
How do you prove that the diagonals of a rhombus are perpendicular?
Answer
444.6k+ 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)
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)
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE
