Chapter 8.3, Problem 11WE

To determine

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

Answer to Problem 11WE

Acute

Explanation of Solution

Given:

The figure:

$AC=13$ , $BC=15$ and $CD=12$

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 8-4: If $c^2<a^2+b^2$ , then $m\angle C<90$ and $△ABC$ is acute.

Consider the given below:

Consider the right triangle $△ADC$ .

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=A{D}^2+C{D}^2\\ {13}^2=A{D}^2+{12}^2\\ 169=A{D}^2+144\end{array}$

  $\begin{array}{l}A{D}^2=169-144\\ A{D}^2=25\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{A{D}^2}=\sqrt{25}\\ AD=5\end{array}$

Now consider the right triangle $△BDC$ .

So, by theorem 8-2,

  $\begin{array}{c}B{C}^2=B{D}^2+C{D}^2\\ {15}^2=B{D}^2+{12}^2\\ 225=B{D}^2+144\end{array}$

  $\begin{array}{l}B{D}^2=225-144\\ B{D}^2=81\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{B{D}^2}=\sqrt{81}\\ BD=9\end{array}$

This implies that,

  $\begin{array}{c}AB=AD+DB\\ =5+9\\ =14\end{array}$

Now, consider the triangle $△ABC$

Let the longest side of the triangle $△ABC$ be c, here $c=BC=15$

Now,

  $\begin{array}{c}{c}^2={15}^2\\ =225\end{array}$

Now find sum of the squares of other two sides $a=AB=14$ , $B=AC=13$ , then

  $\begin{array}{c}{a}^2+b^2={14}^2+{13}^2\\ =196+169\\ =365\end{array}$

It can be observed that,

  $225<365$

That is,

  $c^2<a^2+b^2$

So, by theorem 8-4, $△ABC$ is acute.