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A rhombus is a four sided shape with sides of equal lengths and opposite ones parallel to each other.



1. It’s a parallelogram with equal side lengths.

2. All sides are congruent.

3. The opposite angles are congruent.

4. Opposite sides are parallel.

5. The diagonals bisect each other and are perpendicular.

6. The sum of two adjacent angles is equal to 180°.

7. Diagonals bisect the angles of rhombus.

The Area and Perimeter of Rhombus


The area of rhombus is the plan enclosed by the sides of the rhombus.
The area is found by multiplying the length of the diagonals divided by 2.

Area= (diagonal 1 x diagonal 2) /2

$ \displaystyle A=\frac{{{{d}_{1}}\cdot {{d}_{2}}}}{2}$

The perimeter of a Rhombus is the total length of all sides.

$ \displaystyle P=a+a+a+a=4a$

Where $ \displaystyle a$ is one side of the rhombus.

Example 1: On the figure below the diagonals of the rhombus are $ \displaystyle AO=2x+1$ and $ \displaystyle OC=\frac{1}{2}x+10$. Find x.

From the properties we know that diagonals of rhombus bisect each other into equal parts.
So $ \displaystyle AO=OC$ and $ \displaystyle BO=OD$
Since $ \displaystyle AO=OC$ then $ \displaystyle 2x+1=\frac{1}{2}x+10$
$ \displaystyle 2x-\frac{1}{2}x=10-1$
$ \displaystyle \frac{3}{2}x=9$
$ \displaystyle 3x=18$
$ \displaystyle x=6$

Example 2: On the figure below we are given rhombus $ \displaystyle ABCD$. Find $ \displaystyle x$ and $ \displaystyle y$.

Solution : From the properties all sides are congruent.

$ \displaystyle AB=BC=CD=DA$

$ \displaystyle AB=BC$

$ \displaystyle 2x+5=25$

$ \displaystyle 2x=20$

$ \displaystyle x=10$

From the properties we know that diagonals bisect the angles and also the opposite angles are congruent.

So $ \displaystyle \widehat{B}=\widehat{D}=y$

$ \displaystyle \widehat{{DAC}}=\widehat{{CAB}}$ and $ \displaystyle \widehat{{DCA}}=\widehat{{ACB}}$

Since $ \displaystyle \widehat{A}=\widehat{C}$ then $ \displaystyle \widehat{{CAB}}=\widehat{{BCA}}={{40}^{{}^\circ }}$

From the properties of triangles we know that the sum of angles on a triangle is $ \displaystyle {{180}^{{}^\circ }}$ so:

$ \displaystyle \widehat{{CAB}}+\widehat{{BCA}}+\widehat{{B=}}{{180}^{{}^\circ }}$
$ \displaystyle {{40}^{{}^\circ }}+{{40}^{{}^\circ }}+y={{180}^{{}^\circ }}$
$ \displaystyle {{80}^{{}^\circ }}+y={{180}^{{}^\circ }}$
$ \displaystyle y={{180}^{{}^\circ }}-{{80}^{{}^\circ }}$
$ \displaystyle y={{100}^{{}^\circ }}$

Example 3: Find the diagonal of rhombus if it´s area is $ \displaystyle200c{{m}^{2}}$ and the longest diagonal $ \displaystyle 25cm$.

Solution: Using area of a rhombus formula $ \displaystyle A=\frac{{{{d}_{1}}\cdot {{d}_{2}}}}{2}$

$ \displaystyle 200=\frac{{{{d}_{1}}\cdot 25}}{2}$

$ \displaystyle 400=25{{d}_{1}}$

$ \displaystyle {{d}_{1}}=16$

Example 4: On the figure below we have the rhombus $ \displaystyle ABCD$ with an acute angle $ \displaystyle {{60}^{{}^\circ }}$ and height $ \displaystyle D{{D}_{1}}$.Prove that the area of triangle $ \displaystyle AD{{D}_{1}}$ is one fourth of the area of rhombus.


Solution: On the triangle $\displaystyle \text{AD}{{D}_{1}}$ we have:

$ \displaystyle \widehat{{AD{{D}_{1}}}}={{90}^{{}^\circ }}-\widehat{{DA{{D}_{1}}}}$

$ \displaystyle \widehat{{AD{{D}_{1}}}}={{90}^{{}^\circ }}-{{60}^{{}^\circ }}={{30}^{{}^\circ }}$

We know that the side in front of the angle $ \displaystyle {{30}^{{}^\circ }}$ in a right angle triangle is half the hypotenuse so $ \displaystyle AD=2A{{D}_{1}}$

We label $ \displaystyle A{{D}_{1}}=x$ from where $ \displaystyle AD=2x$

Using Pythagorean theorem:

$ \displaystyle {{\left( {D{{D}_{1}}} \right)}^{2}}={{\left( {AD} \right)}^{2}}-{{\left( {A{{D}_{1}}} \right)}^{2}}$

$ \displaystyle {{\left( {D{{D}_{1}}} \right)}^{2}}={{\left( {2x} \right)}^{2}}-{{\left( x \right)}^{2}}$

$ \displaystyle {{\left( {D{{D}_{1}}} \right)}^{2}}=4{{x}^{2}}-{{x}^{2}}$

$ \displaystyle {{\left( {D{{D}_{1}}} \right)}^{2}}=3{{x}^{2}}$

$\displaystyle D{{D}_{1}}=x\sqrt{3}$

We have:

$ \displaystyle {{A}_{{AD{{D}_{1}}}}}=\frac{{A{{D}_{1}}\cdot D{{D}_{1}}}}{2}$

$ \displaystyle {{A}_{{AD{{D}_{1}}}}}=\frac{{x\cdot x\sqrt{3}}}{2}=\frac{{{{x}^{2}}\sqrt{3}}}{2}$

$ \displaystyle {{A}_{{ABCD}}}=AB\cdot D{{D}_{1}}$

$ \displaystyle {{A}_{{ABCD}}}=2x\cdot x\sqrt{3}=2\sqrt{3}{{x}^{2}}$

Then we observe the ratio:

$ \displaystyle \frac{{{{A}_{{AD{{D}_{1}}}}}}}{{{{A}_{{ABCD}}}}}=\frac{{\frac{{{{x}^{2}}\sqrt{3}}}{2}}}{{2\sqrt{3}{{x}^{2}}}}$

$ \displaystyle \frac{{{{A}_{{AD{{D}_{1}}}}}}}{{{{A}_{{ABCD}}}}}=\frac{1}{4}$

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