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# How do you prove that the diagonals of a rhombus are perpendicular?

Last updated date: 11th Sep 2024
Total views: 399.9k
Views today: 11.99k
Verified
399.9k+ views
Hint:Rhombus has all the four sides equal and opposite two sides of the rhombus are parallel to each other and the diagonals of the rhombus always intersect at right angles. Here first of all draw the figure of rhombus and diagonals.

Complete step by step solution:
Since in the above figure ABCD is a rhombus.
Therefore, all the four sides of the rhombus are equal.
$\therefore AB = BC = CD = DA$
In triangles $\Delta AOB$ and $\Delta COB$ ,
$OA = OC$ (Diagonals of the parallelogram bisect each other where bisection means two equal sections.)
$OB = OB$(Common Sides of the triangles)
$AB = CB$ (Sides of the rhombus are equal)
$\therefore \Delta AOB \cong \Delta COB$(By SSS congruence rule)
By using Congruent part of the congruent triangle,
$\angle AOB = \angle COB$….. (I)
Also, AC is a line,
$\angle AOB + \angle COB = 180^\circ$ (Linear pair)
By using the equation (I)
$\angle AOB + \angle AOB = 180^\circ$
Simplify the above equation,
$\Rightarrow 2\angle AOB = 180^\circ$
Term multiplicative on one side is moved to the opposite side, it goes to the denominator.
$\Rightarrow \angle AOB = \dfrac{{180^\circ }}{2}$
Common factors from the numerator and the denominator cancel each other.
$\Rightarrow \angle AOB = 90^\circ$
From equation (I),
$\begin{gathered} \angle AOB = \angle COB \\ \Rightarrow \angle AOB = 90^\circ \\ \end{gathered}$
Also,
Vertically opposite angles are equal to each other,
$\angle DOC = \angle AOB = 90^\circ$
$\angle AOD = \angle COB = 90^\circ$
Using above equations and its co-relations,
$\angle DOC = \angle AOB = \angle AOD = \angle COB = 90^\circ$
$\Rightarrow AC \bot BD$
Therefore, diagonals of the rhombus are perpendicular to each other.

Note: Remember the difference between different types of the quadrilaterals and follow the properties accordingly. follow the different conditions of the congruence of the triangles to prove these types of solutions such as –
SSS criteria (Side - Side - Side)
SAS criteria (Side – Angle - Side)
ASA criteria (Angle – Side – Angle)
AAS criteria (Angle – Angle – Side)
RHS criteria (Right angle – Hypotenuse – Side)
The text passage provides a detailed geometric proof that the diagonals of a rhombus are perpendicular to each other. It does not contain any mathematical questions or answers to such questions.
no_question
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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# How do you prove that the diagonals of a rhombus are perpendicular?

Last updated date: 11th Sep 2024
Total views: 399.9k
Views today: 11.99k
Verified
399.9k+ views
Hint:Rhombus has all the four sides equal and opposite two sides of the rhombus are parallel to each other and the diagonals of the rhombus always intersect at right angles. Here first of all draw the figure of rhombus and diagonals.

Complete step by step solution:
Since in the above figure ABCD is a rhombus.
Therefore, all the four sides of the rhombus are equal.
$\therefore AB = BC = CD = DA$
In triangles $\Delta AOB$ and $\Delta COB$ ,
$OA = OC$ (Diagonals of the parallelogram bisect each other where bisection means two equal sections.)
$OB = OB$(Common Sides of the triangles)
$AB = CB$ (Sides of the rhombus are equal)
$\therefore \Delta AOB \cong \Delta COB$(By SSS congruence rule)
By using Congruent part of the congruent triangle,
$\angle AOB = \angle COB$….. (I)
Also, AC is a line,
$\angle AOB + \angle COB = 180^\circ$ (Linear pair)
By using the equation (I)
$\angle AOB + \angle AOB = 180^\circ$
Simplify the above equation,
$\Rightarrow 2\angle AOB = 180^\circ$
Term multiplicative on one side is moved to the opposite side, it goes to the denominator.
$\Rightarrow \angle AOB = \dfrac{{180^\circ }}{2}$
Common factors from the numerator and the denominator cancel each other.
$\Rightarrow \angle AOB = 90^\circ$
From equation (I),
$\begin{gathered} \angle AOB = \angle COB \\ \Rightarrow \angle AOB = 90^\circ \\ \end{gathered}$
Also,
Vertically opposite angles are equal to each other,
$\angle DOC = \angle AOB = 90^\circ$
$\angle AOD = \angle COB = 90^\circ$
Using above equations and its co-relations,
$\angle DOC = \angle AOB = \angle AOD = \angle COB = 90^\circ$
$\Rightarrow AC \bot BD$
Therefore, diagonals of the rhombus are perpendicular to each other.

Note: Remember the difference between different types of the quadrilaterals and follow the properties accordingly. follow the different conditions of the congruence of the triangles to prove these types of solutions such as –
SSS criteria (Side - Side - Side)
SAS criteria (Side – Angle - Side)
ASA criteria (Angle – Side – Angle)
AAS criteria (Angle – Angle – Side)
RHS criteria (Right angle – Hypotenuse – Side)
user
The text passage provides a detailed geometric proof that the diagonals of a rhombus are perpendicular to each other. It does not contain any mathematical questions or answers to such questions.

TYPE: NO_QUESTION