Geometric shapes are everywhere in the material world. The basic shapes learned in earlier classes can be found in nearly every object, but not in that exact form. They are usually a blend of many geometric shapes. Some are simple, some really complex. This chapter explores more advanced geometric shapes. It explores, as per Class 10 Maths, the Surface Areas and Volumes of many shapes from the everyday world. It will help you appreciate ordinary things like bowls, tumblers, capsules, etc in the material world.

Introduction |

Surface Area of a Combination of Solids |

Volume of a Combination of Solids |

Conversion of Solid from One Shape to Another |

Frustum of a Cone |

In daily life, there are many solid objects made of a combination of some solids. Sometimes, a new shape is formed when a solid surface is carved out in such a way as if a part of it has been taken out. These new shapes are all made from various combinations of geometric shapes like cones, cylinders, cubes, cuboids, spheres and hemispheres.

A giant water container being carried on a truck looks like a capsule. A capsule looks as if it were made of a hollow cylinder in the middle, and two equal hemispheres at the ends.

Observe a sharpener. It too is an object by design. A basic looking sharpener is cuboidal in shape, but there are two holes inside it. This hollow part is cylindrical in shape. It is as if a cylindrical part was tubed out of the cuboid.

Sometimes, one has to calculate the outer surface area. Sometimes, the surface area inside the object. Sometimes one has to know the volume or capacity of the solid combo object. Other times to calculate the volume of the material needed to construct the object.

In earlier classes, we have seen how to calculate the surface areas and volumes of all these figures. We have seen the formulae for calculating curved surface area, total surface area, volume and how they compare with each other. In NCERT solutions Class 10 Maths Chapter 13, we shall see Surface Area and Volume Formulas for these solid combinations.

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Let us look at the formulae of Surface Area of three-dimensional geometric figures. This is the reference for calculating areas of the advanced shapes.

A cylinder has a height, a curved surface, and two circular surfaces at the ends.

Curved Surface Area (CSA) of Cylinder: 2πrh

Total Surface Area (TSA) of Cylinder: 2πrh + 2πr^{2} = 2πr(h+r)

Removing the base/top of the cylinder reformulates this as 2πrh + πr^{2}

A cone has a circular base and triangular curved top with height ‘h’. Let the slant ‘l’ and radius ‘r’.

CSA of Cone = πrl

TSA of cone = πrl + πr^{2} = πr(l+r)

A cube is a uniform figure, like a 3D square with six square faces/walls. Take ‘l’ as the length of an edge.

Surface Area of a Cube: = sum of the areas of all six walls = 6l^{2}

Vertical Surface Area of a Cube (Remove a surface from the cube) = 5l^{2}

A cuboid is like a 3D rectangle with six rectangular surfaces/walls. It has a length ‘l,’ breadth ‘b’ and height ‘h’.

Surface Area of Cuboids: 2lb+2bh+2lh = 2(lb+bh+lh)

Vertical Surface Area of Cuboids (without the top surface): lb+2bh+2lh

Curved Surface Area of a Hemisphere (half a sphere): 2πr^{2}

Total Surface Area of a Hemisphere: CSA + Circular Base = 2πr^{2} + πr^{2} = 3πr^{2}

Surface Area of a Sphere: 2xCSA(hemisphere) = 4πr^{2}

Recalling the example of the capsule from earlier. A capsule is made of a hollow cylindrical shape in the middle, joining two hemispheres at its ends. Cutting up the surface into its constituent parts and calculating the surface area of each.

TSA of Capsule = CSA(cylinder) + 2xCSA(hemisphere) = 2πrh + 2x2πr^{2} = 2πr(h+2r)

If calculating the area of a test tube, the above formula changes just a bit since the shape has not changed much.

TSA of test tube = CSA(cylinder) + 1xCSA(hemi) = 2πrh + 2πr^{2} = 2πr(h+r) …. (also recall TSA of a cylinder)

Make a toy with a conical top and hemispherical bottom. Their flat surfaces are joined together. This can also remind of the spinning tops that children play and contend with.

TSA of toy = CSA(hemisphere) + CSA(cone) = 2πr^{2} + πrl = πr(2r+l)

To further deepen your understanding of the concepts in Class 10 Math Surface Areas And Volumes, checkout the MSVGo NCERT solutions app for a more interactive learning.

Unlike surface area, during the calculation of volumes of combined surfaces, the volumes of all parts/shapes are added (or subtracted, as the case may be).

Revisit the volume formulas of three-dimensional objects from previous classes.

Volume of a Cube = l^{3}

Volume of Cuboid = (l)(b)(h)

Volume of a Cylinder = πr^{2}h

Volume of a Cone = ⅓πr^{2}h

Volume of a Hemisphere = ⅔πr^{2}h = ⅔πr^{3}

Volume of a Sphere = 4/3 (πr^{2}h) = 4/3 πr^{3}

Take a shape like a bun of bread brought fresh from the market. It appears like a cuboid with the vertical half of a cylinder joining it on the top. The volume can be calculated as follows

Volume of cuboid + ½ Volume of Cylinder = lbh + ½πr^{2}h

Take a glass. The base is curved upwards. It has a cylindrical shape with one part of it carved inside in the shape of a hemisphere. What is its capacity?

Volume of a cylinder - Volume of a hemisphere = πr^{2}h - ⅔πr^{2}h = ⅓πr^{2}h

The radius of the cylinder is the same as that of the hemisphere.

If molten wax is poured equally into separate containers of different shapes, then the wax takes on the shape of those containers: cylindrical, conical, or like a toy. But the volume of wax in each of the new shapes will be the same. This can easily be seen in the case of ice too. Although, note that water expands just a bit when turning into ice.

So far, this chapter has explored the areas and volumes of new shapes formed by the combination of solids. Let’s see its properties if an existing three-dimensional shape is cut and a new shape is formed. By removing the top of a cone, cutting it with a knife parallel to its base, a new shape that is the Frustum of the cone is obtained.

In Indian houses, the steel mug for drinking water usually is an upside-down Frustum. The figure has two circular ends with different radii. The top is smaller, and the base is larger. It has a height, slant length, two radii and two circumferences. Its curved surface area can be calculated by subtracting the CSA of the smaller (removed) cone from the CSA of the larger (original) cone. Let r_{1} and r_{2} be radii of the larger part and the smaller part of the frustum.

In fact, CSA (Frustum) = πl(r_{1}+r_{2}).

Where, l = [h^{2} + (r_{1}-r_{2})^{2} ]^{½}

(The power of ½ outside the bracket means square root)

Volume (Frustum) = ⅓πh(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

TSA (Frustum) = πl(r_{1} + r_{2}) + πr_{1}^{2} + πr_{2}^{2}

Along with many well-solved examples, there are four problem sets, with an optional fifth set for advanced thinking. In total, there are 33 questions about the surface area and volume of the various solid combinations.

In Geometry, area refers to how much space a figure in a two-dimensional plane has, while volume is the space contained inside a three-dimensional figure. Moreover, the curved surface area is an area of a curved surface. Total surface area for a three-dimensional object is the area of its surfaces (which are 2D) and not its internal capacity.

Yes. The NCERT textbooks are available online. The NCERT Solutions for Class 10 Chapter 13 is also available on MSVgo, providing in-depth interactive learning to clarify concepts and for quick revisions.

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