Chapter 8.3, Problem 12WE

To determine

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

Answer to Problem 12WE

Obtuse

Explanation of Solution

Given:

The figure:

$AC=10$ , $BC=17$ and $CD=8$

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,

  $\begin{array}{c}A{C}^2=A{D}^2+C{D}^2\\ {10}^2=A{D}^2+8^2\\ 100=A{D}^2+64\end{array}$

  $\begin{array}{l}A{D}^2=100-64\\ A{D}^2=36\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{A{D}^2}=\sqrt{36}\\ AD=6\end{array}$

Now consider the right triangle $△BDC$ .

So, by theorem 8-2,

  $\begin{array}{c}B{C}^2=B{D}^2+C{D}^2\\ {17}^2=B{D}^2+8^2\\ 289=B{D}^2+64\end{array}$

  $\begin{array}{l}B{D}^2=289-64\\ B{D}^2=225\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{B{D}^2}=\sqrt{225}\\ BD=15\end{array}$

This implies that,

  $\begin{array}{c}AB=AD+DB\\ =6+15\\ =21\end{array}$

Now, consider the triangle $△ABC$

Let the longest side of the triangle $△ABC$ be c, here $c=AB=21$

Now,

  $\begin{array}{c}{c}^2={21}^2\\ =441\end{array}$

Now find sum of the squares of other two sides $a=AC=10$ , $b=BC=17$ , then

  $\begin{array}{c}{a}^2+b^2={10}^2+{17}^2\\ =100+289\\ =389\end{array}$

It can be observed that,

  $441>389$

That is,

  $c^2>a^2+b^2$

So, by theorem 8-5, $△ABC$ is obtuse.