CBSE Class 9 Mathematics Surface areas and Volumes Assignment Set 04

Key Concepts

Section - A
 

Question. If surface areas of two spheres are in the ratio of 4: 9 then the ratio of their volumes is
(a) \( \frac{16}{27} \)
(b) \( \frac{4}{27} \)
(c) \( \frac{8}{27} \)
(d) \( \frac{9}{27} \)
Answer: (c) \( \frac{8}{27} \)
Let the radii of the two spheres be \( r_1 \) and \( r_2 \). The surface area of a sphere is given by the formula \( 4\pi r^2 \). Based on the given problem, the ratio of their surface areas is:
\( \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{4}{9} \)
\( \implies \left(\frac{r_1}{r_2}\right)^2 = \frac{4}{9} \)
\( \implies \frac{r_1}{r_2} = \sqrt{\frac{4}{9}} = \frac{2}{3} \)
Now, we calculate the ratio of their volumes using the volume formula \( V = \frac{4}{3}\pi r^3 \):
\( \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} \)
\( \implies \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 \)
\( \implies \frac{V_1}{V_2} = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \)
Consequently, the correct option is (c).
In simple words: The surface area of a sphere depends on the square of its radius, so the ratio of their radii is the square root of the surface area ratio, which is 2:3. Since volume depends on the cube of the radius, we cube this ratio to get 8:27.

Exam Tip: Keep in mind that for any similar three-dimensional shapes, the ratio of their surface areas represents the square of the ratio of their linear sizes, whereas the ratio of their volumes represents the cube of that same linear ratio.

Question. The surface area of a cube whose edge is 11cm is
(a) \( 725\text{ cm}^2 \)
(b) \( 726\text{ cm}^2 \)
(c) \( 727\text{ cm}^2 \)
(d) \( 728\text{ cm}^2 \)
Answer: (b) \( 726\text{ cm}^2 \)
The edge length of the cube is given as \( s = 11\text{ cm} \). The total surface area of a cube is determined by the formula:
\( \text{Surface Area} = 6s^2 \)
\( \implies \text{Surface Area} = 6 \times (11)^2 \)
\( \implies \text{Surface Area} = 6 \times 121 = 726\text{ cm}^2 \)
Hence, option (b) is the correct choice.
In simple words: A cube has six identical square faces. To find its total surface area, we calculate the area of one face by squaring the side length and then multiply the result by six.

Exam Tip: Do not confuse lateral surface area (\( 4s^2 \)) with total surface area (\( 6s^2 \)) unless a specific instruction is provided in the problem statement.

Question. A match box measures 4cm X 2.5cm X 1.5cm. What will be the volume of a packet containing 12 such boxes?
(a) \( 15\text{ cm}^3 \)
(b) \( 180\text{ cm}^3 \)
(c) \( 90\text{ cm}^3 \)
(d) \( 175\text{ cm}^3 \)
Answer: (b) \( 180\text{ cm}^3 \)
The dimensions of one matchbox are:
Length, \( l = 4\text{ cm} \)
Breadth, \( b = 2.5\text{ cm} \)
Height, \( h = 1.5\text{ cm} \)
We find the volume of a single cuboidal matchbox using:
\( V = l \times b \times h \)
\( \implies V = 4 \times 2.5 \times 1.5 = 15\text{ cm}^3 \)
The total volume of a packet that holds 12 of these boxes is:
\( \text{Total Volume} = 12 \times 15 = 180\text{ cm}^3 \)
This matches option (b).
In simple words: First, find the space occupied by one matchbox by multiplying its length, width, and height. Then, multiply this single volume by 12 to find the total space of the entire packet.

Exam Tip: Always make sure that all the dimensions are in the same unit before performing multiplication to find volume or area.

Question. The curved surface area of a right circular cylinder of height 14cm is 88cm2. Find the diameter of the base of the cylinder.
(a) 1cm
(b) 2cm
(c) 3cm
(d) 4cm
Answer: (b) 2cm
Let the radius of the circular base of the cylinder be \( r \). We are given:
Height of the cylinder, \( h = 14\text{ cm} \)
Curved surface area, \( \text{CSA} = 88\text{ cm}^2 \)
Using the formula for curved surface area:
\( \text{CSA} = 2\pi rh \)
\( \implies 88 = 2 \times \frac{22}{7} \times r \times 14 \)
\( \implies 88 = 88 \times r \)
\( \implies r = 1\text{ cm} \)
The base diameter \( d \) is twice the radius:
\( d = 2r = 2 \times 1 = 2\text{ cm} \)
This corresponds to option (b).
In simple words: Use the formula for the curved area of a cylinder to find its radius first. Since the diameter is twice the radius, multiply the radius by 2 to get the final diameter.

