Class 10 Maths - ODISHA

Surface Areas and Volumes

The chapter Surface Areas and Volumes in Class 10 Mathematics for BSE Odisha students explores the geometry of three-dimensional solids. You will learn to calculate the total surface area, curved surface area, and volume of combined figures like cylinders, cones, spheres, hemispheres, and frustums of a cone. This chapter is vital for board exams as it tests both visualization and multi-step calculation skills. Problems often involve converting one solid shape into another or finding dimensions of objects made by joining two or more basic shapes, carrying significant weight in the final examination.

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Key Concepts

Cuboid and Cube Surface Area

A cuboid has six rectangular faces, and its total surface area depends on its length, breadth, and height, while a cube has six equal square faces.

Cylinder, Cone, and Sphere

These standard solids have specific formulas for curved surface area (CSA), total surface area (TSA), and volume based on their radius and height.

Frustum of a Cone

When a right circular cone is cut by a plane parallel to its base, the lower portion remaining is called a frustum, which has two circular bases of different radii.

Combination of Solids

Real-world objects are often combinations of two or more basic solids, such as a tent being a cylinder surmounted by a cone, requiring careful addition of surface areas.

Conversion of Solids

When a solid shape is melted and recast into another shape, the total volume of the material remains constant throughout the process.

Important Formulas

Total Surface Area of Cuboid = 2(lb + bh + hl)
Curved Surface Area of Cylinder = 2 * pi * r * h
Total Surface Area of Cylinder = 2 * pi * r * (r + h)
Volume of Cylinder = pi * r^2 * h
Curved Surface Area of Cone = pi * r * l
Total Surface Area of Cone = pi * r * (r + l)
Volume of Cone = (1/3) * pi * r^2 * h
Surface Area of Sphere = 4 * pi * r^2
Volume of Sphere = (4/3) * pi * r^3
Curved Surface Area of Hemisphere = 2 * pi * r^2
Total Surface Area of Hemisphere = 3 * pi * r^2
Volume of Hemisphere = (2/3) * pi * r^3
Volume of Frustum of Cone = (1/3) * pi * h * (r1^2 + r2^2 + r1 * r2)
Curved Surface Area of Frustum = pi * l * (r1 + r2)

Board Exam Info

In the BSE Odisha Class 10 Mathematics board exam, this chapter typically carries around 8 to 12 marks. Questions usually include short-answer questions (2-3 marks) involving direct formula application and long-answer questions (5-6 marks) based on the combination of solids or conversion of solid shapes from one form to another.

Frequently Asked Questions

Why do we add only curved surface areas when joining two solids like a cylinder and a hemisphere?

When two solids are joined together, the faces that are glued or attached to each other are hidden from the outside, so we only calculate the exposed outer surface area.

How do I solve problems where one solid is melted and recast into another?

In conversion problems, the volume always remains conserved. Equate the volume of the original solid to the volume of the new solid(s) to find the unknown dimension.

Is the value of pi (22/7 or 3.14) required to be memorized?

Yes, use 22/7 unless a specific value like 3.14 is mentioned in the question paper.

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