Surface Areas and Volumes
Surface Areas and Volumes — Study Notes
NCERT-aligned · 8 notes · 3 shown free
12.1 Introduction
Explanation12.1 Introduction
In Class IX, students are introduced to some basic three-dimensional solids such as cuboids, cones, cylinders, and spheres. They learn to calculate the surface areas and volumes of these solids. This chapter builds upon that foundation by exploring more complex solids formed by combining two or more basic solids. Real-life objects often resemble these combinations rather than simple solids. For example, a truck container carrying oil or water is shaped like a cylinder with hemispherical ends. Similarly, test tubes are combinations of cylinders and hemispheres. Understanding how to find the surface area and volume of such combined solids is essential for practical applications like painting, packaging, and manufacturing. This chapter aims to equip students with the methods to calculate surface areas and volumes of such composite solids, thereby extending their knowledge beyond simple shapes to more realistic and complex objects encountered in daily life.
- Basic solids studied previously include cuboid, cone, cylinder, and sphere.
- Many real-life objects are combinations of these basic solids.
- Examples include truck containers (cylinder with hemispherical ends) and test tubes (cylinder with hemisphere).
- Calculating surface areas and volumes of combined solids is important for practical applications.
- This chapter extends knowledge to handle such composite solids.
- 📌 Solid: A three-dimensional object with length, breadth, and height.
- 📌 Surface area: The total area covering the surface of a solid.
- 📌 Volume: The amount of space occupied by a solid.
12.2 Surface Area of a Combination of Solids
Explanation12.2 Surface Area of a Combination of Solids
When dealing with solids formed by combining two or more basic solids, the surface area calculation requires careful consideration. The total surface area of the combined solid is not simply the sum of the surface areas of the individual solids because some surfaces become internal and are not visible after joining. To find the surface area of such a composite solid, we first break it down into its constituent parts whose surface areas we know how to calculate. For example, a container shaped like a cylinder with two hemispheres attached at the ends has a surface area equal to the sum of the curved surface area of the cylinder and the curved surface areas of the two hemispheres, excluding the circular areas where they join. Similarly, a toy formed by joining a cone and a hemisphere at their bases will have a total surface area equal to the curved surface area of the cone plus the curved surface area of the hemisphere, since the bases are joined and not exposed. The chapter provides detailed examples to illustrate these concepts, including calculating the surface area of a playing top shaped like a cone surmounted by a hemisphere, a decorative block made of a cube and a hemisphere, and a toy rocket shaped by combining a cone and a cylinder with different base diameters. These examples emphasize the importance of identifying which surfaces are exposed and which are hidden when calculating total surface area.
- Surface area of combined solids excludes the joined surfaces.
- Total surface area is sum of curved surface areas of visible parts.
- Example: Cylinder with two hemispheres – sum of curved surface areas only.
- Example: Cone joined with hemisphere – sum of curved surface areas excluding base.
- Careful calculation of slant height and radii is essential.
- Real-life applications include painting and manufacturing.
- 📌 Curved Surface Area (CSA): Area of the curved surface excluding bases.
- 📌 Total Surface Area (TSA): Sum of areas of all the surfaces of a solid.
- 📌 Slant height (l): The length of the slanting side of a cone.
Examples on Surface Area of Combination of Solids
ExplanationExamples on Surface Area of Combination of Solids
This section provides detailed worked examples illustrating the calculation of surface areas for various combined solids. Example 1 involves a playing top shaped like a cone surmounted by a hemisphere with a total height of 5 cm and diameter 3.5 cm
Practice Questions — Surface Areas and Volumes
Includes NCERT exercise questions with answers
Q1.A solid cube of side 12 cm is cut into 8 cubes of equal volume. The side of the new cube is _______
Answer:
6 cm
Explanation:
[{"id": "b620561f-1f45-4614-990b-2e391e224a19", "type": "html", "value": " We know that, Volume of a cube = side³ Given: Side = 12 cm Volume of cube = 12³ = 1728 cm³ Number of small cubes = 8 Volume of small cube = 1728/8 = 216 cm³ Let side of small cube = a Volume of small cube = a³ = 216 cm³ a = ³√ 216 = 6 cm Hence the correct answer is option 2. "}]
Q2.How many spherical lead shots each having a diameter of 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm ?
Answer:
84
Explanation:
[{"id": "5de6fa2c-8eda-4bc2-b7b3-b377a1443225", "type": "html", "value": " We know that, Volume of a cuboid = lbh = 9 x 11 x12 = 1188 cm³ Given : Diameter of lead shot = 3 cm Radius of lead shot = 1.5 cm Volume of one lead shot = 4/3πr³ = 4/3 x 22/7 x 1.5 x 1.5 x 1.5 = 14.14 cm³ Number of lead shots = Volume of cuboid/Volume of one lead shot = 1188 / 14.14 = 84 shots Hence, the correct answer is Option 3 "}]
Q3.The surface area of a solid metallic sphere is 616 cm². It is remelted and cast into a cone of height 28 cm. Find the diameter of the base of the cone so formed.
