Q17. - Geometry

Question

$\odot \mathbf{Q}$ and $\odot \mathbf{R}$ are congruent circles the intersect at $\mathbf{C}$ and $\mathbf{D}$.  $\overline{\mathbf{CD}}$ is called the common chord of the circles. 

1. What kind of quadrilateral is $\mathbf{QDRC}$? Why?
2. $\overline{\mathbf{CD}}$ Must be the perpendicular bisector of $\odot \mathbf{Q}$. Why?
3. If $QC=17$ and $QR=30$, find $CD$.

Step-by-Step Solution

Answer

1. The quadrilateral $\text{QDRC}$ is a rhombus.
2. $CD$ is the perpendicular bisector of $QR$.
3. $CD=16 \text{cm}$.

Step 1. Given information.

Two congruent circles intersect at $C$ and $D$.

$\begin{array}{l}CQ=QR\\ QR=CR\\ CR=RD\end{array}$

Step 2. Concept used.

The sides of the quadrilateral $QDRC$ are equal and all angles of the diagonals bisect each other at right angle, it is a rhombus.

Step 3. Solution.

a.

As the sides of the quadrilateral $QDRC$ are equal and all angles of the diagonals bisect each other at right angle, it is a rhombus.

b.

In $\Delta \text{ COQ}$ and $\Delta \text{ COR}$,

$\begin{array}{c}\angle CQO=\angle CRO\left(\text{given}\right),\text{ and}\\ CO=CO\left(\text{common}\right).\end{array}$

So, by SAS rule,

$\begin{array}{c}\Delta COQ\cong \Delta COR.\\ \end{array}$

By congruent property,

$\begin{array}{c}\angle COQ=\angle COR,\\ \angle QCO=\angle OCR,\\ \angle COQ+\angle COR=180°,\\ 2\angle COQ=180°,\\ \angle COQ=90°.\end{array}$

Therefore, $CD$ is a perpendicular bisector of $QR$.

c.

$CD$ is a perpendicular bisector of $QR$.

$\begin{array}{l}QC=17cm\\ QR=30cm\end{array}$ 

$\begin{array}{c}\text{In ΔQOC:}\\ C{Q}^2=O{Q}^2+O{C}^2,\\ {17}^2={15}^2+O{C}^2,\\ O{C}^2={17}^2-{15}^2,\\ O{C}^2=64,\\ OC=8\text{ cm.}\\ \text{Now, since :}\\ CD=OC+OD\text{ and }\\ OC=OD,\\ \text{we can write: }\\ CD=2OC,\\ CD=16\text{ cm.}\end{array}$

Therefore, $CD=16 \text{cm}$.