Chapter 10 - Geometry - 10.4 Area and Circumference - Exercise Set 10.4 - Page 648: 43

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(a) To calculate the area of the house that needs to be painted, we will find out the total area and subtract the area of the doors and windows that do not need painting. The total area consists of the four walls and two triangular structures with base and heights as \[50\text{ ft}\] and \[10\text{ ft}\] respectively. Compute the total area as follows: \[\begin{align} & {{A}_{1}}=2\left( 40\text{ ft}\times 20\text{ ft} \right)+2\left( 50\text{ ft}\times 20\text{ ft} \right)+2\left( \frac{1}{2}\times 40\text{ ft}\times 10\text{ ft} \right) \\ & =\text{4000 f}{{\text{t}}^{2}} \end{align}\] Compute the area of four windows of dimensions \[8\text{ ft}\]by \[5\text{ ft}\] as follows: \[\begin{align} & {{A}_{2}}=4\left( 8\text{ ft}\times 5\text{ ft} \right) \\ & =\text{16}0\text{ f}{{\text{t}}^{2}} \end{align}\] Compute the area of two windows of dimensions \[30\text{ ft}\]by \[2\text{ ft}\] as follows: \[\begin{align} & {{A}_{3}}=2\left( 30\text{ ft}\times 2\text{ ft} \right) \\ & =\text{12}0\text{ f}{{\text{t}}^{2}} \end{align}\] Compute the area of two doors of dimensions \[80\text{ inch}\]by \[36\text{ inch}\]as follows: \[\begin{align} & {{A}_{4}}=2\left( \frac{80}{12}\text{ft}\times \frac{36}{3}\text{ft} \right) \\ & =40\text{ f}{{\text{t}}^{2}} \end{align}\] Compute the area that is required to be painted as follows: \[\begin{align} & A={{A}_{1}}-\left( {{A}_{2}}+{{A}_{3}}+{{A}_{4}} \right) \\ & =4000\text{ f}{{\text{t}}^{2}}-\left( 160\text{ f}{{\text{t}}^{2}}+120\text{ f}{{\text{t}}^{2}}+40\text{ f}{{\text{t}}^{2}} \right) \\ & =\text{3680 f}{{\text{t}}^{2}} \end{align}\] Hence, the area required to be painted is\[3680\text{ f}{{\text{t}}^{2}}\]. (b) If we apply one coat, the number of cans can be found out by dividing the total area with the area that one can cover. \[\frac{3680\text{ f}{{\text{t}}^{2}}}{500\text{ f}{{\text{t}}^{2}}}=7.36\text{ cans}\] If two coats are applied on the house, the number of cans will be doubled. Therefore, the numbers of cans required are: \[\begin{align} & 7.36\times 2=14.72 \\ & \approx 15\text{ cans} \end{align}\] Hence, the total number of cans required to paint the house with two coats of paint is\[15\]. (c) As we are using \[15\] cans for painting the house, and each can is of one gallon, costing\[\$26.95\], therefore the cost of painting the house will be: \[\begin{align} & \text{Cost}=15\times \$26.95\\&=\$404.25\end{align}\] Hence, the cost of painting the house is\[\$404.25\].