FIND MISSING ANGLES IN RHOMBUS

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Angles in Rhombus

  • Opposite angles are equal in size.
  • Diagonals bisect each other at right angles.
  • Diagonals bisect the angles at the vertex.

Find the value of x in each rhombus :

Problem 1 :

Solution :

By observing the figure,

The opposite angles in a rhombus are equal.

(-6 + 2x)º = 64º

-6 + 2x = 64

2x = 64 + 6

2x = 70

x = 70/2

x = 35

Problem 2 :

Solution :

By observing the figure,

The sum of angles of a linear pair is always equal to 180º.

(-10x + 79) + 41 = 180

-10x + 79 + 41 = 180

-10x + 120 = 180

-10x = 180 - 120

x = -60/10

x = -6

Problem 3 :

The sum of angles of a linear pair is always equal to 180.

130 + (36 + 7x) = 180

130 + 36 + 7x = 180

166 + 7x = 180

7x = 180 – 166

7x = 14

x = 14/7

x = 2

Problem 4 :

Solution :

The opposite angles in a rhombus are equal.

56 = (5x – 34)

56 = 5x – 34

56 + 34 = 5x

90 = 5x

90/5 = x

18 = x

Problem 5 :

Solution :

The sum of angles of a linear pair is always equal to 180.

 (40 - 5x) + 125 = 180

40 – 5x + 125 = 180

165 – 5x = 180

-5x = 180 – 165

-5x = 15

x = -15/5

x = -3

Problem 6 :

Solution :

The sum of angles of a linear pair is always equal to 180.

72 + 9x = 180

9x = 180 – 72

9x = 108

x = 108/9

x = 12

Problem 7 :

Solution :

The opposite angles in a rhombus are equal.

144º = (6x + 96)º

144º = 6xº + 96º

144 - 96 = 6x

48 = 6x

48/6 = x

8 = x

Problem 8 :

Solution :

The sum of angles of a linear pair is always equal to 180.

(x + 66) + 110 = 180

x + 66 + 110 = 180

x + 176 = 180

x = 180 – 176

x = 4

Problem 9 :

Using the properties of each shape to write and solve an algebraic equation for each picture.

Rhombus ABCD

properties-of-rhombus-q8.png

Solution :

Since the shape ABCD is a rhombus, the diagonals will intersect each other with 901 degree measure.

5x = 90

x = 90/5

x = 18

So, the value of x is 18.

Problem 10 :

STUV is a rhombus. What are the values of x and y?

properties-of-rhombus-q9.png

Solution :

In a rhombus, all four sides will be equal.

Property 1 :

Opposite sides will be equal.

2y = 8

y = 8/2

y = 4

Property 2 :

Adjacent sides will be equal.

6x - 4 = 8

6x = 8 + 4

6x = 12

x = 12/6

x = 2

So, the values of x and y are 2 and 4 respectively.

Problem 11 :

Using the properties of rhombuses, write and solve an algebraic equation for each picture.

properties-of-rhombus-q10.png

Property 1 :

Opposite sides and adjacent sides will be equal, then

2x + 13 = 5x + 4

2x - 5x = 4 - 13

-3x = -9

x = 9/3

x = 3

Problem 12 :

properties-of-rhombus-q11.png

Property 1 :

In rhombus, the diagonals will intersect each other at 90 degree.

3x + 60 = 90

3x = 90 - 60

3x = 30

x = 30/3

x = 10

So, the value of x is 10.

Problem 13 :

properties-of-rhombus-q12.png

Property  :

In rhombus, the diagonals will be the angle bisector.

2x + 20 = 3x + 10

2x - 3x = 10 - 20

-x = -10

x = 10

So, the value of x is 10.

Problem 14 :

Shown below are three rhombuses. Find the size of each missing angle.

properties-of-rhombus-q13.png

Solution :

Property 1 :

The opposite angles will be equal.

x = 98

Property 2 :

Co-interior angles will be supplementary.

y + 98 = 180

= 180 - 98

y = 82

z = 82

Problem 15 :

properties-of-rhombus-q14.png

Solution :

Property 1 :

The opposite angles will be equal.

y = 105

Property 2 :

Co-interior angles will be supplementary.

x + 105 = 180

x = 180 - 105

x = 75

z = 75

Problem 16 :

properties-of-rhombus-q15.png

Solution :

Property 1 :

The opposite angles will be equal.

x = 23

Property 2 :

Co-interior angles will be supplementary.

x + y = 180

23 + y = 180

y = 180 - 23

y = 157

z = 157

Problem 17 :

Use each RHOMBUS to find the specified lengths and measures.

properties-of-rhombus-q16.png

a) m∠D =_____

b) m∠DCB =_____

c) m∠1 =____

d) m∠2 =______

e) m∠3 =______

f) m∠4 =___

Solution :

a) Opposite angles are equal, then m∠D = 130

b) m∠DCB = ∠1 + ∠2

Sum of co-interior angles is 180 degree

m∠DCB + m∠CDA = 180

m∠DCB + 130 = 180

m∠DCB = 180 - 130

= 50

c) m∠1 =____

∠2 = ∠3 (equal sides will make equal angles)

130 + ∠2 + ∠3 = 180

130 + 2∠2 = 180

2∠2 = 180 - 130

2∠2 = 50

∠2 = 50/2

∠2 = 25

m∠1 = 25 (angle bisector)

d) m∠2 = 25

e) m∠3 = 25

f) m∠4 = 25

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