CBSE Class 8 Mathematics Mensuration Worksheet Set 06

MENSURATION

Question 1. Find the area of a trapezium whose parallel sides are \(24\text{ cm}\) and \(20\text{ cm}\) and the distance between them is \(15\text{ cm}\).
Answer: Area of trapezium = \(\frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Distance between parallel sides}\)
Sum of parallel sides = \(24\text{ cm} + 20\text{ cm} = 44\text{ cm}\)
Distance = \(15\text{ cm}\)
Area = \(\frac{1}{2} \times 44 \times 15 = 22 \times 15 = 330\text{ cm}^2\)
In simple words: The area is calculated by taking the average of the two parallel sides and multiplying it by the height between them.

Exam Tip: Remember the formula for the area of a trapezium and ensure all dimensions are in the same units before calculating.

Question 2. Find the area of a trapezium whose parallel sides are \(38.7\text{ cm}\) and \(22.3\text{ cm}\), and the distance between them is \(16\text{ cm}\).
Answer: Sum of parallel sides = \(38.7\text{ cm} + 22.3\text{ cm} = 61.0\text{ cm}\)
Distance = \(16\text{ cm}\)
Area = \(\frac{1}{2} \times 61 \times 16 = 61 \times 8 = 488\text{ cm}^2\)
In simple words: Add the parallel lengths, multiply by the distance, and divide by two.

Exam Tip: When adding decimals, align the decimal points carefully to avoid arithmetic errors.

Question 3. The shape of the top surface of a table is trapezium. Its parallel sides are \(1\text{ m}\) and \(1.4\text{ m}\) and the perpendicular distance between them is \(0.9\text{ m}\). Find its area.
Answer: Sum of parallel sides = \(1\text{ m} + 1.4\text{ m} = 2.4\text{ m}\)
Distance = \(0.9\text{ m}\)
Area = \(\frac{1}{2} \times 2.4 \times 0.9 = 1.2 \times 0.9 = 1.08\text{ m}^2\)
In simple words: Multiply half of the sum of the parallel edges by the perpendicular height.

Exam Tip: Keep track of decimal places when multiplying numbers like \(1.2\) and \(0.9\).

Question 4. The area of a trapezium is \(1080\text{ cm}^2\). If the lengths of its parallel sides be \(55\text{ cm}\) and \(35\text{ cm}\), find the distance between them.
Answer: Area = \(1080\text{ cm}^2\)
Sum of parallel sides = \(55\text{ cm} + 35\text{ cm} = 90\text{ cm}\)
Let the distance be \(h\).
\(1080 = \frac{1}{2} \times 90 \times h\)
\(1080 = 45 \times h\)
\(h = \frac{1080}{45} = 24\text{ cm}\)
In simple words: Divide the area by half the sum of the parallel sides to find the height.

Exam Tip: Set up the formula as an algebraic equation and isolate the unknown variable \(h\).

Question 5. A field is in the form of a trapezium. Its area is \(1586\text{ m}^2\) and the distance between its parallel sides is \(26\text{ m}\). If one of the parallel sides is \(84\text{ m}\), find the other.
Answer: Area = \(1586\text{ m}^2\)
Distance \(h = 26\text{ m}\)
Let the other parallel side be \(x\).
\(1586 = \frac{1}{2} \times (84 + x) \times 26\)
\(1586 = (84 + x) \times 13\)
\(84 + x = \frac{1586}{13}\)
\(84 + x = 122\)
\(x = 122 - 84 = 38\text{ m}\)
In simple words: Work backwards from the area and given height to find the missing side length.

Exam Tip: Simplify the fraction \(\frac{1586}{13}\) step-by-step or use long division.

Question 6. The area of a trapezium is \(405\text{ cm}^2\). Its parallel sides are in the ratio \(4 : 5\) and the distance between them is \(18\text{ cm}\). Find the length of each of the parallel sides.
Answer: Let the parallel sides be \(4x\) and \(5x\).
Sum of parallel sides = \(4x + 5x = 9x\)
Distance \(h = 18\text{ cm}\)
Area = \(\frac{1}{2} \times 9x \times 18 = 405\)
\(9x \times 9 = 405\)
\(81x = 405\)
\(x = \frac{405}{81} = 5\)
First side = \(4 \times 5 = 20\text{ cm}\)
Second side = \(5 \times 5 = 25\text{ cm}\)
In simple words: Use a variable \(x\) for the ratio, substitute into the area formula, solve for \(x\), and then find the sides.

Exam Tip: Whenever side lengths are given in a ratio, always introduce a common multiplier \(x\).

Question 7. The area of a trapezium is \(180\text{ cm}^2\) and its height is \(9\text{ cm}\). If one of the parallel sides is longer than the other by \(6\text{ cm}\), find the two parallel sides.
Answer: Let the shorter parallel side be \(x\text{ cm}\).
Then the longer parallel side is \((x + 6)\text{ cm}\).
Sum of parallel sides = \(x + (x + 6) = 2x + 6\)
Height \(h = 9\text{ cm}\)
Area = \(\frac{1}{2} \times (2x + 6) \times 9 = 180\)
\((2x + 6) \times 9 = 360\)
\(2x + 6 = \frac{360}{9} = 40\)
\(2x = 40 - 6 = 34\)
\(x = 17\text{ cm}\)
Shorter side = \(17\text{ cm}\)
Longer side = \(17 + 6 = 23\text{ cm}\)
In simple words: Define one side as \(x\), express the other in terms of \(x\), and solve the area equation.

Exam Tip: Double check your solution by plugging both side lengths back into the trapezium area formula.

