times 3 = 15 \text{ cm}^2\)
In simple words: Multiply the base of the parallelogram by its perpendicular height.

Exam Tip: Ignore side lengths that are not perpendicular to the base when calculating area.

Question 16. Find the total surface area of a cylinder whose base radius is 8 cm and height is 14 cm.
Answer: Radius (\(r\)) = 8 cm
Height (\(h\)) = 14 cm
Total surface area of cylinder = \(2\pi r(r + h)\)
Total surface area = \(2 \times \frac{22}{7} \times 8 \times (8 + 14)\)
Total surface area = \(2 \times \frac{22}{7} \times 8 \times 22\)
Total surface area = \(\frac{7744}{7} = 1106.28 \text{ cm}^2\)
In simple words: Use the total surface area formula which includes both circular bases and the curved wall.

Exam Tip: Total surface area = Curved Surface Area + Area of 2 circular ends.

Question 17. Find the area of a rhombus whose diagonals are of lengths 20 cm and 16 cm.
Answer: Diagonal 1 (\(d_1\)) = 20 cm
Diagonal 2 (\(d_2\)) = 16 cm
Area = \(\frac{1}{2} \times d_1 \times d_2\)
Area = \(\frac{1}{2} \times 20 \times 16 = 10 \times 16 = 160 \text{ cm}^2\)
In simple words: Multiply both diagonal lengths and divide the result by 2.

Exam Tip: Diagonals of a rhombus intersect each other at right angles.

Question 18. Find the height of cuboid whose volume is 490 cm\(^3\) and base area is 35 cm\(^2\).
Answer: Volume = 490 cm\(^3\)
Base area = 35 cm\(^2\)
Height = \(\frac{\text{Volume}}{\text{Base Area}}\)
Height = \(\frac{490}{35} = 14 \text{ cm}\)
In simple words: Divide the volume by the base area to find the height.

Exam Tip: Base area represents \(l \times b\).

Question 19. Find the volume of the cylinder whose height is 7 cm and radius is 20 cm.
Answer: Height (\(h\)) = 7 cm
Radius (\(r\)) = 20 cm
Volume = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times (20)^2 \times 7\)
Volume = \(\frac{22}{7} \times 400 \times 7 = 22 \times 400 = 8800 \text{ cm}^3\)
In simple words: Substitute radius and height into the cylinder volume formula and simplify.

Exam Tip: Notice that the 7 in the denominator cancels out with the height 7.

Question 20. Find the side of a cube whose surface area is 2400 cm\(^2\).
Answer: Surface area of cube = \(6a^2 = 2400\)
\(a^2 = \frac{2400}{6} = 400\)
\(a = \sqrt{400} = 20 \text{ cm}\)
In simple words: Divide the total surface area by 6 faces, then take the square root.

Exam Tip: Each face of a cube is a square.

Question 21. The diagonal of a quadrilateral shaped field is 24 cm and perpendicular dropped on it from the remaining opposite vertices are 6 m and 12 m. Find the area of the field.
Answer: Diagonal (\(d\)) = 24 cm (Note: standard units in this problem combine cm and m; treating consistent units as 24 m or following standard text)
Perpendiculars (\(h_1\) and \(h_2\)) = 6 m and 12 m
Area of quadrilateral = \(\frac{1}{2} \times d \times (h_1 + h_2)\)
Area = \(\frac{1}{2} \times 24 \times (6 + 12)\)
Area = \(12 \times 18 = 216 \text{ m}^2\)
In simple words: A quadrilateral divided into two triangles by a diagonal has an area equal to half the diagonal multiplied by the sum of the two perpendicular offsets.

Exam Tip: This formula is extremely useful for calculating the area of irregular four-sided fields.

Question 22. A godown is in the form of a cuboid of measures 60 m \(\times\) 40 m \(\times\) 20 m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 m\(^3\)?
Answer: Volume of godown = \(60 \times 40 \times 20 = 48000 \text{ m}^3\)
Volume of one box = 0.8 m\(^3\)
Number of boxes = \(\frac{\text{Volume of godown}}{\text{Volume of one box}}\)
Number of boxes = \(\frac{48000}{0.8} = 60000\)
In simple words: Divide the total storage volume by the volume of a single box.

