November 10, 2008

This resource has been contributed by Winpossible, and can also be accessed on their website by clicking here - Surface Area - Prisms, Cylinders, Cones, Pyramids and Spheres.

In this section you'll learn how to calculate the surface area of common solids. It will be presented with the help of some examples, practice questions with solution and using video explanation in own handwriting by the instructor.

Any solid formed by joining the corresponding vertices of two congruent and parallel polygons with line segments is known as a prism. The base of a prism has two congruent polygons; lines connecting the corresponding vertices are lateral edges and they are parallel. The parallelograms formed by lateral edges are called lateral faces. In the real-world common application; as an example, it is a figure most people have personal experience of either wrapping or opening gifts.

In right prism, lateral edges are perpendicular to the bases of the prism and in oblique prism; lateral edges are not perpendicular to the bases of the prism.

The prism surface area formula is:

2(*lw* + *wh* + *hl*)

where*l* is length, *w* is width and *h* is height of the prism.

E.g. to find the surface area of a rectangular prism 5 cm long, 3 cm wide and 2 cm high, substitute 5 for*l*, 3 for *w* and 2 for *h* in the above formula i.e. 2(5 x 3 + 3 x 2 + 2 x 5) which works out to 62cm^{2}.

A right circular cylinder is a solid, generated by the revolution of a rectangle about one of its sides which remain fixed. The curved surface area of a cylinder with radius*r* and height *h* is given by:

Curved surface area of a cylinder = 2?*rh*

Total surface area of a cylinder = 2?*rh* + 2?*r*^{2}

If a cylinder is closed at one end, then

Total surface area of a cylinder = 2?*rh* + ?*r*^{2}

E.g. to find the surface area of a closed right cylinder with radius 7 m and height is 10 m, substitute 7 for*r* and 10 for *h* in the above formula i.e. 2 x ? x 7 x 10 + 2 x ? x 7^{2}, which gives 238? m^{2}.

A right circular cone is a solid generated by the revolution of a right triangle about one of its legs. Slant height of a right circular cone is the distance from any point on the perimeter of the circular base to the vertex of the cone.

Slant height =*l* = ?(*r*^{2} + *h*^{2})

Area of curved surface = ?*rl*

Total surface area of a cone = ?*rl* + ?*r*^{2}

E.g. to find the surface area of a cone, with radius of 7 cm and height is 4 cm, first you have to find the slant height i.e.,*l* = ?(3^{2}+4^{2}) = 5. Now substitute the value of *r* and *l* in the above formula i.e. ? x 3 x 5 + ? x 3^{2}, which gives 24? cm^{2}.

A pyramid is a 3-dimensional figure, whose base is a polygon and triangular lateral faces that meet at vertex. E.g. pyramids that have a square base, have a total of five surfaces.

Surface area of a pyramid = Area of the base + Area of lateral faces E.g. to find the surface area of a square pyramid of base 20 m and slant height 40 m, the area of the base which in this case is a square =20^{2} i.e. 400 and area of lateral faces which is a triangle is =½ x ?20 x ?40 i.e. 400. Therefore,

Surface area of the pyramid = 400 + 4(400) = 2000 m^{2}.

In this section you'll learn how to calculate the surface area of common solids. It will be presented with the help of some examples, practice questions with solution and using video explanation in own handwriting by the instructor.

Any solid formed by joining the corresponding vertices of two congruent and parallel polygons with line segments is known as a prism. The base of a prism has two congruent polygons; lines connecting the corresponding vertices are lateral edges and they are parallel. The parallelograms formed by lateral edges are called lateral faces. In the real-world common application; as an example, it is a figure most people have personal experience of either wrapping or opening gifts.

In right prism, lateral edges are perpendicular to the bases of the prism and in oblique prism; lateral edges are not perpendicular to the bases of the prism.

The prism surface area formula is:

2(

where

E.g. to find the surface area of a rectangular prism 5 cm long, 3 cm wide and 2 cm high, substitute 5 for

A right circular cylinder is a solid, generated by the revolution of a rectangle about one of its sides which remain fixed. The curved surface area of a cylinder with radius

Curved surface area of a cylinder = 2?

Total surface area of a cylinder = 2?

If a cylinder is closed at one end, then

Total surface area of a cylinder = 2?

E.g. to find the surface area of a closed right cylinder with radius 7 m and height is 10 m, substitute 7 for

A right circular cone is a solid generated by the revolution of a right triangle about one of its legs. Slant height of a right circular cone is the distance from any point on the perimeter of the circular base to the vertex of the cone.

Slant height =

Area of curved surface = ?

Total surface area of a cone = ?

E.g. to find the surface area of a cone, with radius of 7 cm and height is 4 cm, first you have to find the slant height i.e.,

A pyramid is a 3-dimensional figure, whose base is a polygon and triangular lateral faces that meet at vertex. E.g. pyramids that have a square base, have a total of five surfaces.

Surface area of a pyramid = Area of the base + Area of lateral faces E.g. to find the surface area of a square pyramid of base 20 m and slant height 40 m, the area of the base which in this case is a square =20

Surface area of the pyramid = 400 + 4(400) = 2000 m

This FREE mini-lesson is a part of Winpossible's online course that covers all topics within Geometry. Click on the video below to go through it. If you like it, you can buy our online course in Geometry by clicking here.

- Mathematics > General
- Mathematics > Geometry
- Education > General

- Grade 9
- Grade 10
- Grade 11
- Grade 12

Solve real-world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms.

use previous knowledge of surface area to make connections to the formulas for lateral and total surface area and determine solutions for problems involving rectangular prisms, triangular prisms, and cylinders;

solve problems involving the lateral and total surface area of a rectangular prism, rectangular pyramid, triangular prism, and triangular pyramid by determining the area of the shape's net.