Question

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

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

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

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

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

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

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

We also know from the givens of the problem that:

AB=BC=CD=DA=18 in\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:

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

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

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

Equation (7) tells us that:

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

The sine of this angle is the following:

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

Therefore, we have

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

And by equation (4), we have:

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

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

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

BD12.901 in\begin{equation}\boxed{\overline{BD} \approx 12.901 \text{ in}}\end{equation}

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