Chapter 8.3, Problem 9WE

To determine

To prove: $△RST$ is a right triangle.

Explanation of Solution

Given:

The figure:

  $m\angle UTS=90$

Concept Used:

Theorem 8-2: The Square of the hypotenuse is equal to the sum of the squares of the base and perpendicular in a right triangle.

Theorem 8-3: If $c^2=a^2+b^2$ , then $m\angle C=90$ and $△ABC$ is right.

Consider the given figure:

Consider the right triangle $△STU$ .

So, by theorem 8-2,

  $\begin{array}{c}S{U}^2=S{T}^2+T{U}^2\\ {13}^2=S{T}^2+{12}^2\\ 169=S{T}^2+144\end{array}$

  $\begin{array}{c}169-144=S{T}^2\\ 25=S{T}^2\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{S{T}^2}=\sqrt{25}\\ ST=5\end{array}$

Now, consider the triangle $△RST$

Let the longest side of the triangle $△RST$ be c, here $c=ST=5$

Now,

  $\begin{array}{c}{c}^2=5^2\\ =25\end{array}$

Now find sum of the squares of other two sides $a=SR=4$ , $b=RT=3$ then

  $\begin{array}{c}{a}^2+b^2=4^2+3^2\\ =16+9\\ =25\end{array}$

It can be observed that,

  $25=25$

That is,

  $c^2=a^2+b^2$

So, by theorem 8-3, $△RST$ is right.

And $m\angle S=90$