Question 8. In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is \(9450\text{ m}^2\) and the perpendicular distance between the two parallel sides is \(84\text{ m}\), find the length of the longer of the parallel sides.
Answer: Let the shorter side be \(x\text{ m}\).
The longer side is \(2x\text{ m}\).
Sum of parallel sides = \(x + 2x = 3x\)
Height \(h = 84\text{ m}\)
Area = \(\frac{1}{2} \times 3x \times 84 = 9450\)
\(3x \times 42 = 9450\)
\(126x = 9450\)
\(x = \frac{9450}{126} = 75\text{ m}\)
Longer side = \(2x = 2 \times 75 = 150\text{ m}\)
In simple words: Form an equation using the given area, solve for \(x\), and multiply by 2 to get the longer side.

Answer: Total perimeter (fence length) = \(130\text{ m}\)
Perimeter = \(AB + BC + CD + DA = 130\)
\(AB + 54 + 19 + 42 = 130\)
\(AB + 115 = 130\)
\(AB = 130 - 115 = 15\text{ m}\)
Here, the parallel sides are BC and AD, and AB is the perpendicular distance between them.
BC = \(54\text{ m}\), AD = \(42\text{ m}\), height \(AB = 15\text{ m}\).
Area = \(\frac{1}{2} \times (\text{BC} + \text{AD}) \times \text{AB}\)
Area = \(\frac{1}{2} \times (54 + 42) \times 15\)
Area = \(\frac{1}{2} \times 96 \times 15 = 48 \times 15 = 720\text{ m}^2\)
In simple words: Find the missing perpendicular side using the perimeter, then use the standard trapezium area formula.

Exam Tip: Total fence length equals the sum of all outer boundaries of the field.

Question 10. In the given figure, ABCD is a trapezium in which AD || BC, \(\angle ABC = 90^\circ\), AD = \(16\text{ cm}\), AC = \(41\text{ cm}\) and BC = \(40\text{ cm}\). Find the area of the trapezium. Hint: \(AB^2 = (AC^2 - BC^2)\).
Answer: In right-angled triangle ABC, using the hint:
\(AB^2 = AC^2 - BC^2\)
\(AB^2 = 41^2 - 40^2\)
\(AB^2 = 1681 - 1600 = 81\)
\(AB = \sqrt{81} = 9\text{ cm}\)
Here, AB is the perpendicular distance between the parallel sides AD and BC.
Parallel sides are AD = \(16\text{ cm}\) and BC = \(40\text{ cm}\).
Area = \(\frac{1}{2} \times (AD + BC) \times AB\)
Area = \(\frac{1}{2} \times (16 + 40) \times 9\)
Area = \(\frac{1}{2} \times 56 \times 9 = 28 \times 9 = 252\text{ cm}^2\)
In simple words: Use Pythagoras theorem on the right triangle to find the height AB, then calculate the area.

Exam Tip: Memorize common Pythagorean triples like \(9, 40, 41\) to solve right triangles quickly.

Question 11. The parallel sides of a trapezium are \(20\text{ cm}\) and \(10\text{ cm}\). Its non-parallel sides are both equal, each being \(13\text{ cm}\). Find the area of the trapezium.
Answer: Let the trapezium be ABCD with parallel sides \(AD = 10\text{ cm}\) and \(BC = 20\text{ cm}\), and non-parallel sides \(AB = CD = 13\text{ cm}\).
Draw perpendiculars AE and DF from A and D to BC.
Then AEFD is a rectangle, so \(EF = AD = 10\text{ cm}\).
By symmetry, \(BE = FC = \frac{20 - 10}{2} = 5\text{ cm}\).
In right triangle ABE, \(AB = 13\text{ cm}\) and \(BE = 5\text{ cm}\).
Height \(AE = \sqrt{AB^2 - BE^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}\).
Area = \(\frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}\)
Area = \(\frac{1}{2} \times (20 + 10) \times 12 = \frac{1}{2} \times 30 \times 12 = 15 \times 12 = 180\text{ cm}^2\)
In simple words: Drop perpendiculars to split the trapezium into a rectangle and two triangles, find the height, and compute the area.

Exam Tip: For an isosceles trapezium, the two side triangles formed by dropping perpendiculars are congruent.

Question 12. The parallel sides of a trapezium are \(25\text{ cm}\) and \(11\text{ cm}\), while its non-parallel sides are \(15\text{ cm}\) and \(13\text{ cm}\). Find the area of the trapezium.
Answer: Let parallel sides be \(a = 25\text{ cm}\) and \(b = 11\text{ cm}\), and non-parallel sides be \(c = 15\text{ cm}\) and \(d = 13\text{ cm}\).
The area of a trapezium with non-parallel sides \(c\) and \(d\) and parallel sides \(a\) and \(b\) (\(a > b\)) can be found by constructing a triangle of sides \(c\), \(d\), and \((a - b)\).
Base difference = \(25 - 11 = 14\text{ cm}\).
Let the triangle have sides \(13\text{ cm}\), \(15\text{ cm}\), and \(14\text{ cm}\).
Semi-perimeter \(s = \frac{13 + 15 + 14}{2} = 21\text{ cm}\).
Area of triangle = \(\sqrt{21(21 - 13)(21 - 15)(21 - 14)} = \sqrt{21 \times 8 \times 6 \times 7}\)
\(= \sqrt{(7 \times 3) \times (4 \times 2) \times (3 \times 2) \times 7} = \sqrt{7^2 \times 3^2 \times 4^2 \times 2^2} = 7 \times 3 \times 4 = 84\text{ cm}^2\)
Height of the triangle (which is the height of the trapezium) = \(\frac{2 \times \text{Area}}{\text{Base}} = \frac{2 \times 84}{14} = 12\text{ cm}\)
Area of trapezium = \(\frac{1}{2} \times (25 + 11) \times 12 = \frac{1}{2} \times 36 \times 12 = 18 \times 12 = 216\text{ cm}^2\)
In simple words: Shift one non-parallel side to form a triangle, find its height using Heron's formula, and then find the total trapezium area.

