a.
To calculate:Thevalue ofxin parallelogram JKMO.
a.

Answer to Problem 9PSA
The value ofxis
Explanation of Solution
Given information:
In a parallelogram JKMO,
OJ= x+ 5,
KM = y - 3,
JK = 2x - 4 .
Calculation:
JKMO is a parallelogram.
The diagonals bisect two
From Equation 1 and Equation 2, we get
b.
To calculate: The value of yin parallelogram JKMO.
b.

Answer to Problem 9PSA
The value of y is
Explanation of Solution
Given information:
In a parallelogram JKMO,
OJ = x+ 5,
KM = y - 3,
JK = 2x−4.
Calculation:
JKMO is a parallelogram.
The diagonals bisect two angles. JKMO is also a rhombus.
From Equation 1 and Equation 2, we get
Substituting the value of x in Equation 1, we get
c.
To calculate: The perimeter of parallelogram JKMO.
c.

Answer to Problem 9PSA
The perimeter is
Explanation of Solution
Given information:
In a parallelogram JKMO,
OJ = x+ 5,
KM = y - 3,
JK = 2x− 4.
Calculation:
JKMO is a parallelogram.
The diagonals bisect two angles. JKMO is also a rhombus.
From Equation 1 and Equation 2, we get
Substituting the value of x in Equation 1, we get
OJ = x+ 5
JK = 2x - 4
Perimeter is sum of all sides.
Chapter 5 Solutions
Geometry For Enjoyment And Challenge
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