Exam Tip: A common mistake is stopping at the radius (\( 1\text{ cm} \)) and choosing option (a). Always double-check if the question asks for the radius or the diameter.

Question. The total surface area of a cone of radius r/2 and length 2l is
(a) \( 2\pi r(l + r) \)
(b) \( \pi r(l + r) \)
(c) \( \pi r\left(l + \frac{r}{4}\right) \)
(d) \( \pi r\left(l + \frac{r}{2}\right) \)
Answer: (c) \( \pi r\left(l + \frac{r}{4}\right) \)
Let the radius of the cone be \( r' = \frac{r}{2} \) and the slant height be \( l' = 2l \). The total surface area of a cone is given by:
\( \text{TSA} = \pi r' (l' + r') \)
Substituting the modified dimensions:
\( \text{TSA} = \pi \left(\frac{r}{2}\right) \left(2l + \frac{r}{2}\right) \)
\( \implies \text{TSA} = \pi \frac{r}{2} \cdot 2\left(l + \frac{r}{4}\right) \)
\( \implies \text{TSA} = \pi r \left(l + \frac{r}{4}\right) \)
Thus, the correct option is (c).
In simple words: Plug the new radius and slant height into the standard total surface area formula, then factor out a 2 from the bracket to simplify the expression and match the options.

Exam Tip: When dealing with variables instead of numbers, write down the standard formula first and then carefully substitute the modified terms to avoid algebraic errors.

Question. The surface area of sphere of radius 10.5cm is
(a) \( 1386\text{ cm}^2 \)
(b) \( 616\text{ cm}^2 \)
(c) \( 1390\text{ cm}^2 \)
(d) \( 10\text{ cm}^2 \)
Answer: (a) \( 1386\text{ cm}^2 \)
The radius of the sphere is \( r = 10.5\text{ cm} = \frac{21}{2}\text{ cm} \). The formula for the surface area of a sphere is:
\( A = 4\pi r^2 \)
Substituting the given values:
\( A = 4 \times \frac{22}{7} \times \left(\frac{21}{2}\right)^2 \)
\( \implies A = 4 \times \frac{22}{7} \times \frac{441}{4} \)
\( \implies A = 22 \times 63 = 1386\text{ cm}^2 \)
This corresponds to option (a).
In simple words: Substitute the radius of 10.5 cm into the sphere's surface area formula. Converting the decimal 10.5 to the fraction 21/2 makes the calculation much easier to solve.

Exam Tip: Whenever the radius has a decimal like .5 or is a multiple of 7, write it as a fraction (like 21/2) to easily cancel out terms with the 7 in the denominator of \( \pi \).

Section - B
 

Question. Find the volume of a sphere whose surface area is 154cm2.
Answer: We are given that the surface area of the sphere is \( 154\text{ cm}^2 \). The surface area formula is:
\( 4\pi r^2 = 154 \)
Using \( \pi = \frac{22}{7} \):
\( 4 \times \frac{22}{7} \times r^2 = 154 \)
\( \implies r^2 = 154 \times \frac{7}{88} \)
\( \implies r^2 = \frac{49}{4} \)
\( \implies r = \frac{7}{2} = 3.5\text{ cm} \)
Now, we calculate the sphere's volume:
\( V = \frac{4}{3}\pi r^3 \)
\( \implies V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 \)
\( \implies V = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} \)
\( \implies V = \frac{539}{3} \approx 179.67\text{ cm}^3 \)
In simple words: First, find the radius using the given surface area. Once the radius is known, use it in the volume formula to calculate the total space inside the sphere.

Exam Tip: Be mindful of units. Surface area is measured in square units (\( \text{cm}^2 \)), while volume must always be expressed in cubic units (\( \text{cm}^3 \)).

