Chapter 3, Problem 18CCSR

To determine

Minimum number of tiles needed to cover the tabletop.

Answer to Problem 18CCSR

Minimum number of tiles to cover tabletop is 9216

Explanation of Solution

Given:

Dimension of tabletop is 2 ft long and 2 ft wide.

Dimension of tile is $\frac{1}{4}\text{in }$ long and $\frac{1}{4}\text{in }$ wide,

It is known that 1 ft = 12 in

Therefore, area of tabletop is,

  $\begin{array}{l}A=2\text{ ft }\times \text{2 ft}\\ \text{=2}\times \left(12\text{ in}\right)\times 2\times \left(12\text{ in}\right)\\ =576{\text{ in}}^2\end{array}$

Area of one tile is,

  $\begin{array}{l}A=\frac{1}{4}\text{ in }\times \frac{1}{4}\text{ in}\\ =\frac{1}{16}{\text{ in}}^2\end{array}$

Let n is the minimum number of tiles required to cover tabletop.

Minimum number of tiles is given by ratio of area of tabletop to area of one tile.

Therefore, minimum number of tiles are,

  $\begin{array}{l}n=\frac{576}{\frac{1}{16}}\\ =576\times 16\\ =9216\end{array}$