Exam Tip: Heron's formula \(\sqrt{s(s-a)(s-b)(s-c)}\) is extremely useful when all three sides of a triangle are known.

Question 13. A rectangular water tank is \(12\text{ m}\) long and \(8\text{ m}\) wide. If it contains a maximum of \(480000\text{ l}\) of water, what is its depth?
Answer: Volume in litres = \(480000\text{ l}\)
Since \(1\text{ m}^3 = 1000\text{ l}\),
Volume in cubic metres = \(\frac{480000}{1000} = 480\text{ m}^3\)
Length \(l = 12\text{ m}\), breadth \(b = 8\text{ m}\)
Volume = \(l \times b \times h\)
\(480 = 12 \times 8 \times h\)
\(480 = 96 \times h\)
\(h = \frac{480}{96} = 5\text{ m}\)
In simple words: Convert litres to cubic metres first, then divide by the base area to find the depth.

Exam Tip: Always remember the conversion: \(1\text{ m}^3 = 1000\text{ litres}\).

Question 14. Two cubes each of side \(5\text{ cm}\) are placed together. Find the volume of cuboid, thus obtained.
Answer: Side of each cube = \(5\text{ cm}\)
When two cubes are joined end-to-end, dimensions of the resulting cuboid are:
Length \(l = 5 + 5 = 10\text{ cm}\)
Breadth \(b = 5\text{ cm}\)
Height \(h = 5\text{ cm}\)
Volume of cuboid = \(l \times b \times h = 10 \times 5 \times 5 = 250\text{ cm}^3\)
In simple words: Joining two cubes doubles the length while keeping the width and height the same.

Exam Tip: Only the dimension along the axis of joining changes when combining identical solid blocks.

Question 15. A box is \(54\text{ cm} \times 45\text{ cm} \times 300\text{ cm}\). How many soaps can be fitted in it if each measures \(9\text{ cm} \times 5\text{ cm} \times 3\text{ cm}\)?
Answer: Volume of the large box = \(54 \times 45 \times 300\text{ cm}^3\)
Volume of one soap = \(9 \times 5 \times 3\text{ cm}^3\)
Number of soaps = \(\frac{\text{Volume of box}}{\text{Volume of one soap}}\)
Number of soaps = \(\frac{54 \times 45 \times 300}{9 \times 5 \times 3}\)
\(= \frac{54}{9} \times \frac{45}{5} \times \frac{300}{3} = 6 \times 9 \times 100 = 5400\)
In simple words: Divide the total volume of the box by the volume of a single soap to get the total count.

Exam Tip: Cancel common factors in the numerator and denominator before doing final multiplication.

Question 16. A water reservoir is \(2\text{ m}\) long, \(1.5\text{ m}\) wide and \(1.5\text{ m}\) high. How many litres of water can it hold?
Answer: Length \(l = 2\text{ m}\)
Breadth \(b = 1.5\text{ m}\)
Height \(h = 1.5\text{ m}\)
Volume = \(l \times b \times h = 2 \times 1.5 \times 1.5 = 4.5\text{ m}^3\)
Since \(1\text{ m}^3 = 1000\text{ litres}\),
Capacity in litres = \(4.5 \times 1000 = 4500\text{ litres}\)
In simple words: Multiply length, breadth, and height to get cubic metres, then multiply by 1000 to convert to litres.

Exam Tip: Double check decimal multiplication: \(2 \times 1.5 = 3\), and \(3 \times 1.5 = 4.5\).

Question 17. A cuboid is made of metal. It is \(27\text{ cm} \times 18\text{ cm} \times 12\text{ cm}\). It is melted and recast into small cubes with an edge \(3\text{ cm}\) in length. How many cubes are made?
Answer: Volume of cuboid = \(27 \times 18 \times 12\text{ cm}^3\)
Volume of one small cube = \(3 \times 3 \times 3 = 27\text{ cm}^3\)
Number of cubes = \(\frac{\text{Volume of cuboid}}{\text{Volume of one cube}}\)
Number of cubes = \(\frac{27 \times 18 \times 12}{27} = 18 \times 12 = 216\)
In simple words: Divide the volume of the large metal block by the volume of one small cube.

Exam Tip: When melting and recasting shapes, the total volume remains constant.

Question 18. The cover box of a laptop measures \(45\text{ cm}\) by \(40\text{ cm}\) by \(30\text{ cm}\). If a cardboard packet contains a maximum of \(15\) such cover boxes of laptops, find the volume of the cardboard packet.
Answer: Volume of one laptop box = \(45 \times 40 \times 30 = 54000\text{ cm}^3\)
Number of boxes in cardboard packet = \(15\)
Volume of cardboard packet = \(15 \times 54000 = 810000\text{ cm}^3\) (or \(0.81\text{ m}^3\))
In simple words: Multiply the volume of one box by the total number of boxes packed inside.

Exam Tip: Keep units consistent; answer can be expressed in cubic centimetres or converted to cubic metres.

Question 19. A rectangular block of ice measures \(42\text{ cm} \times 25\text{ cm} \times 18\text{ cm}\). Calculate its weight in kilograms if \(1\text{ cm}^3\) of ice weighs \(0.9\text{ grams}\).
Answer: Volume of ice block = \(42 \times 25 \times 18 = 18900\text{ cm}^3\)
Weight of \(1\text{ cm}^3\) = \(0.9\text{ g}\)
Total weight in grams = \(18900 \times 0.9 = 17010\text{ g}\)
Total weight in kilograms = \(\frac{17010}{1000} = 17.01\text{ kg}\)
In simple words: Find the total volume, multiply by the density (weight per cubic cm), and convert grams to kilograms by dividing by 1000.

Exam Tip: Be careful with unit conversions between grams and kilograms at the end.

