Most tests often include questions based on the knowledge of the geometries of 3-D objects such as cylinder, cone, cuboid, cube & sphere. The purpose of the article is to help you learn basics of 3-D geometry and encapsulate some of the important formulae and tricks.

The questions on Volume and Surface Area appear in all the competitive exams. Most of the students tend to avoid this topic considering it to be quite complex and calculative. This article would help you not only in memorizing the formulas, but also in understanding direct or indirect applications of these formulas. We strongly advice you go through each and every definition and formula given below to solve questions on Surface Area and Volume.

For any regular solid, Number of faces + Number of vertices = Number of edges + 2. This formula is called __Euler’s formula__.

A prism is a solid, whose vertical faces are rectangular and whose bases are parallel polygons of equal area. A prism is said to be triangular prism, pentagonal prism, hexagonal prism, octagonal prism according to number of sides of the polygon that form the base. In a prism with a base of *n* sides, number of vertices = 2*n*, number of faces = *n* + 2.

Surface area formula of vertical faces of a prism = perimeter of base x height.

1. 60 m^{3}

2. 30 m^{3}

3. 20 m^{3}

4. 15 m^{3}

5. 150 m^{3}

Base area = √s(s-a)(s-b)(s-a) where s = (3+4+5) / 2 = 6

Base area = √6 x 3 x 2 x 1 = 6m

∴ Volume of prism = 6 x 10 = 60m

Must Read 3D Geometry Articles

- Mensuration-Cone & Pyramid
- Mensuration-Solved Examples
- Mensuration- Sphere & Hemisphere
- Mensuration- Right Prism & Cylinder

The surface area of a right circular cylinder with a base of radius 'r' and height ‘h’ is equal to

The volume of cylinder is equal to

In the cylinder given below, surface area is equal to 2(36π) + 2π(6)(9) = 180π, and the volume is equal to 36π(9) = 324π.

1. 120 sq cm

2. 140 sq cm

3. 150.72 sq cm

4. 180.2 sq cm

5. 200 sq cm

Suggested Action:

1. 1.0 m

2. 1.5 m

3. 2.5 m

4. 1.97 m

Also, area of the earth dug out = π(9+2)^{2} - π(9)^{2}

= π[(11+9)(11-9)] = π(20)(2) = 40πm^{2}

∴ Height of circular embankment = (= π*9*9*20 )/( 40π ) = 40.5m.