Answer:
14 cm
Explanation:
[{"id": "6386be0c-2d34-4fde-83f7-0e4f3a07f321", "type": "html", "value": " Surface Area of Sphere = 4πr² = 616 cm² so, r² = 616 / 4π = 49 cm² r = 7 cm Volume of sphere = (4/3) π r³ = (4/3) π 7³ Volume of cone = (1/3) πr² h Volume of cone = Volume of sphere (1/3) πr² h = (4/3) π 7³ r² h = 4 . 7³ r² 28 = 4 . 7³ r² = (4 . 7³) / 28 = 49 r = √49 = 7 cm So, diameter of cone = 2 x 7 = 14 cm Hence the correct answer is Option 2. "}]
Q4.A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field, which is 12m in diameter and 2.5 m deep.If water flows through the pipe at 3.6 km/hr, in how much time will the tank be filled?
Answer:
96 min
Explanation:
[{"id": "13dd3ee7-95c4-43dc-80c3-89dc8292fa5c", "type": "html", "value": " Given : Radius of pipe = 12.5 cm = 0.125m Diameter of cylindrical tank= 12m Radius of Cylindrical tank = 6 m Height of Cylindrical tank = 2.5 m Volume of tank = πr²h = π6²2.5= 90π m³ Speed of water = 3.6 km/hr In 1 hour, length of water column = 3.6 km = 3600 m In 1 minute, length of water column = 3600 /60 m = 60 m Volume of water flowing through pipe in 1 minute = Volume of cylinder with radius 0.125 m cm and height 60 m = πr²h = π (0.125)²60 = π0.9375 m³ Time taken to fill the tank =Volume of tank/Volume of water flowing through pipe in 1 minute = 90π /π0.9375 = 96 minutes. Hence, the correct answer is Option 4. "}]
Q5.Which is the correct formula for calculating the Total Surface Area of the frustum of a cone, if height is h, slant height is l and radii of the circular faces are R,r.?
Answer:
πl(R +r)+πR²+πr²
Explanation:
[{"id": "541213db-31f5-407e-816b-40f39ae0ecfa", "type": "html", "value": " We know that, Total Surface Area = πl(R +r)+πR²+πr² Hence, the correct answer is Option 3. "}]
Q6.The perimeter of the circular ends of a frustum are 48cm and 36 cm. if the height of the frustum is 11 cm , find its volume. ( Take π = 22/7)
Answer:
1554 cm³
Explanation:
[{"id": "eb1e0f49-3375-4458-af49-1d240d0f7b46", "type": "html", "value": " We know that, Perimeter of a circle = 2πr Let radius of Larger end of frustum = R Let radius of smaller end of frustum = r Perimeter of Larger end = 2πR = 48 cm So, R = 48/2π = 24 /π Perimeter of Smaller end = 2πr = 36 cm So, r = 36/2π = 18 /π We know that, Volume of the frustum = 1/3 πh(R² + r² + R r) h = 11cm, R = 24 /π, r =18 /π V = 1/3 x π x 11( (24 /π)² + (18 /π)² + 24 /π x 18 /π) = 11/3π ( 576 + 324 + 432) = 11/3 x (1332) x 7/22 = 1554 cm³ Hence, the correct answer is Option 3 "}]
Q7.ಒಂದು ಸಿಲಿಂಡರ್ ಆಕಾರದ ಪೆನ್ಸಿಲ್ ನ ಒಂದು ತುದಿಯನ್ನು ಚೂಪು ಮಾಡಿದೆ. ಹಾಗಾದರೆ ಅದರಲ್ಲಿ ಈ ಕೆಳಗಿನ ಯಾವ ಘನಾಕೃತಿಗಳು ಜೋಡಿಸಲ್ಪಟ್ಟಿವೆ?
Answer:
ಸಿಲಿಂಡರ್ ಮತ್ತು ಶಂಕು
Q8.ಒಂದೇ ಅಳತೆಯ ವೃತ್ತಾಕಾರದ ಪಾದ ಮತ್ತು ಎತ್ತರಗಳನ್ನು ಹೊಂದಿರುವ ಒಂದು ಶಂಕು ಮತ್ತು ಸಿಲಿಂಡರ್ ಇವುಗಳ ಘನಫಲದ ಅನುಪಾತವು
Answer:
B) 3 : 1
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