Chapter 5.6, Problem 9PSA

a.

To determine

To calculate:Thevalue ofxin parallelogram JKMO.

Answer to Problem 9PSA

The value ofxis$9$ .

Explanation of Solution

Given information:

In a parallelogram JKMO,

  $\overleftrightarrow{JM}$ bisects angle OJK and angle OMK.

OJ= x+ 5,

KM = y - 3,

JK = 2x - 4 .

Calculation:

  

JKMO is a parallelogram.

  $\overline{OJ}\cong \overline{KM}$

  $x+5=y-3----Equation1$

The diagonals bisect two angles.JKMO is also a rhombus.

  $\overline{JK}\cong \overline{KM}$

  $2x-4=y-3----Equation2$

From Equation 1 and Equation 2, we get

  $\begin{array}{l}x+5=2x-4\\ x=9\end{array}$

b.

To determine

To calculate: The value of yin parallelogram JKMO.

Answer to Problem 9PSA

The value of y is $17$ .

Explanation of Solution

Given information:

In a parallelogram JKMO,

  $\overleftrightarrow{JM}$ bisects angle OJK and angle OMK .

OJ = x+ 5,

KM = y - 3,

JK = 2x−4.

Calculation:

  

JKMO is a parallelogram.

  $\overline{OJ}\cong \overline{KM}$

  $x+5=y-3----Equation1$

The diagonals bisect two angles. JKMO is also a rhombus.

  $\overline{JK}\cong \overline{KM}$

  $2x-4=y-3----Equation2$

From Equation 1 and Equation 2, we get

  $\begin{array}{l}x+5=2x-4\\ x=9\end{array}$

Substituting the value of x in Equation 1, we get

  $\begin{array}{l}9+5=y-3\\ y=17\end{array}$

c.

To determine

To calculate: The perimeter of parallelogram JKMO.

Answer to Problem 9PSA

The perimeter is $56$ .

Explanation of Solution

Given information:

In a parallelogram JKMO,

  $\overleftrightarrow{JM}$ bisects angle OJK and angle OMK .

OJ = x+ 5,

KM = y - 3,

JK = 2x− 4.

Calculation:

  

JKMO is a parallelogram.

  $\overline{OJ}\cong \overline{KM}$

  $x+5=y-3----Equation1$

The diagonals bisect two angles. JKMO is also a rhombus.

  $\overline{JK}\cong \overline{KM}$

  $2x-4=y-3----Equation2$

From Equation 1 and Equation 2, we get

  $\begin{array}{l}x+5=2x-4\\ x=9\end{array}$

Substituting the value of x in Equation 1, we get

  $\begin{array}{l}9+5=y-3\\ y=17\end{array}$

OJ = x+ 5

  $OJ=9+5=14$

JK = 2x - 4

  $JK=\left(2\times 9\right)-4=14$

Perimeter is sum of all sides.

  $\begin{array}{l}P=OJ+JK+KM+OM\\ P=14+14+14+14\\ P=56\end{array}$