Chapter 12.1, Problem 45E

(a)

To determine

To find: Thetwo possible values for $x$ when the perimeter of given triangle $ABC$ is $32$ .

Answer to Problem 45E

The two possible values are: $x=9\text{ and }x=8.$

Explanation of Solution

Given:

The triangle $ABC$ is isosceles triangle.

The perimeter of triangle $ABC$ is $32$ and $AB=x$ $\text{and}$ $BC=2x-4$ .

Concept used:

Sum of all side of triangle is perimeter of triangle.

Isosceles triangle has $2$ same sides.

Calculation:

According to the given the triangle is isosceles:

First let $AB=AC=x$ .

Perimeter of the triangle is $32$ .

  $\begin{array}{l}x+x+2x-4=32.\\ 4x=36.\\ x=9.\end{array}$

Second let $BC=AC=2x-4$ .

Perimeter of the triangle is $32$ .

  $\begin{array}{l}2x-4+2x-4+x=32.\\ 5x-8=32.\\ 5x=40.\\ x=8.\end{array}$

Hence, two possible are: $x=9\text{ and }x=8.$

(b)

To determine

To find:The two possible values for $x$ when the perimeter of given triangle $ABC$ is $12$ .

Answer to Problem 45E

There is only one solution for $x$ when perimeter is $12$ .

Explanation of Solution

Given:

The triangle $ABC$ is isosceles triangle.

The perimeter of triangle $ABC$ is $32$ and $AB=x$ $\text{and}$ $BC=2x-4$ .

Concept used:

Sum of all side of triangle is perimeter of triangle.

Isosceles triangle has $2$ same sides.

Calculation:

According to the given the triangle is isosceles:

First let $AB=AC=x$ .

Perimeter of the triangle is $12$ .

  $\begin{array}{l}x+x+2x-4=12.\\ 4x=16.\\ x=4.\end{array}$

Second let $BC=AC=2x-4$ .

Perimeter of the triangle is $12$ .

  $\begin{array}{l}2x-4+2x-4+x=12.\\ 5x-8=12.\\ 5x=20.\\ x=4.\end{array}$

Hence, there is only one solution for $x$ when perimeter is $12$ .