Chapter 9.1, Problem 17WE

a.

To determine

To describe: the quadrilateral QDRC.

Answer to Problem 17WE

The quadrilateral QDRC is a rhombus.

Explanation of Solution

Given information: 

Two congruent circles intersect at C and D.

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.

To determine

To check: whether CD is the perpendicular bisector of QR.

Answer to Problem 17WE

CD is the perpendicular bisector of QR.

Explanation of Solution

Given information: 

  $CQ=QR=CR=RD$

Calculation: 

In triangle COQ and COR,

  $\begin{array}{l}CQ=CR\text{ (radius),}\\ \angle CQO=\angle CRO(\text{given}),\text{ and}\\ CO=CO(\text{common}).\\ \text{So, by SAS rule,}\\ \Delta COQ\cong \Delta COR.\\ \text{By congurent property,}\\ \angle COQ=\angle COR,\\ \Rightarrow \angle QCO=\angle OCR,\\ \Rightarrow \angle COQ+\angle COR=180°,\\ \Rightarrow 2\angle COQ=180°,\\ \Rightarrow \angle COQ=90°.\end{array}$

Hence, CD is a perpendicular bisector of QR.

c.

To determine

To calculate: the length of CD.

Answer to Problem 17WE

  $CD=16\text{cm}$ .

Explanation of Solution

Given information:

  $CQ=QR=CR=RD$

CD is a perpendicular bisector of QR.

  $QC=17\text{ cm}$ and $QR=30\text{ cm}$ .

Calculation: 

In triangle QOC,

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