Chapter 9, Problem 28RP

To determine

Tocalculate:The distance a treasure is far from the old stump.

Answer to Problem 28RP

The treasure is far from old stump by 51 paces.

Explanation of Solution

Given information:

  $\begin{array}{l}DistanceSE=30paces,\\ DistanceNE=20paces,\\ DistanceMW=25paces.\end{array}$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Horizontal distance $=30-6=24$

Vertical distance $=20+25=45$

The final distances are 24 paces east and 45 paces north.

The distance x can be calculated by applying Pythagoras Theorem.

In a right angle $\Delta$ ,we get

  $\begin{array}{l}{\left(x\right)}^2={\left(45\right)}^2+{\left(24\right)}^2\\ x^2=2025+576\\ x^2=2601\\ x=\sqrt{2601}\\ x=51\end{array}$

The treasure is far from old stump by 51 paces.