## Related questions with answers

A rhombus is a quadrilateral with all sides equal in length. Recall that a rhombus is also a parallelogram. Find length AC and length BD in the rhombus shown here.

Solution

VerifiedLet point $P$ be the intersection between $\overline{AC}$ and $\overline{BD}$. By symmetry of the rhombus, we have the following equalities:

$\begin{equation}\overline{BP}=\overline{PD}\end{equation}$

$\begin{equation}\overline{AP}=\overline{PC}\end{equation}$

Equations (1) and (2) give rise to:

$\begin{equation}2 \overline{AP} = \overline{AC}\end{equation}$

$\begin{equation}2 \overline{BP} = \overline{BD}\end{equation}$

We also know from the givens of the problem that:

$\begin{equation}\overline{AB}=\overline{BC}=\overline{CD}=\overline{DA}=18 \text{ in}\end{equation}$

By the symmetry of the rhombus, we also know the following facts:

$\begin{equation}\overline{AC}\bot \overline{BD}\end{equation}$

$\begin{equation}2 m\angle BAP = m \angle BAD\end{equation}$

$\begin{equation}2m \angle ABP = m \angle ABC\end{equation}$

Equation (7) tells us that:

$\begin{equation}m \angle BAP = 21 \text{\textdegree}\end{equation}$

The sine of this angle is the following:

$\begin{equation}\sin(m \angle BAP) = \dfrac{\overline{BP}}{\overline{AB}}\end{equation}$

Therefore, we have

$\begin{equation}\overline{BP} = \overline{AB}\sin(m \angle BAP)\end{equation}$

And by equation (4), we have:

$\begin{equation}\overline{BD} = 2 \overline{AB}\sin(m \angle BAP)\end{equation}$

Substituting equation (7) and equation (5) gives:

$\begin{equation}\overline{BD} = 2 (18)\sin(21 \text{\textdegree})\end{equation}$

$\begin{equation}\boxed{\overline{BD} \approx 12.901 \text{ in}}\end{equation}$

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