Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 2 - Solving Equations - 2-5 Literal Equations and Formulas - Practice and Problem-Solving Exercises - Page 114: 49



Work Step by Step

To find the surface area, we must add the area of each side. The area of the top and bottom is $s\times s$, and the area of each side is $s\times h$. When we add these together we get: $A=s\times s+s\times s+s\times h+s\times h+s\times h+s\times h$ Combine like terms. $A=2(s\times s)+4(s\times h)$ or $A=2s^{2}+4sh$ We are solving for $h$, so we must get $h$ alone. Subtract $2s^{2}$ from both sides. $A-2s^{2}=4sh$ Divide by $4s$. $\frac{A-2s^{2}}{4s}=h$ Plug in the given values: 760 for $A$ and 10 for $s$. $\frac{760-2(10)^{2}}{4(10)}=h$ Solve. $14=h$ Part c tells us that $s=h$. To solve this, substitute $s$ for $h$. $A=2s^{2}+4s^{2}$ Add. $A=6s^{2}$
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