Exam Tip: Ensure all dimensions are in the same unit before finding volume.

Question 23. The internal measures of a cuboidal room are 10 m \(\times\) 8 m \(\times\) 4 m. Find the total cost of whitewashing four walls of a room, if the cost of whitewashing is Rs 5 per m\(^2\).
Answer: Length (\(l\)) = 10 m, Breadth (\(b\)) = 8 m, Height (\(h\)) = 4 m
Area of four walls (Lateral Surface Area) = \(2(l + b)h\)
Area = \(2(10 + 8) \times 4\)
Area = \(2 \times 18 \times 4 = 144 \text{ m}^2\)
Cost of whitewashing = \(\text{Area} \times \text{Rate}\)
Cost = \(144 \times 5 = \text{Rs. } 720\)
In simple words: Find the area of the four walls using the perimeter of the base multiplied by height, then multiply by the cost per square meter.

Exam Tip: Unless specified otherwise, whitewashing of a room includes only the four walls and sometimes the ceiling.

Question 24. Find the area of a rhombus whose side is 5 cm and its altitude is 4 cm. If one of its diagonal is 8 cm long, find the length of the other diagonal.
Answer: A rhombus is also a parallelogram.
Base (side) = 5 cm, Altitude = 4 cm
Area of rhombus = \(\text{Base} \times \text{Altitude} = 5 \times 4 = 20 \text{ cm}^2\)
Now, using the diagonal formula for area:
Area = \(\frac{1}{2} \times d_1 \times d_2\)
\(20 = \frac{1}{2} \times 8 \times d_2\)
\(20 = 4 \times d_2\)
\(d_2 = \frac{20}{4} = 5 \text{ cm}\)
In simple words: Calculate the area using base and height first, then use the diagonal formula to find the missing diagonal.

Exam Tip: Rhombuses possess dual area formulas: base \(\times\) height and half the product of diagonals.

Question 25. In a building there are 4 cylindrical pillars. The radius of each pillar is 21 cm and height is 5 m. Find the curved surface area of four pillars.
Answer: Radius (\(r\)) = 21 cm = 0.21 m
Height (\(h\)) = 5 m
Curved surface area of one pillar = \(2\pi rh\)
Curved surface area of one pillar = \(2 \times \frac{22}{7} \times 0.21 \times 5\)
Curved surface area of one pillar = \(2 \times \frac{22}{7} \times \frac{21}{100} \times 5 = 6.6 \text{ m}^2\)
Curved surface area of 4 pillars = \(4 \times 6.6 = 26.4 \text{ m}^2\)
In simple words: Find the curved surface area of one pillar in meters, then multiply by 4.

Exam Tip: Convert radius from centimeters to meters so units match the height before calculating.

Question 26. A rectangular paper of width 7 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder.
Answer: When a rectangular paper is rolled along its width, the width becomes the height of the cylinder, and the length forms the circumference of the base.
Height (\(h\)) = Width = 7 cm
Radius (\(r\)) = 20 cm
Volume of cylinder = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times (20)^2 \times 7\)
Volume = \(\frac{22}{7} \times 400 \times 7 = 8800 \text{ cm}^3\)
In simple words: The width of the paper becomes the height of the cylinder. Use the given radius to find the volume.

Exam Tip: Pay close attention to whether paper is rolled along its length or width.

Question 27. The perimeter of a trapezium is 52 cm. Its non-parallel sides are 10 cm each and the distance between two parallel sides is 8 cm. Find the area of the trapezium.
Answer: Perimeter = \(\text{sum of all sides} = 52 \text{ cm}\)
Sum of non-parallel sides = \(10 + 10 = 20 \text{ cm}\)
Sum of parallel sides (\(a + b\)) = Perimeter - Sum of non-parallel sides
\(a + b = 52 - 20 = 32 \text{ cm}\)
Distance between parallel sides (\(h\)) = 8 cm
Area of trapezium = \(\frac{1}{2} \times (a + b) \times h\)
Area = \(\frac{1}{2} \times 32 \times 8 = 16 \times 8 = 128 \text{ cm}^2\)
In simple words: Subtract the non-parallel sides from the perimeter to get the sum of the parallel bases, then apply the area formula.

Exam Tip: Perimeter includes all four outer boundaries of the trapezium.