Question. A solid cylinder has a total surface area of 231cm2. Its curved surface area is 2/3 of the total surface area. Find the volume of the cylinder.
Answer: We are given:
Total surface area (\( \text{TSA} \)) = \( 231\text{ cm}^2 \)
Curved surface area (\( \text{CSA} \)) = \( \frac{2}{3} \times 231 = 154\text{ cm}^2 \)
For a solid cylinder, the relationship between TSA and CSA is:
\( \text{TSA} = \text{CSA} + 2\pi r^2 \)
Substituting the values:
\( 231 = 154 + 2\pi r^2 \)
\( \implies 2\pi r^2 = 77 \)
\( \implies 2 \times \frac{22}{7} \times r^2 = 77 \)
\( \implies r^2 = \frac{49}{4} \)
\( \implies r = \frac{7}{2} = 3.5\text{ cm} \)
Next, we find the height \( h \) using the curved surface area formula:
\( \text{CSA} = 2\pi rh = 154 \)
\( \implies 2 \times \frac{22}{7} \times \frac{7}{2} \times h = 154 \)
\( \implies 22h = 154 \)
\( \implies h = 7\text{ cm} \)
Finally, we calculate the volume of the cylinder:
\( V = \pi r^2 h \)
\( \implies V = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 7 \)
\( \implies V = 22 \times \frac{49}{4} = \frac{539}{2} = 269.5\text{ cm}^3 \)
In simple words: Find the curved surface area first, and use it to determine the base area. This gives you the radius, which you can then use with the curved area to find the height, and finally compute the volume.

Exam Tip: Breaking down the total surface area into curved surface area and base areas makes it much simpler to isolate and solve for the radius first.

Question. The diameter of a garden roller is 1.4m and it is 2m long. How much area will it cover in 5 revolutions? (π = 22/7)
Answer: Given data for the cylindrical roller:
Diameter, \( d = 1.4\text{ m} \implies \text{Radius, } r = 0.7\text{ m} \)
Length (height), \( h = 2\text{ m} \)
The area covered in one revolution equals the curved surface area (\( \text{CSA} \)) of the cylinder:
\( \text{Area in 1 revolution} = 2\pi rh \)
\( \implies \text{Area in 1 revolution} = 2 \times \frac{22}{7} \times 0.7 \times 2 = 8.8\text{ m}^2 \)
The area covered in 5 complete revolutions is:
\( \text{Total Area} = 5 \times 8.8 = 44\text{ m}^2 \)
In simple words: The roller only touches the ground with its curved surface. Multiply the curved surface area of the roller by 5 to find the total area it rolls over in 5 turns.

Exam Tip: Remember that a roller covers an area equal to its curved surface area in one revolution. Do not use the total surface area formula for rolling problems.

Question. Three metal cubes whose edge measure 3cm, 4cm and 5cm respectively are melted to form a single cube, find its edge.
Answer: Let the edge lengths of the three cubes be \( s_1 = 3\text{ cm} \), \( s_2 = 4\text{ cm} \), and \( s_3 = 5\text{ cm} \). When melted, their combined volume is equal to the volume of the single large cube formed:
\( V_{\text{new}} = s_1^3 + s_2^3 + s_3^3 \)
\( \implies V_{\text{new}} = 3^3 + 4^3 + 5^3 \)
\( \implies V_{\text{new}} = 27 + 64 + 125 = 216\text{ cm}^3 \)
Let the edge of the new cube be \( S \). Then:
\( S^3 = 216 \)
\( \implies S = \sqrt[3]{216} = 6\text{ cm} \)
In simple words: When materials are melted down and reshaped, the total volume does not change. Add the volumes of the three small cubes together to find the volume of the big cube, then take the cube root to find its edge.

Exam Tip: Memorizing cubes of numbers from 1 to 10 (like \( 6^3 = 216 \)) saves critical time during exams when solving cube root problems.

Question. The dimensions of a cubiod are in the ratio of 1 : 2 : 3 and its total surface area is 88m2. Find the dimensions.
Answer: Let the dimensions of the cuboid be \( l = x \), \( b = 2x \), and \( h = 3x \). The formula for the total surface area of a cuboid is:
\( \text{TSA} = 2(lb + bh + hl) \)
We are given that \( \text{TSA} = 88\text{ m}^2 \). Substituting the values:
\( 2\left(x(2x) + (2x)(3x) + (3x)(x)\right) = 88 \)
\( \implies 2(2x^2 + 6x^2 + 3x^2) = 88 \)
\( \implies 2(11x^2) = 88 \)
\( \implies 22x^2 = 88 \)
\( \implies x^2 = 4 \)
\( \implies x = 2\text{ m} \)
The dimensions are:
Length, \( l = x = 2\text{ m} \)
Breadth, \( b = 2x = 4\text{ m} \)
Height, \( h = 3x = 6\text{ m} \)
In simple words: Represent the dimensions as \( x, 2x, 3x \) and put them into the total surface area formula. Solve for \( x \) to find the actual length, width, and height.