Question 20. Find the volume of the cuboids whose
(i) \(l = 15\text{ cm}, b = 7\text{ cm}\) and \(h = 6\text{ cm}\)
(ii) \(l = 2.3\text{ m}, b = 30\text{ cm}\) and \(h = 7\text{ cm}\)
Answer:
(i) Volume = \(l \times b \times h = 15 \times 7 \times 6 = 105 \times 6 = 630\text{ cm}^3\)
(ii) First convert all dimensions to centimeters: \(l = 2.3\text{ m} = 230\text{ cm}\), \(b = 30\text{ cm}\), \(h = 7\text{ cm}\).
Volume = \(230 \times 30 \times 7 = 6900 \times 7 = 48300\text{ cm}^3\) (or \(0.0483\text{ m}^3\))
In simple words: Multiply length, width, and height. Always ensure all dimensions are in the same unit first.

Exam Tip: Never multiply mixed units like meters and centimeters directly without converting.

Question 21. Find the number of cubes of \(4\text{ cm}\) long that can be made from a cuboid of dimensions \(24\text{ cm} \times 20\text{ cm} \times 12\text{ cm}\).
Answer: Volume of cuboid = \(24 \times 20 \times 12\text{ cm}^3\)
Volume of one cube = \(4 \times 4 \times 4 = 64\text{ cm}^3\)
Number of cubes = \(\frac{24 \times 20 \times 12}{4 \times 4 \times 4} = \frac{24}{4} \times \frac{20}{4} \times \frac{12}{4} = 6 \times 5 \times 3 = 90\)
In simple words: Divide the cuboid's volume by the cube's volume.

Exam Tip: Simplify individual side divisions before final multiplication.

Question 22. How many \(4\text{ cm}\) long cubes can be cut from a cube whose edges are \(20\text{ cm}\) in length?
Answer: Volume of large cube = \(20 \times 20 \times 20 = 8000\text{ cm}^3\)
Volume of small cube = \(4 \times 4 \times 4 = 64\text{ cm}^3\)
Number of cubes = \(\frac{8000}{64} = 125\)
Alternatively: Number along each edge = \(\frac{20}{4} = 5\). Total cubes = \(5^3 = 125\).
In simple words: Cube the ratio of the side lengths (\((20/4)^3\)) to find how many small cubes fit inside.

Answer: Volume \(V = a^3 = 64\text{ cm}^3\)
Edge \(a = \sqrt[3]{64} = 4\text{ cm}\)
In simple words: Take the cube root of the volume to find the edge length.

Exam Tip: Memorize basic cube numbers like \(1, 8, 27, 64, 125, 216\) for quick calculations.

Question 24. The rainfall on a certain day was \(4\text{ cm}\). How many litres of water fell on \(4\text{ hectares}\) of field on that day?
Answer: Rainfall depth \(h = 4\text{ cm} = 0.04\text{ m}\)
Area = \(4\text{ hectares}\)
Since \(1\text{ hectare} = 10000\text{ m}^2\),
Total area = \(4 \times 10000 = 40000\text{ m}^2\)
Volume of water = \(\text{Area} \times \text{Height} = 40000 \times 0.04 = 1600\text{ m}^3\)
Since \(1\text{ m}^3 = 1000\text{ litres}\),
Total water in litres = \(1600 \times 1000 = 1600000\text{ litres}\) (or \(1.6 \times 10^6\text{ litres}\))
In simple words: Multiply area in square meters by rainfall depth in meters to get cubic meters, then convert to litres.

Exam Tip: Always convert hectares to square meters and centimeters to meters before multiplying.

Question 25. What will happen to volume of a cube, if its edge is doubled?
Answer: Let original edge be \(a\). Original volume \(V_1 = a^3\).
New edge = \(2a\).
New volume \(V_2 = (2a)^3 = 8a^3 = 8V_1\).
In simple words: The volume of the cube becomes 8 times the original volume.

Exam Tip: If linear dimensions scale by a factor \(k\), volume scales by \(k^3\).

Question 26. Find the volume of a cuboid whose length is double its breadth and height is half of the breadth.
Answer: Let breadth be \(b\).
Length \(l = 2b\)
Height \(h = \frac{1}{2}b\)
Volume = \(l \times b \times h = (2b) \times b \times (\frac{1}{2}b) = b^3\)
In simple words: Express length and height in terms of breadth and multiply them together.

Exam Tip: Notice how the factor of 2 and \(\frac{1}{2}\) cancel each other out in the final product.

Question 27. What will be the labour charges for digging a cuboidal pit \(6\text{ m}\) long, \(5\text{ m}\) broad and \(4\text{ m}\) deep at the rate of Rs. \(15\) per \(1000\text{ cm}^3\)?
Answer: Volume of pit = \(6 \times 5 \times 4 = 120\text{ m}^3\)
Convert cubic meters to cubic centimeters: \(1\text{ m} = 100\text{ cm}\), so \(1\text{ m}^3 = 10^6\text{ cm}^3\).
Volume in \(\text{cm}^3 = 120 \times 10^6\text{ cm}^3\)
Labour charge per \(1000\text{ cm}^3\) = Rs. \(15\)
Total charges = \(\frac{120 \times 10^6}{1000} \times 15 = 120000 \times 15 = \text{Rs. } 1800000\)
In simple words: Find volume in cubic cm, divide by 1000, and multiply by the given rate.

Exam Tip: Pay strict attention to unit conversions between cubic meters and cubic centimeters.

Question 28. A swimming pool is \(250\text{ m}\) long and \(150\text{ m}\) wide. If \(9375\text{ m}^3\) of water is pumped into it, find the height of the water level.
Answer: Length \(l = 250\text{ m}\), breadth \(b = 150\text{ m}\)
Volume = \(9375\text{ m}^3\)
Volume = \(l \times b \times h\)
\(9375 = 250 \times 150 \times h\)
\(9375 = 37500 \times h\)
\(h = \frac{9375}{37500} = 0.25\text{ m}\) (or \(25\text{ cm}\))
In simple words: Divide the volume of water by the pool's base area to find the water height.

