Rhombus and its Theorems
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In this section we will discuss rhombus and its theorems.Rhombus is a parallelogram with all sides equal and parallel.
Rhombus and its Theorems :
Theorem 1 : The diagonals of a rhombus are perpendicular to each other.
Given : A rhombus ABCD whose diagonals are AC and BD intersect at O.
Prove that : ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90 0
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| 1) ABCD is a rhombus. | 1) Given |
| 2) AB = BC = CD = DA | 2) Properties of rhombus. |
| 3) OB = OD and OA = OC | 3) As Parallelogram is a rhombus so diagonal bisect each other. |
| 4) BO = OD | 4) From (3) |
| 5) BC = DC | 5) Properties of rhombus. |
| 6) OC = OC | 6) Reflexive (common side) |
| 7) ΔBOC ≅ ΔDOC | 7) SSS Postulate. |
| 8) ∠BOC = ≅ ∠DOC | 8) CPCTC |
| 9)∠BOC + ∠DOC = 180 | 9) Linear pair angles are supplementary. |
| 10) 2∠BOC = 180 | 10) Addition property |
| 11) ∠BOC = 90 | 11) Division property |
| 12) ∠BOC = ∠DOC = 90 | 12) As these two angles are congruent. |
Hence, ∠AOB = ∠BOC = ∠COD = ∠DOA = 90 0
Converse of the above theorem is also true ⇒ If the diagonals of a parallelogram are perpendicular, then it is a rhombus.
Example :
1) ABCD is a rhombus with ∠ABC = 56. Determine ∠ACD.
Solution : ABCD is a rhombus.
∠ ABC = ∠ADC ( Opposite angles are equal)
∠ADC = 56 0
∴ ∠ ODC = ½ ∠ADC ( Diagonals of rhombus bisects the angle)
⇒ ∠ODC = ½ x 56
⇒ ∠ODC = 28 0
∠OCD + ∠ODC + ∠COD = 180 ( In ΔOCD, sum of all the angles in a triangle is 180)
∠OCD + 28 + 90 = 180
⇒ ∠OCD + 118 = 180
⇒ ∠OCD = 180 -118
∠OCD = 62 0
Quadrilateral
• Introduction to Quadrilateral
• Types of Quadrilateral
• Properties of Quadrilateral
• Parallelogram and its Theorems
• Rectangle and its Theorems
• Square and its Theorems
• Rhombus and its Theorems
• Trapezoid (Trapezium)and its Theorems
• Kite and its Theorems
• Mid Point Theorem
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