Exam Tip: Be consistent with units. Since the surface area is in square meters (\( \text{m}^2 \)), the resulting dimensions must be in meters (\( \text{m} \)).

Section - C
 

Question. A cuboidal oil tin is 30cm X 40cm X 50cm. Find the cost of the tin required for making 20 such tins if the cost of tin sheet is Rs. 20/m2.
Answer: The dimensions of one cuboidal tin are:
Length, \( l = 30\text{ cm} = 0.3\text{ m} \)
Breadth, \( b = 40\text{ cm} = 0.4\text{ m} \)
Height, \( h = 50\text{ cm} = 0.5\text{ m} \)
The total surface area of one tin is:
\( \text{TSA} = 2(lb + bh + hl) \)
\( \implies \text{TSA} = 2(0.3 \times 0.4 + 0.4 \times 0.5 + 0.5 \times 0.3) \)
\( \implies \text{TSA} = 2(0.12 + 0.20 + 0.15) = 2(0.47) = 0.94\text{ m}^2 \)
The total surface area required for 20 such tins is:
\( \text{Total Area} = 20 \times 0.94 = 18.8\text{ m}^2 \)
Since the rate of the tin sheet is Rs. 20 per \( \text{m}^2 \), the total cost is:
\( \text{Total Cost} = 18.8 \times 20 = \text{Rs. } 376 \)
In simple words: Find the total surface area of one tin in square meters, multiply it by 20 for all the tins, and then multiply by the cost per square meter to get the final price.

Exam Tip: It is usually simpler to convert centimeters to meters at the very beginning when the cost is given per square meter, as converting \( \text{cm}^2 \) to \( \text{m}^2 \) at the end requires dividing by 10,000.

Question. Find the lateral curved surface area of a cylindrical petrol storage tank that is 4.2m in diameter and 4.5m high. How much steel was actually used, if 1/12 of steel actually used was wasted in making the closed tank.
Answer: We are given:
Diameter of the cylinder, \( d = 4.2\text{ m} \implies \text{Radius, } r = 2.1\text{ m} \)
Height, \( h = 4.5\text{ m} \)

(i) Curved Surface Area (CSA):
\( \text{CSA} = 2\pi rh \)
\( \implies \text{CSA} = 2 \times \frac{22}{7} \times 2.1 \times 4.5 = 59.4\text{ m}^2 \)

(ii) Total Steel Actually Used:
We calculate the Total Surface Area (TSA) of the closed cylinder:
\( \text{TSA} = \text{CSA} + 2\pi r^2 \)
\( \implies \text{TSA} = 59.4 + 2 \times \frac{22}{7} \times (2.1)^2 \)
\( \implies \text{TSA} = 59.4 + 27.72 = 87.12\text{ m}^2 \)
Let the actual amount of steel used be \( X \). Since \( \frac{1}{12} \) of it was wasted, the steel incorporated into the tank is:
\( X - \frac{1}{12}X = \frac{11}{12}X \)
Equating this to the total surface area of the tank:
\( \frac{11}{12}X = 87.12 \)
\( \implies X = 87.12 \times \frac{12}{11} \)
\( \implies X = 7.92 \times 12 = 95.04\text{ m}^2 \)
In simple words: First find the curved area of the tank. For the second part, calculate the total surface area of the closed tank, and set it equal to eleven-twelfths of the starting steel to account for the wasted portion.

Exam Tip: Be careful with wastage: if \( \frac{1}{12} \) is wasted, then \( \frac{11}{12} \) of the steel was actually utilized. Do not just add \( \frac{1}{12} \) of the finished tank's area to itself, as that is a common algebraic error.

Question. The radius and height of a cone are in the ratio 4 : 3. The area of the base is 154cm2. Find the area of the curved surface.
Answer: Let the radius of the cone be \( r = 4x \) and the height be \( h = 3x \). The area of the circular base is:
\( \text{Base Area} = \pi r^2 = 154\text{ cm}^2 \)
\( \implies \frac{22}{7} \times r^2 = 154 \)
\( \implies r^2 = 154 \times \frac{7}{22} = 49 \)
\( \implies r = 7\text{ cm} \)
Since \( r = 4x = 7 \), we have:
\( x = \frac{7}{4} \)
The slant height \( l \) of the cone is:
\( l = \sqrt{r^2 + h^2} = \sqrt{(4x)^2 + (3x)^2} = \sqrt{25x^2} = 5x \)
Substituting \( x = \frac{7}{4} \):
\( l = 5 \times \frac{7}{4} = \frac{35}{4} = 8.75\text{ cm} \)
Now, we calculate the curved surface area (CSA) of the cone:
\( \text{CSA} = \pi rl \)
\( \implies \text{CSA} = \frac{22}{7} \times 7 \times \frac{35}{4} = 192.5\text{ cm}^2 \)
In simple words: Find the radius from the base area first. Use the 3:4:5 ratio of a right triangle to find the slant height directly, and then calculate the curved surface area using the cone formula.