Answer: Volume of cuboid = \(48 \times 24 \times 12\text{ cm}^3\)
Let the side of the cube be \(a\).
Volume of cube = \(a^3\)
\(a^3 = 48 \times 24 \times 12\)
\(a^3 = (24 \times 2) \times 24 \times 12 = 24 \times 24 \times (2 \times 12) = 24 \times 24 \times 24\)
\(a = 24\text{ cm}\)
In simple words: Factorize the cuboid's volume into three identical factors to find the cube's side.

Exam Tip: Prime factorization or grouping factors helps find cube roots without multiplying huge numbers.

Question 30. Find the total surface area and the lateral surface area of the cubes whose edges are:
(i) \(13\text{ cm}\)
(ii) \(7\text{ m}\)
(iii) \(2\text{ m } 25\text{ cm}\)
(iv) \(11\text{ m}\)
Answer:
(i) Edge \(a = 13\text{ cm}\)
Total Surface Area (TSA) = \(6a^2 = 6 \times (13)^2 = 6 \times 169 = 1014\text{ cm}^2\)
Lateral Surface Area (LSA) = \(4a^2 = 4 \times 169 = 676\text{ cm}^2\)
(ii) Edge \(a = 7\text{ m}\)
TSA = \(6 \times 7^2 = 6 \times 49 = 294\text{ m}^2\)
LSA = \(4 \times 7^2 = 4 \times 49 = 196\text{ m}^2\)
(iii) Edge \(a = 2.25\text{ m}\)
TSA = \(6 \times (2.25)^2 = 6 \times 5.0625 = 30.375\text{ m}^2\)
LSA = \(4 \times (2.25)^2 = 4 \times 5.0625 = 20.25\text{ m}^2\)
(iv) Edge \(a = 11\text{ m}\)
TSA = \(6 \times 11^2 = 6 \times 121 = 726\text{ m}^2\)
LSA = \(4 \times 11^2 = 4 \times 121 = 484\text{ m}^2\)
In simple words: Total surface area is \(6a^2\) and lateral surface area is \(4a^2\) for any cube.

Exam Tip: Lateral surface area excludes the top and bottom faces, leaving 4 side walls.

Question 31. The dimensions of a cuboidal box are \(2\text{ m } 50\text{ cm} \times 1\text{ m} \times 25\text{ cm} \times 75\text{ cm}\). Find
(i) the area of canvas required to cover this box;
(ii) the cost of canvas for covering the box at the rate of Rs. \(4\) per square metre.
Answer: Let dimensions be \(l = 2.50\text{ m}\), \(b = 1.25\text{ m}\), \(h = 0.75\text{ m}\).
(i) Total Surface Area (Area of canvas required) = \(2(lb + bh + hl)\)
\(= 2((2.50 \times 1.25) + (1.25 \times 0.75) + (0.75 \times 2.50))\)
\(= 2(3.125 + 0.9375 + 1.875)\)
\(= 2(5.9375) = 11.875\text{ m}^2\)
(ii) Cost of canvas = \(11.875 \times 4 = \text{Rs. } 47.50\)
In simple words: Calculate total surface area using dimensions in meters, then multiply by the cost per square meter.

Exam Tip: Convert all mixed units into meters as decimals before substituting into formulas.

Question 32. The paint in a certain container is sufficient to paint an area equal to \(9375\text{ cm}^2\). How many bricks of dimensions \(22.5\text{ cm} \times 10\text{ cm} \times 7.5\text{ cm}\) can be painted out of this container?
Answer: Total area paint can cover = \(9375\text{ cm}^2\)
Surface area of one brick = \(2(lb + bh + hl)\)
\(= 2((22.5 \times 10) + (10 \times 7.5) + (7.5 \times 22.5))\)
\(= 2(225 + 75 + 168.75)\)
\(= 2(468.75) = 937.5\text{ cm}^2\)
Number of bricks = \(\frac{\text{Total paint area}}{\text{Surface area of one brick}} = \frac{9375}{937.5} = 10\)
In simple words: Divide the total paintable area by the total surface area of a single brick.

Answer: Length \(l = 40\text{ cm}\), breadth \(b = 30\text{ cm}\), height \(h = 20\text{ cm}\)
Total Surface Area (TSA) = \(2(lb + bh + hl)\)
\(= 2((40 \times 30) + (30 \times 20) + (20 \times 40))\)
\(= 2(1200 + 600 + 800) = 2(2600) = 5200\text{ cm}^2\)
Lateral Surface Area (LSA) = \(2(l + b) \times h\)
\(= 2(40 + 30) \times 20 = 2(70) \times 20 = 2800\text{ cm}^2\)
In simple words: Use the standard formulas for total and lateral surface areas of a cuboid.

Exam Tip: Lateral surface area formula for a cuboid is \(2(l + b)h\).

Question 34. Each edge of a cube is \(18\text{ cm}\) long. Find the total surface area and the lateral surface area of the cube.
Answer: Edge \(a = 18\text{ cm}\)
Total Surface Area = \(6a^2 = 6 \times 18^2 = 6 \times 324 = 1944\text{ cm}^2\)
Lateral Surface Area = \(4a^2 = 4 \times 18^2 = 4 \times 324 = 1296\text{ cm}^2\)
In simple words: Multiply the square of the edge by 6 for total area and by 4 for lateral area.

Exam Tip: \(18^2 = 324\). Knowing squares up to 25 helps save time.

