Thinking Mathematically (6th Edition)

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

Chapter 10 - Geometry - 10.5 Volume and Surface Area - Exercise Set 10.5 - Page 658: 57

Answer

The volume of the darkly shaded region is \[168\text{ c}{{\text{m}}^{\text{3}}}\].

Work Step by Step

To determine the volume of the darkly shaded region, first thing required to compute the area of the entire cube. Second, compute the volume of the unshaded part that is a pyramid. Finally, compute the shaded part by deducting the volume of the pyramid from the volume of the entire rectangular solid. Compute the volume of the rectangular solid using the equation as shown below: \[\begin{align} & \text{Volume of rectangular solid}=lbh \\ & =\left( 6\text{ cm}\times 6\text{ cm}\times 7\text{ cm} \right) \\ & =252\text{ c}{{\text{m}}^{\text{3}}} \end{align}\] The volume of the rectangular solid is \[8\text{ c}{{\text{m}}^{3}}\]. Now, to compute the volume of the pyramid, compute the area of the square base using the equation as shown below: \[\begin{align} & \text{Area of square base}\left( \text{B} \right)={{a}^{2}} \\ & ={{\left( 6 \right)}^{2}}\text{c}{{\text{m}}^{\text{2}}} \\ & =36\text{ c}{{\text{m}}^{2}} \end{align}\] Thus, the area of the square base is \[36\text{ c}{{\text{m}}^{2}}\]. Now, compute the volume of the unshaded portion that is pyramid using the equation as shown below: \[\begin{align} & \text{Volume of the unshaded pyramid part}=\frac{1}{3}Bh \\ & =\left( \frac{1}{3}\times 36\text{ c}{{\text{m}}^{2}}\times 7\text{ cm} \right) \\ & =84\text{ c}{{\text{m}}^{\text{3}}} \end{align}\] Now, compute the volume of the darkly shaded region using the equation as shown below: \[\begin{align} & \text{Volume of unshaded pyramid part}=\text{Volume of entire rectangular } \\ & \text{solid}-\text{Volume of unshaded } \\ & \text{pyramid part} \\ & =\left( 252\text{ c}{{\text{m}}^{\text{3}}}-84\text{ c}{{\text{m}}^{\text{3}}} \right) \\ & =168\text{ c}{{\text{m}}^{\text{3}}} \end{align}\] Hence, the volume of the darkly shaded region is \[168\text{ c}{{\text{m}}^{\text{3}}}\].
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