Question 28. The parallel sides of a trapezium are 25 cm and 13 cm. Its non-parallel sides are equal, each being 10 cm. Find the area of the trapezium.
Answer: Parallel sides \(a = 25\) cm, \(b = 13\) cm
Non-parallel sides = 10 cm each
To find the height (\(h\)), drop perpendiculars from the ends of the shorter base to the longer base, dividing the base into segments. The middle segment equals 13 cm, and the remaining parts equal \(\frac{25 - 13}{2} = 6\) cm each.
Using Pythagorean theorem on the right triangle: \(6^2 + h^2 = 10^2\)
\(36 + h^2 = 100\)
\(h^2 = 64 \implies h = 8 \text{ cm}\)
Area = \(\frac{1}{2} \times (25 + 13) \times 8\)
Area = \(\frac{1}{2} \times 38 \times 8 = 38 \times 4 = 152 \text{ cm}^2\)
In simple words: Find the height using the Pythagorean theorem by splitting the trapezium into a rectangle and right triangles.

Exam Tip: Constructing perpendiculars is the standard technique for finding unknown heights in isosceles trapeziums.

Question 29. The area of a trapezium is 384 cm\(^2\). Its parallel sides are in the ratio 3:5 and the distance between them is 12 cm. Find the length of each parallel side.
Answer: Let the parallel sides be \(3x\) and \(5x\).
Distance between them (\(h\)) = 12 cm
Area = \(\frac{1}{2} \times (3x + 5x) \times 12 = 384\)
\(\frac{1}{2} \times 8x \times 12 = 384\)
\(48x = 384\)
\(x = \frac{384}{48} = 8\)
First parallel side = \(3 \times 8 = 24 \text{ cm}\)
Second parallel side = \(5 \times 8 = 40 \text{ cm}\)
In simple words: Use variables with the given ratio, set up the area equation, solve for \(x\), and multiply back.

Exam Tip: Whenever sides are given in a ratio, introduce a common variable \(x\).

Question 30. A cylindrical tube, open at both ends is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal in the cylinder.
Answer: Internal diameter (\(d\) ) = 10.4 cm, so Internal radius (\(r\) ) = 5.2 cm
Thickness = 8 mm = 0.8 cm
External radius (\(R\)) = \(r + \text{thickness} = 5.2 + 0.8 = 6.0 \text{ cm}\)
Length (\(h\)) = 25 cm
Volume of metal = Volume of external cylinder - Volume of internal cylinder
Volume = \(\pi R^2 h - \pi r^2 h = \pi h (R^2 - r^2)\)
Volume = \(\frac{22}{7} \times 25 \times (6.0^2 - 5.2^2)\)
Volume = \(\frac{22}{7} \times 25 \times (36 - 27.04)\)
Volume = \(\frac{22}{7} \times 25 \times 8.96 = \frac{4928}{7} = 704 \text{ cm}^3\)
In simple words: Subtract the inner hollow volume from the outer total volume to find the volume of the metal.

Exam Tip: Convert 8 mm to 0.8 cm first so all measurements are in centimeters.

Question 31. The top surface of a box is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Answer: From the figure, a regular octagon of side 5 cm and parallel width 11 cm with height 4 cm can be split into two identical trapeziums and a central rectangle.
Trapezium parallel sides: 11 cm and marked base; height = 4 cm.
Central rectangle dimensions: 11 cm by 5 cm.
Area of central rectangle = \(11 \times 5 = 55 \text{ cm}^2\)
Area of one trapezium = \(\frac{1}{2} \times (11 + \text{side}) \times \text{height}\)
Total area = Area of two trapeziums + Area of rectangle = \(2 \times \left[\frac{1}{2} \times (11 + 5) \times 4\right] + 55\)
Total area = \((16 \times 4) + 55 = 64 + 55 = 119 \text{ cm}^2\)
In simple words: Divide the regular octagon into two trapeziums on the top and bottom and a rectangle in the middle, then add their areas.

Exam Tip: Breaking complex polygons into standard shapes (rectangles and trapeziums) makes area calculation easy.