Exam Tip: Recognizing that the radius, height, and slant height of a cone form a Pythagorean triple (3:4:5 ratio) helps you find the slant height immediately without complex square root calculations.

Question. A sphere, cylinder and cone are of the same radius and same height. Find the ratio of their curved surfaces.
Answer: Let the common radius of the sphere, cylinder, and cone be \( r \). For these three shapes to share the same height, the height must equal the diameter of the sphere:
\( h = 2r \)
We find the curved surface areas as follows:
1. Sphere:
\( S_1 = 4\pi r^2 \)
2. Cylinder:
\( S_2 = 2\pi rh = 2\pi r(2r) = 4\pi r^2 \)
3. Cone:
The slant height \( l \) is:
\( l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (2r)^2} = \sqrt{5r^2} = r\sqrt{5} \)
So, \( S_3 = \pi rl = \pi r(r\sqrt{5}) = \sqrt{5}\pi r^2 \)

The ratio of their curved surface areas is:
\( S_1 : S_2 : S_3 = 4\pi r^2 : 4\pi r^2 : \sqrt{5}\pi r^2 \)
\( \implies S_1 : S_2 : S_3 = 4 : 4 : \sqrt{5} \)
In simple words: A sphere's height is its diameter \( 2r \), so the cylinder and cone also have height \( 2r \). Find the curved areas of all three using this height, and then divide out the common \( \pi r^2 \) term to get the ratio.

Exam Tip: A sphere's maximum height is always equal to its diameter (\( 2r \)). Using this key relationship allows you to express all heights in terms of a single variable, \( r \).

Question. A hemispherical bowl of internal diameter 36cm contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3cm and height 6cm. How many bottles are required to empty the bowl?
Answer: Hemispherical Bowl:
Diameter = \( 36\text{ cm} \implies \text{Radius, } R = 18\text{ cm} \)
The volume of liquid inside the bowl is:
\( V_{\text{bowl}} = \frac{2}{3}\pi R^3 = \frac{2}{3}\pi (18)^3\text{ cm}^3 \)

Cylindrical Bottle:
Radius, \( r = 3\text{ cm} \), Height, \( h = 6\text{ cm} \)
The volume of one bottle is:
\( V_{\text{bottle}} = \pi r^2 h = \pi (3)^2 (6) = 54\pi\text{ cm}^3 \)

Number of Bottles Required:
\( N = \frac{V_{\text{bowl}}}{V_{\text{bottle}}} \)
\( \implies N = \frac{\frac{2}{3}\pi \times 18 \times 18 \times 18}{54\pi} \)
\( \implies N = \frac{12 \times 18 \times 18}{54} = 72 \)
Thus, 72 bottles are required.
In simple words: Calculate the volume of liquid in the large bowl and divide it by the volume that a single small bottle can hold. Keeping \( \pi \) in the equation allows it to cancel out easily.

Exam Tip: Never waste time calculating the exact decimal value of \( \pi \) in volume-casting or bottle-filling questions, as \( \pi \) will always cancel out during division.

Question. A hemisphere of lead of radius 8cm is cast into a right circular cone of base radius 6cm. Determine the height of the cone.
Answer: Given data:
Radius of the hemisphere, \( R = 8\text{ cm} \)
Base radius of the cone, \( r = 6\text{ cm} \)
Since the hemisphere is recast into a cone, their volumes are equal:
\( V_{\text{cone}} = V_{\text{hemisphere}} \)
\( \implies \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi R^3 \)
\( \implies r^2 h = 2R^3 \)
Substituting the radius values:
\( (6)^2 \times h = 2 \times (8)^3 \)
\( \implies 36h = 2 \times 512 \)
\( \implies 36h = 1024 \)
\( \implies h = \frac{1024}{36} = \frac{256}{9} \approx 28.44\text{ cm} \)
In simple words: Set the volume of the cone equal to the volume of the hemisphere. Cancel out \( \pi \) and the fraction \( 1/3 \) from both sides, then solve directly for the cone's height.

