Chapter 8.2, Problem 51HP

To determine

To draw:

Right triangle with side lengths that form a Pythagorean triple for given condition.

Explanation of Solution

Given information:

Condition:

If you double the length of each side, is the resulting triangle acute, right or obtuse. If you length of each side.

Sketch:

  

Interpretation:

A Pythagorean triple is a set of three nonzero whole numbers $a,b,c$ such that $a^2+b^2=c^2$ .

Common Pythagorean triples are:

  $\left(3x,4x,5x\right),\left(5x,12x,13x\right),\left(8x,15x,17x\right),\left(7x,24x,25x\right)$

So with the help of the triple $\left(3,4,5\right)$ above figure is drawn.

If the dimensions are doubled then the side lengths will be $6,8$ and $10$ .

If the sum of the squares of the lengths of the shortest sides of a triangle is equal to the square of the length of the longest side, then the triangle is a right triangle.

Now, classify the triangle by comparing the square of the longest side to sum of the squares of the other two sides.

Now, find the value of $c^2$

  $\begin{array}{l}{c}^2={10}^2\\ =100\end{array}$

Now, find the value of $a^2+b^2$

  $\begin{array}{l}{a}^2+b^2=6^2+8^2\\ =36+64\\ =100\end{array}$

If the dimensions are halved, then the side lengths will be $1.5,2,2.5$

Now, find the value of $c^2$

  $\begin{array}{l}{a}^2+b^2={1.5}^2+2^2\\ =2.25+4\\ =6.25\end{array}$

Since, $c^2=a^2+b^2$ , converse of Pythagorean Theorem shows that the new triangle is a right triangle.