In our day to day life, we came across various solids having different shape and size, in which we can calculate both surface area and volume. We need to calculate surface area and volume of the objects around us. But what if these basic shapes combine and form some different shape other than the original one. Now the question is how you will calculate the volume, surface and areas of new objects. While calculating the surface area and volumes of these new shapes we need to observe the new form. A deep discussion is given below that will create a more clear image about these objects and its calculation.

The surface area of an object is given by the total area of the surface that an object occupies, or we can say the total area of all the surfaces of any three-dimensional figure. The surface area of figures other than cube or cuboids can be calculated as the lateral area of the figure plus its every base, in case of prism and cylinder are the same then we can take it as twice the area of the base. The surface area of any given figure can be calculated with the help of the example of a gift as a three-dimensional figure and let the surface area be the wrapping paper, so the amount of wrapping paper used to cover the gift is the surface area of the given three-dimensional figure. Surface area can be given by the following formula:

Surface area = Lateral area + (n * base)

Where, n = no. of bases present (n = 2 for prism/ cylinder, n = 1 for pyramids/ cones, and n = 0 for spheres/ circles)

Surface area can be further divided into 2 types such as:

1. Total surface area – The area including the base and the curved part is called a total surface area.

Volume can be calculated with the help of following formulas for different figures:

Volume of sphere (V) = (4/3) π x (radius)3

Volume of prism or cylinder (V) = base area x height

Volume of pyramid or cone (V) = (1/3) x base area x height

**Combination of solids:**

To calculate the surface area or the volume of these types of solids, first, we have to see the number of solid shapes that form these shapes, as these three-dimensional structures contain various one-dimensional shapes having an example of a cube formed with the help of six squares which is a one-dimensional shape. The surface area of a given composite shape is the sum of the area of all the faces in that solid. To understand the combination of solids we can take one example of ice cream filled cone, which is the fusion of a cone and the hemisphere-shaped ice cream. So, the total surface area of the ice cream filled cone is equal to the sum of the curved surface area of the hemisphere and the curved surface area of the cone.

The curved surface area of cone = πrl,

And the curved surface area of hemisphere = 2πr3

So, the total surface area of ice cream filled cone = 2πr3 + πrl

To calculate the volume of the combinational shapes we have to first figure out the different shapes involved in it to form the composite shape. The volume of the combinational shapes can be calculated by calculating the volume of the specific shapes through which new combinational shape is formed and add them to form the total volume of the composite shape.

Similarly in case of volume, the volume of the ice cream filled cone can be found out by individually calculating the volume of the hemisphere and the volume of the cone and then adding up to form the volume of ice cream filled cone.

Volume of cone = 1/3 πr^{2}h

Volume of hemisphere = 2/3 πr^{3}

The surface area of an object is given by the total area of the surface that an object occupies, or we can say the total area of all the surfaces of any three-dimensional figure. The surface area of figures other than cube or cuboids can be calculated as the lateral area of the figure plus its every base, in case of prism and cylinder are the same then we can take it as twice the area of the base. The surface area of any given figure can be calculated with the help of the example of a gift as a three-dimensional figure and let the surface area be the wrapping paper, so the amount of wrapping paper used to cover the gift is the surface area of the given three-dimensional figure. Surface area can be given by the following formula:

Surface area = Lateral area + (n * base)

Where, n = no. of bases present (n = 2 for prism/ cylinder, n = 1 for pyramids/ cones, and n = 0 for spheres/ circles)

Surface area can be further divided into 2 types such as:

1. Total surface area – The area including the base and the curved part is called a total surface area.

2. Curved surface area – The area of the curved part excluding the base is called as the curved surface area.

Volumes of any given object can be said as the amount of liquid it can contain in it. Basically, the quantity enclosed by the given three-dimensional objects is called the volume of that object. The volume of the one dimensional (for e.g. lines), as well as the two-dimensional object (for e.g. squares), are considered zero as the volume is considered as quantity. The basic properties to find the volume of any given object are as follows:

1. Any given object has the volume of length * breadth * height (V = lwh).

2. The total volume of any given object is the sum of all non-overlapping regions.

3. Exactly the same when superimposing figures have the same volume.

4. Depending on the unit cube, every polyhedral region has a unique volume.

Volume can be calculated with the help of following formulas for different figures:

Volume of sphere (V) = (4/3) π x (radius)3

Volume of prism or cylinder (V) = base area x height

Volume of pyramid or cone (V) = (1/3) x base area x height

To calculate the surface area or the volume of these types of solids, first, we have to see the number of solid shapes that form these shapes, as these three-dimensional structures contain various one-dimensional shapes having an example of a cube formed with the help of six squares which is a one-dimensional shape. The surface area of a given composite shape is the sum of the area of all the faces in that solid. To understand the combination of solids we can take one example of ice cream filled cone, which is the fusion of a cone and the hemisphere-shaped ice cream. So, the total surface area of the ice cream filled cone is equal to the sum of the curved surface area of the hemisphere and the curved surface area of the cone.

The curved surface area of cone = πrl,

And the curved surface area of hemisphere = 2πr3

So, the total surface area of ice cream filled cone = 2πr3 + πrl

To calculate the volume of the combinational shapes we have to first figure out the different shapes involved in it to form the composite shape. The volume of the combinational shapes can be calculated by calculating the volume of the specific shapes through which new combinational shape is formed and add them to form the total volume of the composite shape.

Similarly in case of volume, the volume of the ice cream filled cone can be found out by individually calculating the volume of the hemisphere and the volume of the cone and then adding up to form the volume of ice cream filled cone.

Volume of cone = 1/3 πr

Volume of hemisphere = 2/3 πr

Shape | Equation | Variables | Volume | Perimeter |

Cube | 6s^{2} | s = side length | V = a^{3} | 6a |

Cuboid | 2 (lw + lh + wh) | l = length, w = width, h= height | V = abc | 4 (l + b+ h) |

Triangular prism | bh + l(a + b + c) | b = base length of triangular, h = height of triangular, l = distance between triangular bases, a, b, c = sides of triangular | - | - |

All prisms | 2B + Ph | B = the area of one base, P = the perimeter of one base, h= height | - | - |

Sphere | 4πr^{2} = πd^{2} | r = radius of sphere, d = diameter | V = 4/3πr^{3} | - |

Spherical lune | 2r^{2}θ | r = radius of sphere, θ = dihedral angle | - | - |

Torus | (2πr) (2πR) = 4π^{2}Rr | r = minor radius (radius of the tube), R = major radius (distance from the center of the tube to center of tortus) | V = 2π^{2 }Rr^{2} | |

Closed cylinder | 2πr^{2 } + 2πrh = 2πr ( r + h ) | r = radius of the circular base, h = height of the cylinder | V = πr^{2}h | - |

Lateral surface area of a cone | πr (√r^{2} + h^{2}) = πrs | S = (√r^{2} + h^{2}S = slant height of the cone.R = radius of the circular base,h = height of the cone | - | - |

Pyramid | B + PL/2 | B = area of base, P = perimeter of base, l = slant height | - | - |

Tetrahedron | √3a^{2} | a = side length | - | - |

Square pyramid | b^{2 }+ 2bs = b^{2 }+ 2b √(b/2)^{2 }+ h^{2 } | b = base length, s = slant height, height = vertical height | - | - |

Rectangular pyramid | Lw + l√ (w/2)^{2 }+ h^{2 }+ w √ (l/2)^{2 }+ h^{2 } | L = length, w = width, h = height | - | - |