Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 9 - Section 9.1 - Prisms, Area, and Volume - Exercises - Page 407: 20

Answer

$V = 1080~in^3$ $A = 636~in^2$

Work Step by Step

We can let the dimensions of the closed box be: $W = 9~in$ $L = 10~in$ $H = 12~in$ We can find the volume: $V = W \times L \times H = 9~in \times 10~in \times 12~in$ $V = 1080~in^3$ We can find the surface area: $A = (2)(W \times L)+(2)(W \times H)+(2)(H \times L)$ $A = (2)(9~in \times 10~in)+(2)(9~in \times 12~in)+(2)(12~in \times 10~in)$ $A = 180~in^2+216~in^2+240~in^2$ $A = 636~in^2$
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