Question 32. A pool is 20 m long, 15 m broad and 4 m deep. Find the cost of cementing its floor and its walls at the rate of Rs. 12 per square metre.
Answer: Length (\(l\)) = 20 m, Breadth (\(b\)) = 15 m, Depth (\(h\)) = 4 m
Area of floor = \(l \times b = 20 \times 15 = 300 \text{ m}^2\)
Area of four walls = \(2(l + b)h = 2(20 + 15) \times 4 = 2 \times 35 \times 4 = 280 \text{ m}^2\)
Total area to be cemented = Area of floor + Area of walls = \(300 + 280 = 580 \text{ m}^2\)
Cost of cementing = \(580 \times 12 = \text{Rs. } 6960\)
In simple words: Find the area of the floor and the four walls (leaving the top open), add them together, and multiply by the rate.

Exam Tip: A swimming pool has no ceiling, so only the floor and four walls are cemented.

Question 33. A cylindrical container of radius 28 cm contains sufficient water to submerge a rectangular solid of dimensions 32 cm \(\times\) 22 cm \(\times\) 14 cm. Find the rise in the level of water, when the solid is completely submerged.
Answer: Volume of rectangular solid = \(32 \times 22 \times 14 = 9856 \text{ cm}^3\)
Radius of cylindrical container (\(r\)) = 28 cm
Let the rise in water level be \(h\).
Volume of water displaced = Volume of cylinder of radius 28 cm and height \(h\)
\(\pi r^2 h = 9856\)
\(\frac{22}{7} \times (28)^2 \times h = 9856\)
\(\frac{22}{7} \times 784 \times h = 9856\)
\(2464 \times h = 9856\)
\(h = \frac{9856}{2464} = 4 \text{ cm}\)
In simple words: The volume of the submerged solid equals the volume of the raised water in the cylinder. Equate both volumes to find the height.

Exam Tip: Archimedes' principle states that the volume of an immersed object equals the volume of displaced liquid.

Question 34. A rectangular piece of iron sheet is 44 m long and 20 m broad. It is rolled along its length to form a cylinder. Find the volume of the cylinder so formed.
Answer: Rolled along its length means:
Circumference of base (\(2\pi r\)) = Length of sheet = 44 m
Height of cylinder (\(h\)) = Breadth of sheet = 20 m
\(2 \times \frac{22}{7} \times r = 44\)
\(\frac{44}{7} \times r = 44 \implies r = 7 \text{ m}\)
Volume of cylinder = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times (7)^2 \times 20\)
Volume = \(\frac{22}{7} \times 49 \times 20 = 22 \times 7 \times 20 = 3080 \text{ m}^3\)
In simple words: The length forms the circumference to find the radius, and the breadth becomes the height. Then calculate volume.

Exam Tip: Rolling along length versus rolling along breadth swaps which dimension becomes the radius versus height.

Question 35. The cost of papering the wall of a room, 12 m long, at the rate of Rs. 1.35 per square meter is Rs. 340.20. The cost of matting the floor at Re. 0.85 per square metre is Rs. 91.80. Find the height of the room.
Answer: Length of room (\(l\)) = 12 m
Area of floor = \(\frac{\text{Total cost of matting}}{\text{Rate}} = \frac{91.80}{0.85} = 108 \text{ m}^2\)
Since Area of floor = \(l \times b\), we have \(12 \times b = 108 \implies b = 9\) m.
Area of four walls = \(\frac{\text{Total cost of papering}}{\text{Rate}} = \frac{340.20}{1.35} = 252 \text{ m}^2\)
Area of four walls = \(2(l + b)h\)
\(2(12 + 9) \times h = 252\)
\(2(21) \times h = 252\)
\(42h = 252 \implies h = \frac{252}{42} = 6 \text{ m}\)
In simple words: Find floor area and breadth first, then use the wall papering cost to find the wall area and calculate room height.

Exam Tip: Divide total cost by rate per square unit to find the underlying surface area.

Question 36. A tin is in a cylindrical shape whose base has a diameter of 14 cm and height 20 cm. A label is placed around the surface of the container. If the label is placed 2 cm from top and bottom, what is the area of the label?
Answer: Diameter (\(d\)) = 14 cm, so Radius (\(r\)) = 7 cm
Total height of tin = 20 cm
The label leaves 2 cm from the top and 2 cm from the bottom.
Height of the label (\(h'\)) = Total height - 2(2 cm) = \(20 - 4 = 16 \text{ cm}\)
Area of the label = Curved surface area with height 16 cm = \(2\pi r h'\)
Area = \(2 \times \frac{22}{7} \times 7 \times 16\)
Area = \(2 \times 22 \times 16 = 704 \text{ cm}^2\)
In simple words: Subtract the top and bottom margins from the total height to find the height of the label, then calculate its curved surface area.