Exam Tip: Simplify the equations first by canceling out common terms (like \( \frac{1}{3}\pi \)) on both sides before plugging in the numbers to minimize calculation errors.

Section - D
 

Question. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 6cm and its height is 4cm. Find the cost of painting the toy at the rate of Rs. 5 per 1000cm2.
Answer: Diameter of the cone's base, \( d = 6\text{ cm} \implies \text{Radius, } r = 3\text{ cm} \). The radius of the hemisphere is also \( r = 3\text{ cm} \). Height of the cone, \( h = 4\text{ cm} \).
First, find the slant height \( l \) of the cone:
\( l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = 5\text{ cm} \)
The total surface area of the toy to be painted is the sum of the curved surface areas of the cone and the hemisphere:
\( \text{Total Area} = \pi r l + 2\pi r^2 = \pi r(l + 2r) \)
\( \implies \text{Total Area} = 3.14 \times 3 \times (5 + 2 \times 3) \)
\( \implies \text{Total Area} = 9.42 \times 11 = 103.62\text{ cm}^2 \)
The rate of painting is Rs. 5 per 1000 \( \text{cm}^2 \):
\( \text{Cost} = 103.62 \times \frac{5}{1000} = 103.62 \times 0.005 \approx \text{Rs. } 0.51 \)
In simple words: Find the outer surface areas of both the cone and the hemisphere. Add them together and multiply the total area by the rate per square centimeter to get the final cost.

Exam Tip: Since the base of the cone is joined to the hemisphere, its flat base area is not exposed. Do not include the base area of either shape in your surface area calculations.

Question. Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14cm.
Answer: For the largest cone to fit inside a cube of edge \( s = 14\text{ cm} \):
- The base diameter of the cone must equal the edge of the cube: \( d = 14\text{ cm} \implies \text{Radius, } r = 7\text{ cm} \).
- The height of the cone must also equal the edge of the cube: \( h = 14\text{ cm} \).
The volume of this cone is:
\( V = \frac{1}{3}\pi r^2 h \)
\( \implies V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14 \)
\( \implies V = \frac{1}{3} \times 22 \times 7 \times 14 \)
\( \implies V = \frac{2156}{3} \approx 718.66\text{ cm}^3 \)
In simple words: The largest cone inside a cube will have a height equal to the cube's side and a base diameter equal to the cube's side. Use these measurements to calculate its volume.

Exam Tip: Remember that for any shape inscribed in a cube, the maximum dimensions are limited by the cube's edge length. For a cone, the height is \( s \) and the diameter is \( s \).

Question. A cone of height 24cm and slant height 25cm has a curved surface area 550cm2. Find its volume use π = 22/7
Answer: We are given:
Height of the cone, \( h = 24\text{ cm} \)
Slant height, \( l = 25\text{ cm} \)
First, find the radius \( r \) using the relationship:
\( r = \sqrt{l^2 - h^2} \)
\( \implies r = \sqrt{25^2 - 24^2} = \sqrt{625 - 576} = \sqrt{49} = 7\text{ cm} \)
We can verify with the curved surface area:
\( \text{CSA} = \pi rl = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2 \)
Now, we calculate the volume of the cone:
\( V = \frac{1}{3}\pi r^2 h \)
\( \implies V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 \)
\( \implies V = \frac{1}{3} \times 22 \times 7 \times 24 \)
\( \implies V = 22 \times 7 \times 8 = 1232\text{ cm}^3 \)
In simple words: First find the radius using the height and slant height in the Pythagorean theorem. Once the radius is known, plug it along with the height into the cone's volume formula.

Exam Tip: When the curved surface area is given alongside height and slant height, it acts as a verification check. Solving for the radius first is the standard approach.

Question. The radius and height of a cone are 6cm and 8cm respectively. Find the curved surface area of the cone.
Answer: Given data:
Radius of the cone, \( r = 6\text{ cm} \)
Height of the cone, \( h = 8\text{ cm} \)
First, find the slant height \( l \):
\( l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ cm} \)
Now, calculate the curved surface area (CSA):
\( \text{CSA} = \pi rl \)
\( \implies \text{CSA} = \pi \times 6 \times 10 = 60\pi\text{ cm}^2 \)
In simple words: Find the slant height using the Pythagorean theorem with the given radius and height. Then multiply the radius, slant height, and \( \pi \) to get the curved surface area.

Exam Tip: If the options or question answer key leaves the answer in terms of \( \pi \), do not waste time substituting \( 22/7 \) or \( 3.14 \) unless explicitly asked to do so.