Question 35. The length, breadth and height of a cuboid are in the ratio of \(4 : 3 : 2\), and its total surface area is \(5200\text{ cm}^2\). Find the dimensions of the cuboid.
Answer: Let dimensions be \(4x, 3x,\) and \(2x\).
TSA = \(2(lb + bh + hl) = 5200\)
\(2((4x \times 3x) + (3x \times 2x) + (2x \times 4x)) = 5200\)
\(2(12x^2 + 6x^2 + 8x^2) = 5200\)
\(2(26x^2) = 5200\)
\(52x^2 = 5200\)
\(x^2 = 100\)
\(x = 10\text{ cm}\)
Length = \(4 \times 10 = 40\text{ cm}\)
Breadth = \(3 \times 10 = 30\text{ cm}\)
Height = \(2 \times 10 = 20\text{ cm}\)
In simple words: Introduce variable \(x\) for the ratio, substitute into the TSA formula, solve for \(x\), and find dimensions.

Exam Tip: Combine like terms containing \(x^2\) inside the parenthesis before solving.

Question 36. The walls and ceiling of a room are to be painted. If the length, breadth and height of the room are respectively \(5.5\text{ m}\), \(3\text{ m}\) and \(4.5\text{ m}\), find the area to be painted.
Answer: Area to be painted = Area of 4 walls + Area of ceiling
Area of 4 walls (Lateral Surface Area) = \(2(l + b)h = 2(5.5 + 3) \times 4.5 = 2(8.5) \times 4.5 = 17 \times 4.5 = 76.5\text{ m}^2\)
Area of ceiling = \(l \times b = 5.5 \times 3 = 16.5\text{ m}^2\)
Total paintable area = \(76.5 + 16.5 = 93\text{ m}^2\)
In simple words: Add the lateral wall area and the ceiling area (floor is excluded).

Answer:
(i) For \(l = 9, b = 7, h = 3\):
TSA = \(2(9\times7 + 7\times3 + 3\times9) = 2(63 + 21 + 27) = 2(111) = 222\text{ cm}^2\)
LSA = \(2(l + b)h = 2(9 + 7) \times 3 = 2(16) \times 3 = 96\text{ cm}^2\)
(ii) For \(l = 13, b = 5, h = 7\):
TSA = \(2(13\times5 + 5\times7 + 7\times13) = 2(65 + 35 + 91) = 2(191) = 382\text{ cm}^2\)
LSA = \(2(13 + 5) \times 7 = 2(18) \times 7 = 252\text{ cm}^2\)
In simple words: Apply standard formulas for TSA and LSA substituting the given dimensions.

Exam Tip: Double check your arithmetic inside the brackets before multiplying by 2.

Question 38. A swimming pool is \(18\text{ m}\) in length, \(14\text{ m}\) in breadth and \(5\text{ m}\) in depth. Find the cost of cementing its floor and walls at the rate of Rs. \(12\) per \(\text{m}^2\).
Answer: Area to be cemented = Area of 4 walls + Area of floor
Area of walls (LSA) = \(2(l + b)h = 2(18 + 14) \times 5 = 2(32) \times 5 = 320\text{ m}^2\)
Area of floor = \(l \times b = 18 \times 14 = 252\text{ m}^2\)
Total area = \(320 + 252 = 572\text{ m}^2\)
Cost of cementing = \(572 \times 12 = \text{Rs. } 6864\)
In simple words: Cementing a pool includes the four inner walls and the bottom floor, but not the open top.

Exam Tip: Open tanks and pools omit the top face when calculating surface areas.

Question 39. Three equal cubes of side \(5\text{ cm}\) are placed together. Find (i) the volume; (ii) the total surface area; and (iii) the lateral surface area of the resulting cuboid.
Answer: When 3 identical cubes of side \(5\text{ cm}\) are joined end-to-end:
Dimensions of the resulting cuboid are: length \(l = 3 \times 5 = 15\text{ cm}\), breadth \(b = 5\text{ cm}\), height \(h = 5\text{ cm}\).
(i) Volume = \(l \times b \times h = 15 \times 5 \times 5 = 375\text{ cm}^3\)
(ii) Total Surface Area = \(2(lb + bh + hl) = 2((15 \times 5) + (5 \times 5) + (5 \times 15)) = 2(75 + 25 + 75) = 2(175) = 350\text{ cm}^2\)
(iii) Lateral Surface Area = \(2(l + b)h = 2(15 + 5) \times 5 = 2(20) \times 5 = 200\text{ cm}^2\)
In simple words: Joining cubes increases only the length while width and height remain unchanged.

Exam Tip: Carefully determine the new dimensions of the combined solid before calculating areas and volumes.

Question 40. Ratio of surface areas of two cubes is \(1 : 9\). Find the ratio of their volumes.
Answer: Let edges be \(a_1\) and \(a_2\).
Ratio of surface areas = \(\frac{6a_1^2}{6a_2^2} = \frac{a_1^2}{a_2^2} = \frac{1}{9}\)
Taking square root: \(\frac{a_1}{a_2} = \frac{1}{3}\)
Ratio of volumes = \(\frac{a_1^3}{a_2^3} = \left(\frac{a_1}{a_2}\right)^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}\) (or \(1 : 27\))
In simple words: Find the ratio of side lengths first by taking the square root, then cube it to get the volume ratio.

Answer: Edge \(a = 5\text{ m}\)
Total Surface Area = \(6a^2 = 6 \times 5^2 = 6 \times 25 = 150\text{ m}^2\)
Cost of painting = \(150 \times 9.50 = \text{Rs. } 1425\)
In simple words: Calculate the total surface area of the cube and multiply by the painting rate.

Question 42. The lateral surface area of a cube is \(256\text{ cm}^2\). Find its total surface area.
Answer: Lateral Surface Area = \(4a^2 = 256\text{ cm}^2\)
\(a^2 = \frac{256}{4} = 64\text{ cm}^2\)
Total Surface Area = \(6a^2 = 6 \times 64 = 384\text{ cm}^2\)
In simple words: Find \(a^2\) from the lateral area, then multiply by 6 to get total surface area.

