Chapter 8.3, Problem 14WE

To determine

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

Answer to Problem 14WE

Right

Explanation of Solution

Given:

The figure:

$AD=2$ , $DB=8$ and $CD=4$

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 below:

Consider the right triangle $△ADC$ .

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=A{D}^2+C{D}^2\\ A{C}^2=2^2+4^2\\ A{C}^2=4+16\\ A{C}^2=20\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{A{C}^2}=\sqrt{20}\\ AC=2\sqrt{5}\end{array}$

Now consider the right triangle $△BDC$ .

So, by theorem 8-2,

  $\begin{array}{c}B{C}^2=B{D}^2+C{D}^2\\ B{C}^2=8^2+4^2\\ B{C}^2=64+16\\ B{C}^2=80\end{array}$

Taking square root both sides,

  $\begin{array}{c}\sqrt{B{C}^2}=\sqrt{80}\\ BC=4\sqrt{5}\end{array}$

Now it can be observed that,

  $\begin{array}{c}AB=AD+DB\\ =2+8\\ =10\end{array}$

Now, consider the triangle $△ABC$

Let the longest side of the triangle $△ABC$ be c, here $c=AB=10$

Now,

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

Now find sum of the squares of other two sides $a=AC=2\sqrt{5}$ , $b=BC=4\sqrt{5}$ then

  $\begin{array}{c}{a}^2+b^2={\left(2\sqrt{5}\right)}^2+{\left(4\sqrt{5}\right)}^2\\ =20+80\\ =100\end{array}$

It can be observed that,

  $100=100$

That is,

  $c^2=a^2+b^2$

So, by theorem 8-5, $△ABC$ is right.