Question. A well with 10m inside diameter is dug 14m deep. Earth taken out of it is spread all around to a width of 5m to form an embankment. Find the height of embankment.
Answer: Well Dimensions:
Diameter = \( 10\text{ m} \implies \text{Radius, } r = 5\text{ m} \)
Depth, \( h = 14\text{ m} \)
The volume of earth dug out of the well is:
\( V = \pi r^2 h = \pi \times 5^2 \times 14 = 350\pi\text{ m}^3 \)

Embankment Dimensions:
The earth is spread around the well to form a circular ring embankment.
Inner radius, \( r = 5\text{ m} \)
Width = \( 5\text{ m} \)
Outer radius, \( R = r + \text{width} = 5 + 5 = 10\text{ m} \)
The area of the embankment base is:
\( \text{Area} = \pi(R^2 - r^2) = \pi(10^2 - 5^2) = 75\pi\text{ m}^2 \)
Let the height of the embankment be \( H \). The volume of soil in the embankment matches the dug-out volume:
\( 75\pi \times H = 350\pi \)
\( \implies 75H = 350 \)
\( \implies H = \frac{350}{75} = \frac{14}{3} \approx 4.67\text{ m} \)
In simple words: Find the volume of earth dug out from the cylindrical well. Set this volume equal to the volume of the hollow cylindrical embankment to find its height.

Exam Tip: Be careful when calculating the outer radius of the embankment. It is the sum of the well's radius and the embankment's width, which is \( 5 + 5 = 10\text{ m} \), not just the width of \( 5\text{ m} \).

Question. A metallic sheet is of the rectangular shape with dimensions 48cm X 36cm. From each one of its corners, a square of 8cm is cutoff. An open box is made of the remaining sheet. Find the volume of the box.
Answer: Original sheet dimensions: \( 48\text{ cm} \times 36\text{ cm} \).
Side of the square cut from each corner = \( 8\text{ cm} \).
When folded to make an open box, the dimensions of the box are:
Length, \( L = 48 - 2(8) = 32\text{ cm} \)
Breadth, \( B = 36 - 2(8) = 20\text{ cm} \)
Height, \( H = 8\text{ cm} \)
The volume of the box is:
\( V = L \times B \times H \)
\( \implies V = 32 \times 20 \times 8 = 5120\text{ cm}^3 \)
In simple words: Cutting out squares from the corners reduces both the length and width of the sheet by twice the side of the cut square. The height of the box is the side of the cut square.

Exam Tip: Remember to subtract the corner square length twice (once for each end) from both the length and breadth of the sheet to get the dimensions of the box base.

Self Evaluation
 

Question. Water in a canal, 30dm wide and 12dm deep is flowing with a velocity of 20km per hour. How much area will it irrigate in 30min. if 9cm of standing water is desired? (10dm = 1 meter)
Answer: Let's first convert all given dimensions into meters:
Width, \( w = 30\text{ dm} = 3\text{ m} \)
Depth, \( d = 12\text{ dm} = 1.2\text{ m} \)
Flow speed = \( 20\text{ km/h} = 20,000\text{ m/h} \)

The cross-sectional area of the canal is:
\( \text{Area of cross-section} = 3 \times 1.2 = 3.6\text{ m}^2 \)

Length of water flowing in 30 minutes (\( 0.5\text{ hours} \)):
\( \text{Length} = 20,000 \times 0.5 = 10,000\text{ m} \)

Thus, the volume of water flowing in 30 minutes is:
\( V = 3.6 \times 10,000 = 36,000\text{ m}^3 \)

Let the area of the field to be irrigated be \( A \). The required depth of standing water is \( 9\text{ cm} = 0.09\text{ m} \). Since the volume of water remains constant:
\( V = A \times \text{depth} \)
\( \implies 36,000 = A \times 0.09 \)
\( \implies A = \frac{36,000}{0.09} = 400,000\text{ m}^2 \)
In simple words: Find the volume of water that flows through the canal in half an hour. Set this volume equal to the area of the field times the desired depth of standing water to solve for the area.

Exam Tip: Pay extremely close attention to the units in this question: decimeters (dm), kilometers per hour (km/h), and centimeters (cm) must all be converted to a uniform unit, preferably meters, before calculation.

