Chapter 8.3, Problem 13WE

To determine

To find: if the triangle $△ABC$ is acute, right or obtuse.

Answer to Problem 13WE

Obtuse

Explanation of Solution

Given:

The figure:

$AC=13$ , $BC=\sqrt{34}$ and $CD=3$

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 8-5: If $c^2>a^2+b^2$ , then $m\angle C>90$ and $△ABC$ is obtuse.

Consider the given below:

Consider the right triangle $△ADC$ .

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=A{D}^2+C{D}^2\\ {13}^2=A{D}^2+3^2\\ 169=A{D}^2+9\end{array}$

  $\begin{array}{l}A{D}^2=169-9\\ A{D}^2=160\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{A{D}^2}=\sqrt{160}\\ AD=4\sqrt{10}\end{array}$

Now consider the right triangle $△BDC$ .

So, by theorem 8-2,

  $\begin{array}{c}B{C}^2=B{D}^2+C{D}^2\\ {\left(\sqrt{34}\right)}^2=B{D}^2+3^2\\ 34=B{D}^2+9\end{array}$

  $\begin{array}{l}B{D}^2=34-9\\ B{D}^2=25\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{B{D}^2}=\sqrt{25}\\ BD=5\end{array}$

This implies that,

  $\begin{array}{c}AB=AD+DB\\ =4\sqrt{10}+5\\ =12.65+5\\ =17.65\end{array}$

Now, consider the triangle $△ABC$

Let the longest side of the triangle $△ABC$ be c, here $c=AB=17.65$

Now,

  $\begin{array}{c}{c}^2={\left(17.65\right)}^2\\ =311.52\end{array}$

Now find sum of the squares of other two sides $a=AC=13$ , $b=BC=\sqrt{34}$ then

  $\begin{array}{c}{a}^2+b^2={13}^2+{\left(\sqrt{34}\right)}^2\\ =169+34\\ =203\end{array}$

It can be observed that,

  $311.52>203$

That is,

  $c^2>a^2+b^2$

So, by theorem 8-5, $△ABC$ is obtuse.