Chapter 4.6, Problem 46STP

To determine

To choose:the approximate distance.

Answer to Problem 46STP

The correct answer is C 39.6 ft.

Explanation of Solution

Given information :

A surveyor determined the distance across a field was $\sqrt{1568}$ feet

The first perfect square less than 1568 is 1521

The first perfect square greater than 1568 is 1600.

  $\begin{array}{l}1521<1568<1600\\ {39}^2<1568<{40}^2\\ \sqrt{{39}^2}<1568<\sqrt{{40}^2}\\ 39<\sqrt{1568}<40\end{array}$

The square root of 1568 is between the integers 39 and 40.Since 1568 is closer to 1521 than to 1600.

Therefore, $\begin{array}{l}\sqrt{1568}\approx 39\\ \sqrt{1568}\approx 39.6\left[\text{use calculator}\right]\end{array}$

Hence, the correct answer is C 39.6 ft.