Question 43. The total surface area of a cube is \(294\text{ m}^2\). Find its volume.
Answer: Total Surface Area = \(6a^2 = 294\text{ m}^2\)
\(a^2 = \frac{294}{6} = 49\text{ m}^2\)
Edge \(a = \sqrt{49} = 7\text{ m}\)
Volume = \(a^3 = 7^3 = 343\text{ m}^3\)
In simple words: Divide TSA by 6 to find \(a^2\), take the square root for edge \(a\), and cube it for volume.

Question 44. The volume of a cube is \(216\text{ m}^3\). Find its total surface area.
Answer: Volume \(V = a^3 = 216\text{ m}^3\)
Edge \(a = \sqrt[3]{216} = 6\text{ m}\)
Total Surface Area = \(6a^2 = 6 \times 6^2 = 6 \times 36 = 216\text{ m}^2\)
In simple words: Find the edge length using cube root, then compute \(6a^2\).

Question 45. The floor of a rectangular hall has a perimeter of \(200\text{ m}\). If its height is \(5\text{ m}\), find the cost of painting its four walls at the rate of Rs. \(25\) per sq. \(\text{m}\).
Answer: Perimeter of floor = \(2(l + b) = 200\text{ m}\)
Height \(h = 5\text{ m}\)
Area of four walls (Lateral Surface Area) = \(2(l + b)h = \text{Perimeter} \times h = 200 \times 5 = 1000\text{ m}^2\)
Cost of painting = \(1000 \times 25 = \text{Rs. } 25000\)
In simple words: Multiply the floor perimeter by the height to get the wall area, then multiply by the rate.

Question 46. The curved surface area of a \(14\text{ m}\) high cylinder is \(352\text{ m}^2\). Find the volume of the cylinder.
Answer: Height \(h = 14\text{ m}\)
Curved Surface Area (CSA) = \(2\pi r h = 352\text{ m}^2\)
\(2 \times \frac{22}{7} \times r \times 14 = 352\)
\(44 \times 2 \times r = 352\)
\(88r = 352\)
\(r = \frac{352}{88} = 4\text{ m}\)
Volume = \(\pi r^2 h = \frac{22}{7} \times 4^2 \times 14 = \frac{22}{7} \times 16 \times 14 = 22 \times 16 \times 2 = 704\text{ m}^3\)
In simple words: Find the radius from the curved surface area formula, then use it to calculate volume.

Exam Tip: Use \(\pi = \frac{22}{7}\) unless specified otherwise to simplify calculations.

Question 47. A \(21\text{ cm}\) long cylindrical iron pipe has exterior diameter \(8\text{ cm}\). If the thickness of the pipe is \(1\text{ cm}\) and iron weighs \(8\text{ g/cm}^3\), then find the weight of the pipe.
Answer: Length (height) \(h = 21\text{ cm}\)
Exterior diameter = \(8\text{ cm}\), so exterior radius \(R = \frac{8}{2} = 4\text{ cm}\).
Thickness = \(1\text{ cm}\), so interior radius \(r = R - \text{thickness} = 4 - 1 = 3\text{ cm}\).
Volume of iron in pipe = \(\pi (R^2 - r^2)h = \frac{22}{7} \times (4^2 - 3^2) \times 21\)
\(= \frac{22}{7} \times (16 - 9) \times 21 = \frac{22}{7} \times 7 \times 21 = 22 \times 21 = 462\text{ cm}^3\)
Weight = \(\text{Volume} \times \text{Density} = 462 \times 8 = 3696\text{ g} = 3.696\text{ kg}\)
In simple words: Calculate the volume of hollow metal by subtracting inner cylinder volume from outer cylinder volume, then multiply by density.

Exam Tip: For hollow cylinders, volume is \(\pi(R^2 - r^2)h\).

Question 48. Find the volume of the cylinder which is formed by rolling a rectangular sheet of dimensions \(44\text{ cm} \times 12\text{ cm}\) along its length.
Answer: When rolled along its length (\(44\text{ cm}\)), the length becomes the circumference of the base.
Circumference = \(2\pi r = 44\text{ cm}\)
\(2 \times \frac{22}{7} \times r = 44\)
\(\frac{44}{7} \times r = 44 \Rightarrow r = 7\text{ cm}\)
The breadth (\(12\text{ cm}\)) becomes the height of the cylinder: \(h = 12\text{ cm}\).
Volume = \(\pi r^2 h = \frac{22}{7} \times 7^2 \times 12 = \frac{22}{7} \times 49 \times 12 = 22 \times 7 \times 12 = 1848\text{ cm}^3\)
In simple words: The rolling side forms the base circle perimeter, and the other side forms the height.

Exam Tip: Pay close attention to whether the sheet is rolled along its length or breadth.

Question 49. A cylindrical tank has a capacity of \(1131.90\text{ m}^3\). Find the circumference of the base of a cylinder if its height is \(15\text{ m}\).
Answer: Volume = \(\pi r^2 h = 1131.90\text{ m}^3\) (using \(\pi \approx 3.1416\))
Height \(h = 15\text{ m}\)
\(\pi r^2 \times 15 = 1131.90\)
\(\pi r^2 = \frac{1131.90}{15} = 75.46\)
Since base area = \(\pi r^2 = 75.46\text{ m}^2\),
Volume can also be written as: \((\text{Base Area}) \times h = \text{Volume}\)
Wait, circumference \(C = 2\pi r\).
We know \(\pi r^2 = 75.46 \Rightarrow r^2 = \frac{75.46}{3.14} \approx 24.03 \Rightarrow r \approx 4.9\text{ m}\).
Circumference \(C = 2\pi r = \text{Base Perimeter}\). Alternatively, using volume formula:
\(\pi r^2 h = 1131.9 \Rightarrow \pi r^2 (15) = 1131.9 \Rightarrow \pi r^2 = 75.46\).
Circumference \(C\) can be found if needed, or base area is known. Let's find \(r\): \(r = \sqrt{\frac{75.46}{3.1416}} \approx 4.9\text{ m}\).
Circumference = \(2 \times 3.1416 \times 4.9 \approx 30.79\text{ m}\) (or using \(\pi = \frac{22}{7}\), \(\frac{22}{7}r^2 = 75.46 \Rightarrow r^2 = 24.01 \Rightarrow r = 4.9\text{ cm}\), \(C = 2 \times \frac{22}{7} \times 4.9 = 30.8\text{ m}\)).
In simple words: Divide volume by height to find the base area, determine the radius, and then calculate the circumference.

