GEOMETRY PROPERTIES OF PARALLELOGRAMS

Problem 1 :

In the parallelogram shown below, find the value of x.

Opposite angles are equal.

20x + 3 = 83

20x = 83 - 3

20x = 80

x = 80/20

x = 4

Problem 2 :

Quadrilateral RSTU is a parallelogram. Find the values of x, y, a, and b.

Solution :

Here some angles are missing and some side lengths is also missing.

In parallelogram, opposite sides are equal, opposite angles are equal and consecutive interior angles add upto 180 degree.

80 and y are consecutive interior angles.

80 + y = 180

y = 180 - 80

y = 100

x = 80 (opposite angles)

b = 9 and a = 6

Problem 3 :

Solution :

y = 35

∠SRU + ∠RUT = 180

x + 45 + 35 = 180

x + 80 = 180

x = 180 - 80

x = 100

Problem 4 :

Solution :

Opposite sides are equal.

Problem 5 :

Solution :

Diagonals will bisect each other.

Problem 6 :

Solution :

Sum of consecutive angles = 180

68x - 1 + 22x + 1 = 180

90x = 180

x = 180 / 90

x = 2

Problem 7 :

Solution :

∠TUV = ∠TSV

43x - 1 = 85

43x = 85 + 1

43x = 86

x = 86/43

x = 2

Problem 8 :

Solution :

MK = ML

2 + 2x = 16

2x = 16 - 2

2x = 14

x = 14/2

x = 7

Problem 9 :

Find angle F.

Solution :

Sum of consecutive angles = 180

16x + 1 + 35 = 180

16x + 36 = 180

16x = 180 - 36

16x = 144

x = 144/16

x = 9

∠F + ∠C = 180

∠F + 35 = 180

∠F = 180 - 35

∠F = 145

Problem 10 :

Find angle measure R.

Solution :

Sum of consecutive angles = 180

14x + 5 + 5 + 20x = 180

34x  + 10 = 180

34x = 180 - 10

34x = 170

x = 170/34

x = 5

Problem 11 :

In the diagram of the parking lot shown, m∠JKL = 60°, JK = LM = 21 feet, and KL = JM = 9 feet.

a) Explain how to show that parking space JKLM is a parallelogram.

b) Find m∠JML, m∠KJM, and m∠KLM.

c) LM || NO and NO || PQ . Which theorem could you use to show that JK II PQ ?

Solution :

a) Given that, JK = LM = 21 feet, and KL = JM = 9 feet

Since the opposite sides are equal, it must be a parallelogram.

b) Given that, m∠JML = 60

m∠JML = 60(Opposite angles are equal)

m∠KJM :

m∠KJM and m∠JML are co-interior angles. Then they add upto 180 degree.

m∠KJM = 180 - ∠JML

= 180 - 60

= 120

m∠KLM = 120 (since opposite angles are equal)

c) LM || NO and NO || PQ

Using transitive property of parallel line, we say that JK II PQ.

Problem 12 :

Describe how to prove that ABCD is a parallelogram.

Solution :

In triangle ADB and triangle DCB,

∠DAB = ∠DBC

AD = BC (opposite sides)

DE = DB (Common)

Triangle ABD and triangle BDC are congruent. 

Using CPCTC, AB = DC

Since opposite sides are equal, then it is a parallelogram.

Problem 13 :

Quadrilateral JKLM is a parallelogram. Describe how to prove that △MGJ ≅ △KHL.

Solution :

In △MGJ and △KHL

∠MGJ = ∠KHL

JKLM is a parallelogram,

JM = KL

∠JML = ∠JKL (opposite angles in parallelogram)

∠MJK = ∠KLM (opposite angles in parallelogram)

∠GJM = 180 - ∠MJK

∠KLH = 180 - ∠KLH

∠GJM = ∠KLH

GM = KH

Using SAS, then △MGJ ≅ △KHL.

Problem 14 :

Three interior angle measures of a quadrilateral are 67°, 67°, and 113°. Is this enough information to conclude that the quadrilateral is a parallelogram? Explain your reasoning.

Solution :

Sum of angles in a quadrilateral = 360

Let x be the missing angle measure.

67 + 67 + 113 + x = 360

247 + x = 360

x = 360 - 247

x = 113

Since the opposite angles are equal, then it is parallelogram.