6.4: The Area of a Rhombus

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The area of a rhombus can be found by using the formula for the area of a parallelogram, $A=bh$, since a rhombus is a special kind of parallelogram (Figure $\PageIndex{1}$). However, if the diagonals are known the following formula can be used instead (Figure $\PageIndex{2}$):

Figure $\PageIndex{1}$: The area of Thombus $ABCD$ is $bh$. Figure $\PageIndex{2}$: The area of rhombus $ABCD$ is $\dfrac{1}{2} d_1d_2$

Theorem $\PageIndex{1}$

The area of a rhombus is one-half the product of the diagonals.

\[A=\dfrac{1}{2} a_{1} a_{2}\]

Proof

Referring to Figure $\PageIndex{2}$,

Area of $\triangle ABC$ = $\dfrac{1}{2} bh = \dfrac{1}{2} (AC)(BE) = \dfrac{1}{2} d_1 (\dfrac{1}{2} d_2) = \dfrac{1}{4} d_1d_2$.

Area of $\triangle ADC$ = $\dfrac{1}{2} bh = \dfrac{1}{2} (AC)(DE) = \dfrac{1}{2} d_1 (\dfrac{1}{2} d_2) = \dfrac{1}{4} d_1d_2$.

Area of rhombus $ABCD$ = Area of $\triangle ABC$ + Area of $\triangle ADC$ = \dfrac{1}{4}d_1d_2 + \dfrac{1}{4} d_1d_2 = \dfrac{1}{2} d_1d_2\).

Example $\PageIndex{1}$

Find the area of the rhombus:

$clipboard_e584434d2945eb134a723eed58ff71683.png$

Solution

$A=\dfrac{1}{2} a_{1} d_{2}=\dfrac{1}{2}(8)(6)=\dfrac{1}{2}(48)=24$

Answer: 24.

Example $\PageIndex{2}$

Find the area and perimeter of the rhombus:

$clipboard_e389bf70d17872bd59d6516c720933434.png$

Solution

The diagonals of a rhombus are perpendicular so $\triangle CDE$ is a right triangle. Therefore we can apply the Pythagorean theorem.

\[\begin{array} {rcl} {5^2 + x^2} & = & {(x+1)^2} \\ {25 + x^2} & = & {x^2 + 2x + 1} \\ {24} & = & {2x} \\ {12} & = & {x} \end{array}\]

$d_1 = 12 + 12 = 24$. $d_2 = 5 + 5 =10$. $A = \dfrac{1}{2}d_1d_2 = \dfrac{1}{2} (24)(10) = 120$.

$CD = x + 1 = 12 + 1= 13$.

Perimeter = 13 + 13 + 13 + 13 = 52.

Answer: $A = 120$, $P = 52$.

Example $\PageIndex{3}$

Find the area of the rhombus:

$clipboard_ed16a7f2abde52cac8d98ad8df60fe1c8.png$

Solution

As in Example 4.5.6 of section 4.5, we obtain $AC = 4\sqrt{3}$ and $BD = 4$, Area = $\dfrac{1}{2} d_1d_2 = \dfrac{1}{2} (AC)(BD) = \dfrac{1}{2}(4\sqrt{3})(4) = 8\sqrt{3}$.

Answer: $A = 8\sqrt{3}$.