Exam Tip: Remember to subtract margins from both the top and the bottom (multiply margin by 2).

Question 37. A rectangle piece of metal sheet 11 m \(\times\) 4 m is folded without overlapping to make a cylinder of height 4 m. Find the volume of the cylinder.
Answer: Height of cylinder (\(h\)) = 4 m
The other dimension (11 m) forms the circumference of the base.
\(2\pi r = 11\)
\(2 \times \frac{22}{7} \times r = 11\)
\(\frac{44}{7} \times r = 11 \implies r = \frac{11 \times 7}{44} = \frac{7}{4} = 1.75 \text{ m}\)
Volume of cylinder = \(\pi r^2 h\)
Volume = \(\frac{22}{7} \times \left(\frac{7}{4}\right)^2 \times 4\)
Volume = \(\frac{22}{7} \times \frac{49}{16} \times 4 = \frac{22 \times 49 \times 4}{7 \times 16} = \frac{4312}{112} = 38.5 \text{ m}^3\)
In simple words: Use the 11 m side as the circumference to find the radius, and 4 m as the height, then compute the volume.

Exam Tip: Keeping fractions unsimplified until the final step helps prevent rounding errors.

Question. Three cubes of metal whose edges are in the ratio 3 : 4 : 5 are melted down into a single cube whose diagonal is 12 3 cm. Find the edge of smallest of the three given cubes.
(a) 6 cm
(b) 7 cm
(c) 8 cm
(d) 9 cm

Answer: A

Question. The area of two circular fields is in the ratio 16 : 49. If the radius of the latter is 14 m, what is the radius of the former?
(a) 32 m
(b) 18 m
(c) 8 m
(d) 4 m

Answer: A

Question. The radii of the two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. What is the ratio of their volumes?
(a) 20 : 27
(b) 18 : 27
(c) 15 : 21
(d) 11 : 15

Answer: D

Question. The volume of the metallic cylindrical pipe is 748 cm³ . Its length is 14 cm and its external radius is 9 cm. Find its thickness.
(a) 4
(b) 3
(c) 2
(d) 1

Answer: D

Question. Find the number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder whose height is 8 cm and diameter 6 cm.
(a) 620
(b) 640
(c) 660
(d) 680

Answer: B

Question. The inside circumference of a circular field is 1188 m. A road 7 m wide is constructed on the outside. Find its area.
(a) 8070 m²
(b) 8270 m²
(c) 8370 m²
(d) 8470 m²

Answer: D

Question. A piece of metal in the form of a cone of radius 3 cm and height 7 cm melted and cast into a cube. Find the side of cube (nearest to a whole number)
(a) 5 cm
(b) 4' cm
(c) 3 cm
(d) 2 cm

Answer: B

Question. A room is 15 metres long, 4 metres broad and 3 metres high. Find the cost of white washing its four walls at 50 P. per m² . '
(a) Rs. 60
(b) Rs. 57
(c) Rs.55
(d) Rs. 52

Answer: B

Question. How much paint is required for painting the outer surface of the water tank 2 m × 4 m × 3 m if it needs 1/6 of a litre for every m2?
(a) 16 litres
(b) 15 litres
(c) 12 litres
(d) 10 litres

Answer: C

Question. Three solid spheres of gold whose radii are 1 cm, 6 cm and 8 cm respectively are melted into a single solid sphere. Find the radius of the sphere.
(a) 6 cm
(b) 7 cm
(c) 8 cm
(d) 9 cm

Answer: D

Question. The vertical height of a conical tent is 42 dm and its diameter of the base is 54 dm. How many persons can it accommodate if each person is to be allowed 2916 dm³ of space? (π = 22/7)
(a) 11
(b) 10
(c) 9
(d) 8

Answer: A

Question. If the radius of the base of the right circular cylinder is reduced by 50%, keeping the same height, what is the ratio of the volume of the reduced cylinder to that of the original.
(a) 1 : 9
(b) 1 : 8
(c) 1 : 4
(d) 1 : 2