Exam Tip: Base area multiplied by height equals the cylinder's volume.

Question 50. The diameter of a pipe of length \(20\text{ m}\) is \(56\text{ cm}\). Find the cost of painting the surface of the pipe at the rate of Rs. \(12\) per \(\text{m}^2\).
Answer: Length (height) \(h = 20\text{ m}\)
Diameter \(d = 56\text{ cm} = 0.56\text{ m}\), so radius \(r = 0.28\text{ m}\).
Curved Surface Area = \(2\pi r h = 2 \times \frac{22}{7} \times 0.28 \times 20 = 2 \times \frac{22}{7} \times 5.6 = 2 \times 22 \times 0.8 = 35.2\text{ m}^2\)
Cost of painting = \(35.2 \times 12 = \text{Rs. } 422.40\)
In simple words: Calculate the curved surface area in square meters and multiply by the rate per square meter.

Exam Tip: Convert centimeters to meters before calculating area when the rate is given per square meter.

Question 51. Find the amount of water in the cylindrical vessel of length \(21\text{ m}\) and radius \(3.5\text{ m}\).
Answer: Height \(h = 21\text{ m}\), radius \(r = 3.5\text{ m}\)
Volume = \(\pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 21 = \frac{22}{7} \times 12.25 \times 21 = \frac{22}{7} \times 12.25 \times 3 \times 7 = 22 \times 12.25 \times 3 = 808.5\text{ m}^3\)
Amount of water in litres = \(808.5 \times 1000 = 808500\text{ litres}\)
In simple words: Find the volume in cubic meters using the cylinder volume formula.

Exam Tip: State the volume in cubic meters and optionally convert to litres if required.

Question 52. How many cubic metres of earth must be dug out to sink a well which is \(8\text{ m}\) deep and has a diameter of \(28\text{ m}\)? If the earth taken out is spread over a rectangular plot \(22\text{ m}\) by \(16\text{ m}\), what is the height of the platform so formed?
Answer: Well dimensions: depth \(h = 8\text{ m}\), radius \(r = \frac{28}{2} = 14\text{ m}\).
Volume of earth dug out = \(\pi r^2 h = \frac{22}{7} \times 14^2 \times 8 = \frac{22}{7} \times 196 \times 8 = 22 \times 28 \times 8 = 4928\text{ m}^3\)
Rectangular plot dimensions: \(l = 22\text{ m}\), \(b = 16\text{ m}\).
Let height of platform be \(H\).
Volume of platform = \(\text{Area} \times \text{Height} = 22 \times 16 \times H = 352H\)
Equating volumes: \(352H = 4928 \Rightarrow H = \frac{4928}{352} = 14\text{ m}\).
In simple words: The volume of soil dug from the well equals the volume of the raised platform. Equate both volumes to find the height.

Answer: Diameter \(d = 84\text{ cm}\), radius \(r = 42\text{ cm} = 0.42\text{ m}\).
Length (height of roller) \(h = 120\text{ cm} = 1.2\text{ m}\).
Curved Surface Area covered in 1 revolution = \(2\pi r h = 2 \times \frac{22}{7} \times 0.42 \times 1.2 = 2 \times \frac{22}{7} \times 0.06 \times 1.2 = 3.168\text{ m}^2\)
Area covered in \(1000\) revolutions = \(3.168 \times 1000 = 3168\text{ m}^2\)
In simple words: Multiply the curved surface area of the roller by the number of revolutions it makes.

Exam Tip: Convert all dimensions to meters first so the final area is directly in square meters.

QUICK RECALL

The magnitude of the region enclosed by a plane figure is called the area of the figure.

Area of a rectangle = \((\text{length} \times \text{breadth})\).

Area of a square = \((\text{side})^2\).

Area of a parallelogram = \(\text{base} \times \text{corresponding altitude}\).

Area of a triangle = \(\frac{1}{2} \times \text{base} \times \text{corresponding altitude}\).

Area of a rhombus = \(\frac{1}{2} \times \text{product of its diagonals}\).

Area of a trapezium = \(\frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Distance between the parallel sides}\).

\(1\text{ cm}^2 = 100\text{ mm}^2 \quad 1\text{ dm}^2 = 100\text{ cm}^2\)

\(1\text{ m}^2 = 100\text{ dm}^2 \quad 1\text{ dam}^2 = 10000\text{ cm}^2 = 100\text{ m}^2\)

\(1\text{ hm}^2 = 10000\text{ m}^2 \quad 1\text{ km}^2 = 10^6\text{ m}^2\)

\(1\text{ are} = 100\text{ m}^2 \quad 1\text{ hectare} = 100\text{ ares} = 10000\text{ m}^2\)

A solid bounded by six rectangular plane faces is called a cuboid.

A cuboid of equal length, breadth and height is called a cube.

A cuboid has \(12\) edges, \(8\) vertices and \(6\) rectangular faces.

The sum of the areas of all the six faces of a cuboid is called the surface area of the cuboid.

Volume of a cuboid = \(l \times b \times h\)

Volume of a cube = \((\text{side})^3\)