Chapter 8.5, Problem 19PSB

a.

To determine

To find the perimeter of the triangle.

Answer to Problem 19PSB

The perimeter of the triangle is 24 cm.

Explanation of Solution

Given information:

One side of a triangle is 4cm longer than another side. The ray bisecting the angle formed by these sides divides the opposite side into 5 cm and 3 cm segments.

Formula used:

The below theorem is used:

Angle bisector theorem states that an angle bisector of a triangle will divide the opposite side into two segments that are proportional to the other two sides of the triangle.

Calculation:

  

Using Angle Bisector Theorem, we get

  $\begin{array}{l}\frac{y}{3}=\frac{4+y}{5}\\ 5y=12+3y\\ 2y=12\\ y=6\end{array}$

Perimeter of ΔABC

  $\begin{array}{l}=y+(y+4)+(3+5)\\ =6+(6+4)+(3+5)\\ =24\end{array}$

b.

To determine

To find the perimeter of the triangle in terms of x.

Answer to Problem 19PSB

The perimeter of the triangle is $4x+8$ .

Explanation of Solution

Given information:

One side of a triangle is x cm longer than another side. The ray bisecting the angle formed by these sides divides the opposite side into 5 cm and 3 cm segments.

Formula used:

The below theorem is used:

Angle bisector theorem states that an angle bisector of a triangle will divide the opposite side into two segments that are proportional to the other two sides of the triangle.

Calculation:

  

Using Angle Bisector Theorem, we get

  $\begin{array}{l}\frac{y}{3}=\frac{y+x}{5}\\ 3(y+x)=5y\\ 3y+3x=5y\\ 3x=2y\\ y=\frac{3x}{2}\end{array}$

Perimeter of ΔABC

  $\begin{array}{l}=y+(y+x)+(3+5)\\ =\frac{3x}{2}+(\frac{3x}{2}+x)+(3+5)\\ =4x+8\end{array}$