Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 10 - Geometry - 10.4 Area and Circumference - Exercise Set 10.4 - Page 648: 38

Answer

The cost of plastering \[60\text{ft}\text{.}\] of wall in a house with \[9\text{ft}\text{.}\] of ceiling is\[\$1080\].

Work Step by Step

To find the cost of plastering the wall, we need to find the area of the wall (in square yards) and multiply it with the cost per square yard. It is known that 1 feet is equal to\[{1}/{3\text{ yd}}\;\] and same can be written in fraction as follows: \[\text{1 ft}=\frac{1}{3}\text{ yd}\] Convert 60 ft into yd as follows: \[\begin{align} & 60\text{ ft}=\frac{60\text{ ft}}{1}\times \frac{1\text{ yd}}{3\text{ ft}} \\ & =\frac{60\text{ ft}}{3\text{ ft}} \\ & =20\text{ ft} \end{align}\] Convert 9 ft into yd as follows: \[\begin{align} & \text{9 ft}=\frac{\text{9ft}}{1}\times \frac{1\text{ yd}}{3\text{ ft}} \\ & =\frac{\text{9 ft}}{3\text{ ft}} \\ & =3\text{ ft} \end{align}\] Therefore, the area of the wall (in square yards) will be as follows: \[\begin{align} & A=lb \\ & =\text{20 yd}\times 3\text{ yd} \\ & =\text{60 yd}{{\text{.}}^{\text{2}}} \end{align}\] The cost of plastering (denoted by C) the wall will be: \[\begin{align} & C=A\times \$18\\&=60\text{y}{{\text{d}}^{2}}\times\$18\\&=\$1080\end{align}\] Hence, the cost of plastering \[60\text{ft}\text{.}\] of wall in a house with\[9\text{ft}\text{.}\] of ceiling.
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