Rhombuses

Consider the rhombus below. Prove that for the given rhombus, AC ⊥ BD.

Solution:

By the definition of a rhombus:

$AB=BC=CD=DA$

Now consider the triangles $AOB$ and $COB$:

The side $OB$ is a side of both triangles. Now, $AB=BC$, since $ABCD$ is a rhombus. We also have $OA=OC$, since $ABCD$ is also a parallelogram and the diagonals of a parallelogram bisect each other. Therefore, triangle $AOB$ is congruent to triangle $COB$. In other words, they are exactly the same triangles, just rotated in different positions.

This implies that:

$\angle AOB=\angle COB$

Additionally, $\angle AOB$ and $\angle COB$ lie on the same straight line. Thus,

$\angle AOB+\angle COB=180°$

Therefore,

$\angle AOB=\angle COB=180÷2=90°$

Similarly we can show that:

$\angle AOD=\angle COD=90°$

Thus, $AC$ and $BD$ are perpendicular. That is, $AC\perp BD$.

A rhombus has diagonals of lengths $10cm$ and $15cm$. What is the area of the rhombus?

Solution:

Using the specific formula for the area of a rhombus, we have:

$Area\mathit{}\mathit{=}\mathit{}\frac{\mathit{1}}{\mathit{2}}{d}_{\mathit{1}}{d}_{\mathit{2}}$

Substituting in $d_1=10$and $d_2=15$, we have:

$$\begin{gathered} \mathrm{Area}=\frac{1}{2}\times 10cm\times 15cm \\ =75c{m}^2 \end{gathered}$$

Thus, the area of this rhombus is $75c{m}^2$.

Consider the below rhombus. Given that $\angle ACB=35°$, find $\angle DBA$.

Rhombus example - StudySmarter Originals

Solution:

Recall that in a rhombus, each diagonal bisects, which means that we have a pair of equal angles. We also know that opposite angles in a rhombus are equal.

Thus,

$\angle ACB=\angle OAB=35°$

Now, we mentioned earlier that the diagonals in a rhombus are perpendicular to each other. Therefore,

$\angle AOB=90°$

Since $OAB$ forms a triangle, and angles in a triangle sum to $180°$, we can work out $\angle DBA$:

$\angle DBA+\angle AOB+\angle OAB=180°$

So, substituting in the known angles, we have

$\angle DBA+90°+35°=180°$

which implies that

$\angle DBA+125°=180°$

Subtracting $180°$ from both sides, we obtain:

$\angle DBA=180°-125°=55°$

So, we have that $\angle DBA=55°$.

Consider the following rhombus depicted below. Given that $\angle ADC=120°$, find $\angle OAD$.

Solution:

Recall that diagonal $BD$ bisects $\angle ADC$. Therefore, we have,

$\angle ADO=120°÷2=60°$.

We also have that $\angle DOA=90°$ due to the perpendicular bisector.

So, since angles in a triangle sum to $180°$, we have:

$\angle OAD=180°-90°-60°=30°$.

Thus, $\angle OAD=30°$.