Answer: C

Question. It needs 50 ml paint for painting a picture 50 cm × 25 cm. How much paint is needed to paint a similar picture 100 cm × 50 cm?
(a) 100 ml
(b) 400 ml
(c) 750 ml
(d) 200 ml

Answer: D

Question. What is the length of the greatest rod that can be placed in a room whose length is 10 m, breadth 8 m and height 6 m. Given that 2 = 1.42 .
(a) 13.2
(b) 14.2
(c) 15.2
(c) 16.2

Answer: B

Question. The sides of a rectangular plot are in the ratio of 4 : 3 and its area is 1452 m² . Find the cost of fencing it at Rs 2.50 P per metre.
(a) Rs. 480
(b) Rs. 385
(c) Rs.375
(d) Rs. 365

Answer: B

Question. The area of four walls of a room is 120 m² and its length is twice the breadth, the height being 4 metres. Find the area of the floor.
(a) 50 m²
(b) 60 m²
(c) 75 m²
(d) 100 m²

Answer: A

Question. The edge of three iron cubes are 6 cm, 8 cm, 10 cm respectively. A new cube was made by melting them. Find the edge of the new cube.
(a) 8
(b) 12
(c) 14
(d) 10

Answer: B

Question. Find the cost of fencing a circular field at the rate of 50 P. per metre if its area is 13860 m² . (π = 22/7)
(a) Rs. 660
(b) Rs.540
(c) Rs. 700
(d) Rs. 800

Answer: A

Question. Two cubes, each of edge 12 cm are joined end to end. Find the surface area of resulting cuboid.
(a) 1450 cm²
(b) 1440 cm²
(c) 1420 cm²
(c) 1410 cm²

Answer: B

Question. What is the radius of a circular field whose area is equal to the sum of the areas of the three circular fields with radii 4 m, 4.5 m and 6 m respectively?
(a) 9 m
(b) 10.5 m
(c) 10 m
(d) 8.5 m

Answer: D

Question. A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, what is the ratio of their base to their heights?
(a) 3 : 4
(b) 5 : 4
(c) 6 : 7
(d) 9 : 8

Answer: A

Question. If all the sides of a triangle be increased by 200 percent what is the corresponding increase in its area?
(a) 300%
(b) 400%
(c) 600%
(d) 800%

Answer: D

Question. How many metres of carpet 50 cm wide will be required to cover the floor of a room 30 m × 20 m?
(a) 1000
(b) 600
(c) 2400
(d) 1200

Answer: D

Question. Three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted to fo a single cube. Find the surface area of the new cube.
(a) 210 cm²
(a) 213 cm²
(c) 224 cm²
(d) 216 cm²

Answer: D

Question. The radii of the two circular field is in the ratio 3 : 5. The area of the first field is what percent less than the area of the second?
(a) 50%
(b) 60%
(c) 40%
(d) 64%

Answer: D

Q 1 Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm.

Q 2 Find the altitude of a trapezium, the sum of the lengths of whose bases is 6.5 cm and whose area is 26 cm².

Q 3 Find the height of a cuboid whose volume is 275 cm³ and base area is 25 cm².

Q 4 Find the area of a rhombus whose diagonals are of measurements 6 cm and 8 cm.

Q 5 Find the volume of the cylinder whose base diameter is 14 cm and height is 10 cm.

Q 6 Find the area of a triangle whose base is 4 cm and altitude is 6 cm.

Q 7 Find the total surface area of a cube whose volume is 343 cm³.

Q 8 Find the side of a cube whose surface area is 2400 cm².

Q 9 How many bricks will be required for a wall which is 8 m long, 6m high and 22.5 cm thick, if each brick measures 25 cm X 11.25 cm X 6 cm?

Q 10 The diameter of garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions?

Q 11 Find the volume of a cuboid whose length is 8 cm, width is 3 cm and height is 5 cm.

Q 12 A cylindrical tank has a capacity of 5632 m3. If the diameter of its base is 16 m, find its depth.

Q 13 Find the volume of 64 cubes whose one side is 4 cm.

Q 14 Find the volume of a cylinder whose base radius is 14 cm and height is 35 cm.