5.C: Surface Area

Last updated
Save as PDF

Page ID: 31464

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Overview

The purpose of this lesson is to learn how to find the surface area of a solid.

This lesson will address the following CCRS Standard(s) for Geometry:

7.G.6: 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

Directions

Take notes while watching videos below
Go to http://wamap.org and log into our course to complete assignment 5.B with 80% or better.

Watch

Surface Area [9:38]

Do

Complete assignment 5.B with 80% or better at http://wamap.org

Summary

In this lesson we have learned:

If P is the perimeter of the base of a prism (or circumference of the base of a cylinder), and B is the area of the base, then the surface area of the prism (or cylinder) is $A = 2 B + P l$ (where l is the slant height)
If P is the perimeter of the base of a pyramid (or circumference of the base of a cone), and B is the area of the base, then the surface area of the pyramid (or cone) is $A = B + \frac{1}{2} P l$ (where l is the slant height)
The surface area of a sphere is $A = 4 π$ </mi> <msup> <mi>r</mi> <mn>2</mn> </msup> </math>' data-equation-content="A=4\pi r^2">