Question. Three cubes of each side 4cm are joining end to end. Find the surface area of resulting cuboid.
Answer: When three identical cubes of side \( 4\text{ cm} \) are joined end to end, the dimensions of the resulting cuboid are:
Length, \( L = 4 + 4 + 4 = 12\text{ cm} \)
Breadth, \( B = 4\text{ cm} \)
Height, \( H = 4\text{ cm} \)
The total surface area of this cuboid is:
\( \text{TSA} = 2(LB + BH + HL) \)
\( \implies \text{TSA} = 2(12 \times 4 + 4 \times 4 + 4 \times 12) \)
\( \implies \text{TSA} = 2(48 + 16 + 48) = 2 \times 112 = 224\text{ cm}^2 \)
In simple words: When you connect three cubes in a row, the total length becomes three times a single side, but the width and height stay the same. Plug these new dimensions into the standard cuboid area formula.

Exam Tip: A faster way is to count the exposed faces. Since there are 3 cubes, there are originally \( 3 \times 6 = 18 \) faces. When joined, 4 faces are hidden at the joints, leaving 14 exposed faces. Total Area = \( 14 \times (4 \times 4) = 224\text{ cm}^2 \).

Question. A hollow cylindrical pipe is 210cm long. Its outer and inner diameters are 10cm and 6cm respectively. Find the volume of the copper used in making the pipe.
Answer: Given values:
Length of the pipe, \( h = 210\text{ cm} \)
Outer radius, \( R = \frac{10}{2} = 5\text{ cm} \)
Inner radius, \( r = \frac{6}{2} = 3\text{ cm} \)
The volume of copper used is the difference between the outer and inner cylindrical volumes:
\( V = \pi R^2 h - \pi r^2 h = \pi(R^2 - r^2)h \)
\( \implies V = \frac{22}{7} \times (5^2 - 3^2) \times 210 \)
\( \implies V = 22 \times (25 - 9) \times 30 \)
\( \implies V = 22 \times 16 \times 30 = 10,560\text{ cm}^3 \)
In simple words: Find the total volume using the outer radius, and subtract the empty space inside using the inner radius to get the volume of the copper metal.

Exam Tip: Factoring out \( \pi h \) before calculation makes the problem simpler as you only need to multiply \( R^2 - r^2 \) instead of working out two large separate volumes.

Question. A semi circular sheet of metal of diameter 28cm is bent into an open conical cup. Find the depth and capacity of cup.
Answer: Semicircular sheet details: Diameter = \( 28\text{ cm} \implies \text{Radius, } R = 14\text{ cm} \).
When bent into an open conical cup:
1. The sheet radius becomes the slant height of the cone: \( l = R = 14\text{ cm} \).
2. The perimeter boundary of the semicircle matches the cone's base circumference:
\( 2\pi r = \pi R \)
\( \implies 2r = 14 \implies r = 7\text{ cm} \)

(i) Depth of the cup (\( h \)):
\( h = \sqrt{l^2 - r^2} \)
\( \implies h = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} = 7\sqrt{3} \approx 12.12\text{ cm} \)

(ii) Capacity of the cup (\( V \)):
\( V = \frac{1}{3}\pi r^2 h \)
\( \implies V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 7\sqrt{3} = \frac{1078\sqrt{3}}{3}\text{ cm}^3 \)
Using \( \sqrt{3} \approx 1.732 \):
\( V \approx \frac{1078 \times 1.732}{3} \approx 622.26\text{ cm}^3 \)
In simple words: Bending a semicircle into a cone turns its radius into the cone's slant height, and its circular boundary into the cone's base circumference. Use these to find the base radius, depth, and volume of the cup.

Exam Tip: A classic exam problem! Remember the golden rule for folding sheets into cones: the radius of the sector always becomes the slant height of the cone, and the arc length of the sector becomes the base circumference.

Question. If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of second sphere?
Answer: Let the radius of the first sphere be \( r \). Its volume is:
\( V_1 = \frac{4}{3}\pi r^3 \)
If the radius is doubled, the new radius becomes \( 2r \). The volume of this second sphere is:
\( V_2 = \frac{4}{3}\pi (2r)^3 = \frac{4}{3}\pi \times 8r^3 \)
Now, we calculate their volume ratio:
\( \frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi \times 8r^3} \)
\( \implies \frac{V_1}{V_2} = \frac{1}{8} \)
Thus, the ratio is \( 1:8 \).
In simple words: Since the volume of a sphere is proportional to the cube of its radius, doubling the radius increases the volume by a factor of eight (\( 2^3 = 8 \)).

Exam Tip: For quick multiple-choice questions, remember that if any 3D shape is scaled by a factor of \( k \), its volume scales by \( k^3 \). Here, \( k = 2 \), so the volume ratio is \( 1:2^3 = 1:8 \).